Number system Proc 1
1.How many natural numbers between 1 and 500 are not divisible by 7 and leave a remainder of 4 when divided by 4? (a)125 (b)107 (c)106 (d)124 1.b. All the numbers which leave a remainder of 3 when divided by 4 will be of the form 4n+3, 4n+3, when n is whole number. For n=0, 1, 2, 3…… 124. We have 1< 4n + 3< 500 Therefore, the number of values of n is given range =124 = 0 +1 = 125. Now observe that, 4n +3 is divisible by 7 if and n = 7k + 1, where k is a whol e number. For k= 0, 1, 2, 3……17. 3……1 7. We have 1 < 4(7k + 1) +3 < 500 Therefore, the number of values of k in the given range = 17 =0 +1 = 18 Hence, the required number of elements =125 -18 = 107.
4. Find the number of odd factors of 36036. (a) 72 (b) 46 (c) 24 (d) 36 5.c. 36036=36x001=22x32x7x11x13 Number of odd factors of 36036=Number of factors of (32x7x11x13)=(2+1)x(1+1)x(1+1)x(1+1)=24.
24. A real number ‘N’ decreased by times itself following which
is subtracted subtracted to obtain
N1’ where ‘a’ is natural number greater than 1. Same operations when performed on N1 result in N2 and so on. If Nn = 0, then find N.
(a)
(b)
(c)
(d)
24.b Nn+1 = Nn (1- )-
Nn =
If N1 = 0, N =
If N2 = 0, then N 1 =
N=
Generalizing this, it can be deduced that
=
N=
25. If a b and c are non-zero real numbers such that
following is necessarily true? (a) a + b + c = 0 (b) + bc + ca (c) a = b = c = 1 (d) Either (a) or (b)
which which of the
25. d
We know that - 3 abc = (a + b + c)( – ab – ab – – bc bc – – ca) ca) – ab – ab – – bc bc – – ca) ca) = 0 (a + b + c)( (a + b + c) = 0 or ( – ab – ab – – bc bc – – ca) ca) = 0. 26. N is a three-digit natural number divisible by 11 such that the sum of its unit and hundreds digits is greater than its tens digit. What is the number of possible values of N? (a) 76 (b) 36 (c) 45 (d) 44
26. b Let the number be 100a+ 10b + c. a + c – c – b b > 0 a + c – c – b b =0 or a multiple of 11 Combining these two we get. a + c – c – b =11 (a + c < 18) a + c =11 +b For b = 0, a + c = 11 a, c can be (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3) or (9, 2) 8 solutions For b = 1, the number of solutions = 7 For b = 2, the number of solutions = 6 For b = 3, the number of solutions = 5 For b = 4, the number of solutions = 4 For b = 5, the number of solutions = 2 For b = 7, the number of solutions = 1 Hence, the total number of solutions =
1. Let a, b, c, d, e, f, g, h, I and j be natural number less than 6. If LCM of
(a)
which of the following cannot be the value of
1. d LCM
(b)
(c) 2
=
=
(d) 3
and
Since all the ten number are less than 6 and their HCF is a multiple of 5, the only possible HCF is 5. B= d =f =f h =j =5
∴
LCM (a, c , e, g, i) =6
This implies that the values of a, c, e, g and I should be among 1, 2 and 3 such that at least one of them . is equal to 2 and at least one of them is equal to 2 and at least one of them is equal to 3. Therefore,
lies lies between and
(Including both).
Hence, 3 is not a possible value. 2.What is the remainder when 997996997996997996 is divided by 997? (a) 994 (b) 906 (c) 3 (d) 91 2.b N=997 x 1015 + 996 x 10 12 + 997 x 10 9 + 996 x 10 5 + 997 x 10 3 + 996 x 10 0 If N is divided by 997, we find the remainder by parts.
∴
Remainder = 0 x35 +(-1) x 34 + 0 x 3 3 + (-1) x 3 2 + 0 x 3 1 + (-1)= -91( 103 when divided by 997 will leave a remainder of of 3) Hence, required remainder = 997 – 997 – 91 91 = 906. 3. If k is a natural number such that 1 < k < 78, for how many values of k is the tens digit of 31k equal to 6? (a) 8 (b) 15 (c) 7 (d) 16 k
k
k
k
2
k
k-1
k
k
3.a 31 = (1+30) =C0+ C1x30+ C2x30 + ... + Ck-1x30 + Ck x 30
The third term onwards, all the terms would be multiples of 100 hence would n ot contribute to the tens digit of the number.
∴
Tens digit if 31k = Tens digit of ( kC0++ kC1x30)= Tens digit of (1+30k) = units digit of 3k=6
digit of 31 k=Tens digit of k is 2 ⇒ k = 2, 12, 22, …. , 72. Hence, the number of possible values of k is 8. ⇒Units
4. If P=6!+7!+8!+9!+10!, which of the following is factor of P? (a) 4206 (b) 4199 (c) 4277 (d) 4212 4.a P=(6!+7!+8!+9!+10!)
=6!x(1+7+56+504+5040) =720 x120 x8 x 701 =120 x 8 x 4206 Hence, 4206 is a factor of P. None of the other options can be a factor of P as all of them are multiples of 13. 16. Find the number of zeroes at the end of 57!. (a) 13 (b) 14 (c) 15 (d )12
16.a In order to count the the number of zeroes zeroes at the end end of 57!, we need to find the highest power of 5 in 57!.
Highest power of 5 in 57!= 57!= Where
⌈ x⌉
denotes the greatest interger less than or equal to x. denotes
Proc 2 1.Let P = {n!+1, {n! +1, n!+2, …, n!+n} such that ‘n’ is a natural number greater than 42. At the most, how many elements in set P can be prime numbers. (a)0 (b)1 (c)2 (d)More than 2
∈
1.b (n!+a) is always divisible by a fro a < n, a N. If a =1. Then n! + 1 is divisible by 1, but n! + 1 can be a prime number. Hence, there can be at the most 1 prime number in set P. 6. b
N =3a3 and N < 200, therefore, possible values of N are 3, 24, 81 and 192.
Also, N = 6b2 =
These conditions are satisfied only by N = 24. Hence, there is only 1 possible value of N. 7. Three natural numbers a, b and c are such that a 2 + b2 = c2, If a = 20, how many ordered pairs are possible ?
7.d c2-b2=202
Therefore, both
2
3
have to be even factors of 400, as c and b are natural numbers.
400 can be written as 2 x 200, 4 x 100, 8 x 50, 10 x 40, 20 x 20. But, 20 x 20 is not admissible since b ≠ 0. Hence, the ordered pairs are
11. When a3+ b3+ c3+ d3 + e3 is divided by 7, the remainder is ‘r’. If a, b, c are positive integers and none of them is divisible by 7, what is the sum of a ll the possible values of ‘r’?
10
11.d Any positive integer x, which is not divisible divisible by 7, can be respresented as 7n k, where k= 1,2,3,4,5, or 6.
*+ *+ *+ *+ * + * + * + * + * + * + =
=7n1+k3
* + *+
Therefore, the remainder when
Rem
=1, =1, Rem
=1, Rem
Rem
=1, =1, Rem
=6, =6, Rem
is divided by 7 is equal to that of
3
3
3
3
when divided by 7.
3
So the possible remainders when a +b +c + d + e is divided by 7 are:
Rem Rem Rem Rem Rem Rem
Hence, the sum of all the possible values of ‘r’ = 1 + 2 + 3+ 4 +5 + 6 =21 12. If all the four-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, 7 and 8 without repetition are arranged in ascending order, wh at will be the rank of the number 5283?
None of these
12.a The total number of four-digit number formed using the the given digits which are less less than 5000 = 4 x 7 x 6 x 5 = 840 Similarly, the number of four -digit numbers between 5000 and 5200 = 6 x 5 = 30; and the number of four-digit numbers between 5200 and 5280 = 5 x 5 = 25. Hence, the rank of 5283 is 840 + 30+ 25+ 2 = 897. 13.If [X] represents the greatest integer less than or equal to ‘x’ and non negative negative integer values of ‘x’are possible? (a) 3
(b) 4
(c) 5
*+ *+
,how ,how many
(d) 6
*+ *+ *+ *+ ∴ x *x+ *x+
13.d If If
=0, then x =0, 1, 2 or 3. =0,
=1, =1, then x =6, or 7.
No other value of ‘ is possible.
16.Bells B1 and B2 ring 8 times and 48 times in a m inute respectively. If they start ringing simultaneously, after how much time (in seconds) will B 2 ring exactly 10 times more than B 1 ? (a)8 (b) 10 (c)15 (d) 20 16.c B2 will ring 48-8=40 times more than B 1 in a minute. Hence, to ring 10 times more, m ore, it will take
17. Find the number of digits in 3 150 (Given log3=0.4771) (a) 71 (b) 72 (c) 66
17.b Let =3150. Taking log on both sides, Log =150 log 3 =71.565 Hence, the number of digits is 71 + 1 = 72.
24. Consider two sequences S1 and S2: S1={1, 2, 6, 10, 15, …… 15 terms} S2={1, 5, 14, 30, 55, ….. 10 terms} The number of common elements between S1 and S2 is
(d) 67
(a) 3 (b) 2 (c) 4 (d) 5 24.a The terms of sequence S1 represent the sum of first ‘n’ natural numbers.Therefore, the 15th term of S1 will be
The terms of sequence S2 represent represent the sum of the squares of first ‘n’ natural numbers. Therefore, the 10th terms of S 2 will be
w
compared to 120.
It gives us a hint that the common terms are much before the 10 th termd of S2. Now we can count the common terms manually. The three common terms are 1, 55 and 91.
Proc 4
2.
What will be the remainder when 45! Is divided by 47?
(a) 45
(b) 1
(c) 0
(d)46
2b. We know that when (P-1)! (P-1)! Is divided by ‘P’ where ‘p’ is a prime number, we get (p-1) as the remainder. Thus, 46!=47k+46. Or 46 x 45!=47k+46 Let the remainder remainder obtained when 45! Is divided by 47 be ‘r’. On equating remainder after dividing the above equation by 47 on both sides, we get. 46 x r=46
→
r=1
Hence, the remainder=1
Le– ‘xyz’ ‘yxz’ yzx’yxz w‘zyx’ ee ume u peeevuume u e ‘xyz’ A ‘xyz’ one fifth of (xyz(a) 64
(b) 56
that (yxz-yzx) is
(c)72
yxz’, ‘yzx’ and ‘zyx’ are distinct, x, y and z have to be distinct natural numbers less than 10. According to the question: (100y+10x+z)-(100y+10z+x)
xyz xyz yxz yxz ⇒ xy =
9x-9z=
⇒
2y=x+z Hence, both x and z are either odd or even.
∴
Case I: When x and z are odd, xyz can be formed in sP2=20 ways, [ This amounts to nothing more than finding the number of ways in which an ordered pair can be formed from 1, 3, 5, 7 and 9]
∴
Case II: When x and z are even, xyz can be formed in 4P2=12 ways. [ This amounts to nothing more than finding the number of ways in which an ordered pair can be formed from 2, 4, 6 and 8] Hence, the number of possible values of ‘xyz’ =20+12=32.
11.From the set of first 10 natural numbers. A picks some numbers and calculates their product to be ‘T’ The product of the remaining numbers in the equal to ‘S’ if ‘T’ is completely divisi ble by ‘S’ the minimum value of (T/S) is (a)7
(b)14
(c)21
(d)16
11. From the set of first 10 natural numbers. A picks some numbers and calculates their product to be ‘T’. the product of the remaining numbers in the set is equal to ‘s’. If ‘T’ is completely divisible by ‘S’, the minimum value of (T/S) is (a) 7 (b)14 (c)21 (d)16 = 8 c9 c 11.a Since (T/S) needs to be minimized, we will try to minimize to value of T and maximize the value of S. But T is completely divisible by S which which means 7 has to be a part of T.( 7 is a prime number and doesn’t have any multiple less than 10 other than itself.) Product of all the numbers other than 7=1 x 2 x 3 x 4 x 5 x 6 x 8 x 9 x 10 = 720 x 720 Or, 1 x 2 x 3 x 4 x 5 x 6=8 x 9 x 10 S0,we have T = 7 x720and S 720 Hence, minimum value of (T/S) = 7
∴
12.There exist three positive integes P, Q and R such that P is not grater than Q, Q is not greater than R and the sum of P, Q and R is not more than 10. How many distinct sets of the values of P, Q and R are possible? (a)120 (b)43 (c)11 (d)31 12.d According to the question : P < Q < R and P+Q+R< 10
Case I: If P=1: For Q=1, R can take 8 values (from 1 to 8) For Q=2, R can taken 6 values (from 2 to 7) Case II: If P = 2: For Q = 2, R can taken 5 values (from 2 to 6) For Q =3, R can taken 3 Values (form 3 t 5) For Q=4, R can taken only 1 value (R=4) Observe that there are no solutions for Q > 4 Total number of solutions in this case = 5 =3+1=9. Case III: If P =3: For Q=3, R can taken 2 values (3 and 4) Observe that there are no solutions for Q > 4. Total number of solutions in this case=2
∴
Total number of distinct sets of the values of P,Q and R=20+9+2=31.
13. One of two natural numbers A and B, one of two natural numbers C and D and one of two natural numbers E and F are multiplied together. The sum of all such possible products is equal to 105. How many distinct values are possible for A?
(a)6
(b) 4
(c) 12
(d) 7
13.a Possible products are ACE, ADE, ADF, BCF, BDE and BDR. Their Sum=A(CE+CF+DE+DF)+B(CE+CF+DE+DF)=(A+B)(C+D)(E+F)=105=3 X 5 X 7.
S0, It can be deduced that (A+B) is one of 3, 5 and 7. If (A+B) =3, A can take the value 1 and 2. If (A+B) =5, A can take the values 1, 2, 3 and 4. If (A+B) =6, A can take the values 1, 2, 3,4, 5 and 6. Hence, A can take the values 1, 2, 3, 4. 5 and 6.
23. In a certain colony, the number of people who drink only tea, only coffee, both tea and coffee and neither tea nor coffee are x, 2x, people who drink tea can be (a)37
(b) 111
respectively. The number of
(c) 40
(d) None of these
23.c Since the number of people is always an integer, we can deduce that ‘x’ and ‘3x’ have to be factors of 111. Therefore, the possible values of ‘x’ are 1 and 37.
So, the number of people who drink tea is equal to x+ 40 (if x =37).
which can be 112 (if x=1) or which
25.If a, b, and d are four odd positive integer in non- decreasing order, such that (ab+cd)-(a+b+c+d)=6, how many values are possible for c?
(a)20
(b)21
(c)22 (d)23
25. (ab+cd)-(a+b+c+d)=6
⇒ ⇒
b(a-1)+d(d-1)-a-c=6 b(a-1)+d(c-1)-a+1-c+1=8
Since a, b. c and d are odd positive integers in non-decreasing order, we can conclude that (a-1), (b-1), (c-1) and (d-1) are non-negative even integers. If (a-1)=0, (c-1)(d-1)=8
⇒
c=3 and d=5 and hence, b is either 1 or 3.
If (a-1)=2, then b=c=d=3.
It can be easily observed that no other case is possible, Hence, there is only one possible value for c. Proc 5 1.If one of the digits of Z, a two-digit number, is added to these square of the other digit, Z is obtained. How many values of Z are possible? (b)2 (c)3 (d)More than 3 (a)1 1.a Let the number be 10a+b. Then, Then, according according to the question: question:
⇒
Case 1: 10a+b 1: 10a+b – – a a2=b a-or a-or 10. This is not possible as ‘a’ can be neither ‘o’ nor a two-digit two -digit number.
⇒
Case 2: 10a+b=a+b2 b2-b=9a Or b(b-1)=9a.The only possible set of values which satisfies this question is b=0, a=8. So the only possible number is 89. (c×d)
(e×f)
(a×b)
11. If (a×b) =(e×f); (c×d) =(a×b) and (e×f) =(c×d) where a, b, c, d, e and f are nonzero integers, then the sum of an the possible values of (a+b+c+d+e+f) is (a)8
(c)0
(b)6
11.c (e×f)
(a×b)
=(c×d) or[ (a×b)
(c×d) (a×b)
]
(d)16
-(c×d)
Similarly, (a×b)(a×b×c×d×e×f)=(a×b) And (e×f)
(a×b×c×d×e×f)
(e×f)
We can conclude that (a×b×c×d×e×f)=1and (a×b)=(c×d)=(e×f)=1 or(a×b×c×d×e×f)=-1 and (a×b)=(c×d)=(e×f)=-1[As they are all non – non –zero zero integers] If (a×b×c×d×e×f)=1 and and (a×b)=(c×d)=(e×f)=1, the possible sets of values of a,b,c.d.eand f are (1,1,1,1,1), (-1,-1,1,1,1,.1), (1,1,1,1,-1,-1), (-1,-1,-1,-1,1,1), (-1,-1,1,1,-1,-1),(1,1,-1,-1,-1,1) and (-1,-1,-1,-1,-1,-1), So the possible values of (a+b+c+d+e+f)are 6,2,-2 and -6. If (a×b×c×d×e×f)=-1 and and (a×b)= (c×d)=(e×f)=-1, (c×d)=(e×f)=-1, then in in each of of the pairs (a, b), (c, (c, d), and (e, f), one number will be ‘‘-1’ and the other will will be ‘1’. So So the only possible value of (a+b+c+d+e+f) is O. Thus, the sum of all the possible values values of (a+b+c+d+e+f)=6+2+(-2)+(-6)+0=0. 17. How many natural numbers num bers ‘n’ satisfy the inequality 5100 < n150 < (150n)5? (a) 11
(c) 10
(b) 9
(d) 15
17.c The inequality can be rewritten as (52)50 < (n3)50 < (150n)50 2 3 Or 5 < n < 150n Or n3 > 25 and n2 < 150 The possible values of n which satisfy the condition are 3, 4, 5 ………., 12.So there are 10 25. How many five-digit numbers with distinct digits can be formed using 1, 3, 5, 7 and 9 such that ‘3’ appears before ‘7 ? (a) 120
(b)90
(c) 24
(d) 60
25. d There are 5 odd numbers, numbers, We can select two places in 5C2 ways and the number of ways of putting e and 7 in that places is only 1.Remaining places can be filled in 31 ways. So the total number of way will be 5C2 x 1 x 31 =10 x 6 =60.
27. A number of that that can be expressed as the square as well as the cube of a natural number, and has 13 factors, is called a Dante’s number. What is the total number of Dante’s numbers less than 106 ? (a)1
(b)2
(c)3
(d)4 12
27.b Since the number number has exactly 13 factors, the number must be of the form of a (where ‘a’ is prime). Hence, the possible values are 2 12, 312, 512 ……, out of which only 2 12, 12 6 3 are less than 10 .
Proc 8
1.
The number of consecutive consecutive zeros at the end of k!, where k is a natural number, is ‘a’. lf 4 < a S 6, then the number of values that ‘k’ can assume is (a) 0 (b) 5 (c) 10 (d) 15
1. b
The number of consecutive consecutive zeros at the end end of kl will be equal equal to the highest power of 5 contained in k!. Here, 4
4.
If E = 3 + 8 +15 + 24 + +195, then what is the sum of the the prime factors factors of E? (a) 29 (b) 31 (c) 33 (d) 23
5. b
E=3+6+15+24+...+195 E=3+6+15+24+...+195 = 1 x 3 + 2 x 4 + 3 x 5 + 4 x 6 +...+ 13 x 15 .’. Tn = n(n + 2) and n = 13 .’. E = =
∑ ∑
= =
+2x
+ 2 x
= 1001 = 7 x 11 x 13
Hence, the sum ofthe prime factors of E = 7 + 11 + 13 = 31. 13.
Let S = {1,2,3,4,5,6,7,8,9,10,11,12}. The number of subsets of S comprising composite number(s) only and that of those comprising prime number(s) only are N1 and N2 respectively. What is the absolute difference between N 1 and N2? (a) 0 (b) 32 (c)48 (d) 24 13. c The given given set S has 6 composite and 4 prime numbers. numbers. 6 The number of subsets of S comprising composite composite numbers only =2 -1 4 The number of subsets of S comprising prime numbers numbers only= 2 -1 6 4 Hence, the required differences = (2 -1) – -1) – (2 (2 -1) = 48 18.
‘ab’ is a two digit prime number such that one of its digit is 3. If the absolute difference between the digits of the number is not a factor of 2, then how many values can ‘ab’ assume? (a)5 (b)3 (c)6 (d) 8
18.b Since ‘ab’ is a two digit prime number and one of its digit is 3, it van assume any of the values among 13, 23, 31, 37, 43, 53, 73 and 83. As the absolute difference between the digits of the number is not a factor of 2, the number among the obtained numbers that satisfy the aforementioned condition are 37, 73 and 83. Hence, the number of values that ‘ab’ can assume is 3. 23.
The number of factors of the square of a natural number is 105. The number of factors of the cube of the same number is ‘F’. Find the maximum possible value of ‘F’. (a)208 (b)217 (c)157 (d)280
23.d Let the number be N. 3 2 In order to maximize the number of factors of N , N must be expressed as a product of as many prime factors as possible.
2
No. of factors of N =105 = 3 x 5 x 7 = (2+1)(4+1)(6+1) 2 2 4 6 .’. N = (a) (b) (c) , where a, b and care prime numbers. 3 3 6 9 .’. N = (a) (b) (c) 3 Hence, the number of factors of N = (3+1) x (6+1) x (9+1) = 4 x 7 x 10 = 280 77
28.
33
99
135
What is the unit digit of (3768 – (3768 – 63 63 + 45 – 76 – 76 + 118 )? (a)1 (b)3 (c)7 (d)None of these 68
28.d Unit digit of 37 = 1 77 Unit digit of 63 = 3 33 Unit digit of 45 = 5 99 Unit digit of 76 = 6 135 Unit digit of 118 = 2 Hence, the required unit digit = (1 + 5 + 2) – (3 – (3 + 6) = 9. 29.
12 8
2
13
The coefficient of a b in the expansion of (a + b) is
(a)
(b)
(c)
(d) None of these 2 6
8
29.d The expression expressio n a12b8can be rewritten as (a ) b . 2 2 6 8 We can observe that the sum of the powers of a and b in (a ) b is (6 + 8) 2 2 i.e. 14. But in the expansion of (a + b)13 the sum of the powers of a and b must be 13 in each of the terms. Hence, the given term does not exist in the expansion i.e. the required coefficient is zero.
Proc 9
1.
Let f(x)=*x+, where *x+ denotes the greatest integer less than or equal to x. If ‘a’ and ‘b’ are two real number such that f(3b -2) = a-2 and f(a+2) = b+6, then find the sum of a and b. (a) 6
(b) 4
(c) 8
(d) None of these
1 (c) Since [x] denotes an integer value, we can say that (a - 2) and (b + 6) are integers. Therefore, (3b - 2) and (a + 2) will also be integers. The greatest integer of an integer is the integer itself. f(3b - 2) = (a - 2) => 3b - 2 = a - 2 ....(i) f(a + 2) = b + 6 => a + 2 = b + 6 ....(ii) Solving (i) and (ii) we get a = 6 and b = 2 Hence, the sum of of a and b = 6 + 2 = 8 12.‘D’ 12. ‘D’ is the absolute difference between a three - digit natural number and the sum of the digit of the number. If ‘D’ is a multiple of 12 and greater than 504, then how many values can ‘D’ assume? (b) 12
(b) 53
(c) 13
(d) 14
12(b) Let the three-digit three- digit natural number be ‘abc’. .’. D = 100a 10b c – (a – (a + b + c) = 9 (11a + b) Since D > 504 (= 9 x 56) and it is also a multiple of 12. (11a + b) has to be a multiple of 4 and greater than 56. The possible values of a and b can be tabulated as: A B 5 5,9 6 2,6 7 3,7 8 0,4,8 9 1,5,9 Hence, D can assume 12 values. 20.
What is the remainder when 13 (a) 53
20(a) Rem
(b) 23
2013
is divided by 100?
(c) 13
(d) None of these
*+ *+ *+ * + * + * +
= Rem
= = Rem
= = Rem
= = Rem
= = Rem
= Rem = Rem
23.
[ ] *+ = = 53.
35x4y9z, a seven-digit seven-digit number, is completely divisible by 44. If ‘x’, ‘y’and ‘z’ are single digit whole numbers, find the sum of all the distinct possible values of (y+z-x). (b) 12
(b) 60
(c) 72
(d) 66
26(d) Since 35x4y9z is divisible divisible by 44, it must be divisible by by 4 as well as 11. According to the divisibility rule for 4 i.e. the number formed by the last two digits has to be divisible by 4, z must be either 2 or 6. As the number is divisible by 11, (3 + x + y + z) – (5 – (5 + 4 + 9) i.e. (x + y + z 15) must be of the form 11k, where k is an integer. Case I: When I: When z = 2 x + y + 2 – 2 – 15 15 = 11k x + y – y – 13 13 – – 0 or -11 (since x + y ≤ 18) x + y = 13 or 2 The possible values of x, y and (y + z - x) can be tabulated as: x y y + z – x – x 9 4 -3 8 5 -1 7 6 1 6 7 3 5 8 5 4 9 7 0 2 4 1 1 2 2 0 0 Case II: When II: When z = 6
x + y + 6 – 6 – 15 15 = 11k => x + y – y – 9 9 – – 0 0 (since x + y ≤ 16) => x + y = 9 The possible values of x, y and (y + z - x) can be tabulated as: x 9 8 7 6 5 4 3 2 1 0
Y 0 1 2 3 4 5 6 7 8 9
y + z – x – x -3 -1 1 3 5 7 9 11 13 15
Proc 10
a
1.‘P' 1.‘P' is the product of ten consecutive two -digit natural numbers. lf 2 is one of the factors of P i then the maximum value that ‘a’ can assume is (a) 11 1. c
(b) 12
(c) 13
(d) 14
In order to maximize maximize the the highest highest power of 2 in the product, one of the ten numbers has to be 64 as this is the highest two- digit number of the form k 2 , where k is a natural number; there has to the maximum number of multiples of 8 among the ten numbers. In a set of ten consecutive natural numbers, there can be a maximum of two numbers that will be a multiple of 8. The possible sets of ten consecutive natural numbers that satisfy the aforementioned conditions are 55 to 64, 56 to 65, 63 to 72 and 64 to 73. The highest power of 2 in the product of any of these sets of ten numbers will be 13.
x
2.
f(x) = 2 – 1, – 1, if x is an odd natural number x 2 + 3, if x is an even natural number What is the remainder when f(1) + f(2) + f(3) + ... + f(73) is divided by 9? (a) 2 (b) 6 (c) 1 (d) 7
2.c
f(1) = 1, f(2) = 7, f(3) = 7, f(4) = 19 and so on.
* * ++ * + * + * + * + * + * +
Rem
= = Rem
= Rem
= = -1
= Rem
= = Rem
= Rem
3.
3. b
= Rem
= = Rem
= 1.
Find the rightmost non-zero dlgltln11! (a) 4 (b) 8 (C) 2
(d) 6
8
4
2
2
2
6
4
11! = 2 3 5 7 11 = (2 5 ) (2 3 7 11) 2 2 Since (2 5 ) ends in zero, the rightmost non-zero digit in 11! will be units 6 4 digit of (2 3 7 11) i.e. 8. 4.
How many natural numbers less than 25 have a composite number of factors? (a) 13 (b) 10 (c) 11 (d) 5
4. c
We know that 1 has only one factor which is neither prime nor composite. So, we are left with only 23 numbers. We can find out the numbers which have composite number of factors by excluding those numbers which have a prime number of factors from the rest 23 numbers. k The numbers of the form a - 1, where a and k are prime numbers, have a prime number of factors. Therefore, there are 12 numbers, each less than 2 25, which have a prime number of factors and these numbers are 2, 3, 2 , 5, 2 4 7, 3 , 11, 13, 2 , 17, 19 and 23. Hence, the required number of numbers = 23 -12 = 11.
16.
F(a,b) = F
=
If ‘a’, ‘b’ and ‘c’ are distinct positive integers such that any two of them are co-prime, then what is the sum of a, b and c? (a) 6 (b) 7 (c)8 (d) 9 20. c a, b and c are distinct distinct positive integers such such that any two of them are co prime. .’. HCF (a,b) =1 and LCM (a,b) = ab
* +
.’. F(a,b) = Now, F
= F(ab,c) =
=
Abc = 12 = 1 x 3 x 4 Hence, the sum of of a, b and c = 1 + 3 + 4 = 8.
26.
‘x’, ‘y’ and ‘z’ are real numbers such that x > 6, y < 0 and z > -2. How many of the following statements are definitely true? (i) x+y+z>0 (ii) xy + yz + zx = 0 2 4 (iii) x + y + z < 0 (a) (b) (c) 2 (d) 3
26. a Given, x > 6, y <0 and and z > -2. (i) x + y + z will always not be greater than zero, since y < 0. Ex: x = 7, y = -9, z = 1 (ii) xy + yz + zx will always not be zero. Ex : x =7, y = -2, z = -1. 2 4 (iii) x + y + z will always be greater than zero, since x > 6. Hence, none of the three statements is true.
UNProctored 1
1. What is the remainder remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + … + 12 × 12! is is divided by 13? (a) 1 (b) 12 (c) 11 (d) 0 Solution: 1. b
1! + 2 × 2! + 3 × 3! + 4 × 4! +…+ 12 × 12! = (2! – (2! – 1!) 1!) + (3! – (3! – 2!) 2!) + (4! – (4! – 3!) 3!) +…+ (13! – 12!) 12!) = 13! – 13! – 1 1 So the remainder = – = – 1 i.e. 12
( )
11. What is the highest possible value of ‘n’ for which (a) 13 (b) 10 (c) 11
– 1 1 is divisible by (d) 12
?
Solution:
+1)(
= +1) +1) ( = +1)( +1)( = +1)( +1) ( = +1)( now +1 when divided by , always gives a reminder of 2. Therefore, the highest power of 2 which can divided each of thems +1)( 1024
9
2
1
Hence, 3 1 is divisible by 2 × 2 × 2 and so the highest power of 2 by which it is divisible = 9 + 2 + 1 = 12
17. A school having 270 students provides facilities facilities for playing four games – games – Cricket, Football, Tennis and Badminton. There are a few students in the school who do not play any of the four games. It is known that for every student in the school who plays at least N games, there are two students who play at least (N – (N – 1) 1) games, for N = 2, 3 and 4. If the number of students who play all the four games is equal to the number of students who play none, then how many students in the school play exactly two of the four games? (a) 30 (b) 60 (c) 90 (d) 120
Solution: 17. b Let Let there be ‘a’ students who play exactly four games. Then there will be ‘2a’ students who play at least three games i.e. either three or four games. Hence, we can say that ‘a’ students play exactly three games. Similarly, there are ‘2a’ students who play play exactly two games and ‘4a’ students who play exactly one game. Also, since ‘a’ students play all four games, the number of students playing none of the four games should also be ‘a’. Hence, a + 4a + 2a + a + a = 270 i.e. a = 30. So the number of students who play exactly two games is 2a i.e. 2 x 30 = 60. 21. Some numbers can be expressed as the sum of three of their factors. E.g. 12 can be expressed as the sum of 2, 4 and 6. How many other such numbers are there which are less than 100? (a) 16 (b) 15 (c) 7 (d) 2 Solution:
21. b Let N be one such number. The The factors of N in the decreasing order can be listed as
If N is not divisible by 2, then the three largest factors of N can be
,Since their sum is less than N, ,Since
must be a factor of N.
Similarly, if N is not divisible by 3, then the next two
largest factors of N after can be and Again,
since their sum is less than N, must be a factor of N. Now,
and must be added to to get N. Hence, N
must be divisible by 6. The total number of multiples of 6 below 100 are 16, including the number 12. So there are 15 such numbers other than 12.
UNP 3
2.
Seven students appeared for an exam in which the maximum marks were 100. The marks scored by the students students in the exam were found found to be seven distinct integer values. The arithmetic mean of their marks was 88 whereas the median of their marks was 92. What W hat could be the maximum difference between the marks of any two students? (a) 54 (b) 46 (c) 57 (d) 60
2. a
The median of their marks was 92, therefore, one student must have scored 92, three must have scored more than 92 and three less than 92. If the difference in the marks scored by any two students is to be maximized, then the sum of the marks of the top six students must be made as high as possible in order to arrive at the lowest possible marks for the seventh student. Therefore, marks of the top six students can be 100, 99, 98, 92, 91 and 90. Marks of the seventh student = 88 × 7 – (100 + 99 + 98 + 92 + 91 + 90) = 46 Hence, the required difference = 100 – 46 = 54
∴
3.
Find the smallest number which when divided by 3, 5, 7 and 11 leaves remainders 2, 4, 6 and 1 respectively. (a) 419 (b) 104 (c) 209 (d) None of these
3. a
The form of a number number which when divided by 3, 5 and 7 leaves remainders 2, 4 and 6 respectively = k × (LCM of 3, 5 and 7) – 1, where k is a natural number = 105k – 1 Now, (105 k – 1) leaves remainder 1 when divided by 11, if k = 4, 15, 26, … Hence, the smallest such number = 105 × 4 – 1 = 419.
10.
If a number is written as 100100100100100100100 in Base 2, how will it be written in Base 8? (a) 4444444 (b) 7777777 (c) 2222222 (d) None of these
12.
Find the nu number mber of possible solutions of m! = a! + b! + c!, where m, a, b and c are positive integers. (a) 2 (b) 3 (c) 7 (d) 1
12. d Given m! = a! + b! + c! ...(i) Since m must be greater than each of a, b and c, the maximum value that each of the three can take is (m – 1). The minimum value of a! + b! + c! is 3, hence, m cannot
be less than 3. For m > 3, m! > 3 × (m – 1)! a! + b! + c! When m = 3, the only possible solution is a = b = c = 2. Hence, there is only one solution to the given equation.
19.
A natural number N has ‘k’ disti nct prime factors. If the total number of factors of N is 72, then what is the product of all the possible values of k? (a) 180 (b) 60 (c) 24 (d) 120
21.
Two distinct numbers are picked randomly from the first ten natural numbers. Find the probability that the L.C.M. of the two numbers is 6.
(a)
21. c
(b)
(c)
UNP 4
A = {1, 4, 7, 10, ..., term} B = {9, 16, 23, 30, ..., term} If S = , how many elements are there in S?
(a) 3722 1.a.
(d) None of these
The total number of ways in which two numbers can be picked from the first ten natural numbers = The pairs of two distinct natural numbers from 1 to 10 whose LCM is 6 are (6, 1), (6, 2), (6, 3) and (3, 2). Hence, the probability =
1.
(b) 3732
(c) 3914
A = {1, 4, 7, 10, ..., term} B = {9, 16, 23, 30, ..., term} We first need to find n(A B).
(d) 3924
The term of A=1+(n-1)3=3n-2 The term of B=9+(m-1)7=7m+2. n(A B). will be equal to the number of solutions of 3n-2=7m+2, wher n and m are natural numbers less than 2005.
n=
If m=2, n=6, m=5, n=13, . . . m=857, n=2001.
∴ )=
Hence, n(
4.
)=2004+2004-286= 3722
∈
How many terms of the sequence given below are integers?
(a) 2
4.c
) =n(A)+n(B)-
(b) 3
(c) 4
(d) 5
, where m is an integer.
For m to be an integer,(2n-1) should be 1,3,7 or 21 i.e. one of the factors of 21. Hence, the number of integers in the sequence is 4.
6.
If y
(a) – 13 6. d
what is the maximum possible integer value of y such that x < 0?
(b) – 14
(c) – 16
(d) – 17
Had x been a positive real number, then
( A.M.
G.M.)
⇒ √ ⇒ Since x
< 0,
Hence, the maximum possible integer values of y is -17.
15.
What is the maximum possible integer value of p if , where n is a natural number? (a) 30 (b) 31 (c) 32 (d) 33
* + * +
15.b 7 8 9 10 ….. 135=n =n
In order to find the maxium possible value of p, we have to calculate the highest power of
in
Highest power of
in
Highest power of 5 in
=
=31
=32
[33-1] =32
Where [ ] denotes the greatest integer function. Therefore, the highest power of 720 in
is 31.
Hence, the maxium possible values of p is 31. 19.
India fielded ‘n’ (> 3) bowlers in a test match, and they operated in pairs. If a particular bowler did not bowl in pair with at least two other bowlers in the team, then at most how many bowlers could have bowled in pair with every other bowler in the team? (a) n – 3 (b) n – 1 (c) n – 2 (d) None of these
19.a let the bowlers be represented by
If a particular bowler (say) did not bowl in pair with two other bowlers, and (say), then and also did not bowl in pair with every bowler in the team. Therefore, atleast atleast three bowlers must must not have bowled in in pair with every other bowler in the team. Hence, the maximum number of bowlers who could have bowled in pair with every other bowler in the team is n-3.
22.
If = 1, 2, 3, 4, ... , 23 and = 207, 208, 209, 210, 211, ... , 691 , how many elements of the set are divisible by at least four distinct prime numbers that are elements of the set ? (a) 9 (b) 8 (c) 11 (d) 12
22.c if a number of the set is divisible by atleast four distinct prime numbers of the set , then it will be divisible by their product as well. The number of numbers in divisible by the product of 2,3,5 and 7, i.e. 210=3 The number of numbers in divisible by the product of 2,3,5 and 11 i.e. 330=2 The number of numbers in divisible by the product of 2,3,5 and 13, i.e. 390=1 The number of numbers in divisible by the product of 2,3,5 and 17, i.e. 510=1 The number of numbers in divisible by the product of 2,3,5 and 19, i.e. 570=1 The number of numbers in divisible by the product of 2,3,5 and 23, i.e. 690=1 The number of numbers in divisible by the product of 2,3,5 and 11, 11 , i.e. 462=1 The number of numbers in divisible by the product of 2,3,5 and 13, i.e. 546=1 There is no other combination of four or mor e prime numbers in set that divides any of the elements of set Hence , the required number of elements =11
28.
A florist sells only two kinds of flowers – Rose and Tulip. On a particular day, he sold 70 Roses and 90 Tulips. If none of his customers bought more than one flower of each type, what is the minimum number of customers that must have visited his shop on that day? (a)160 (b) 90 (c) 120 (d) 70
28.b
his shop must have been visited by at least 90 customers. The number of customers who bought both Rose and Tulip=70. The rest 20 customers must have bought just 1 Tulip each.
29.
The question given below is followed by two statements, A and B. Mark the answer using the following instructions:
Mark (a) if the question can be answered by using one of the statements alone, but cannot be answered by using the other statement alone. Mark (b) if the question can be answered by using either statement alone. Mark (c) if the question can be answered by using both the statements together, but cannot be answered using either statement alone. Mark (d) if the question cannot be answered even by using both the statements together. Q. x and y are natural numbers. What is the product of x and y? A. The sum of x, y and their LCM is 11. B. The sum of x and y is 5. 29.c
From statement A: As A: As the sum of x,y nd their LCM is 11, the possible pairs of the values of x and y will be (1,5) and (2,3). Therefore, the product of x and y ei ther 5 or 6. Hence, statement A alone is the not sufficient. From statement B: As x+y=5, the possible pairs of the values of x and y will be (1,4) and (2,3). Therefore, the product of x and y is either 4 or 6. Hence, statement B alone is not sufficient. From statement A and B: combining the two statements, statements, we get the product of x and y as 6
UNPROC 5 2.
2. c
If = 3, (a) 255
= 7 and (b) 1023
for n
(c) 511
3, then the value of (d) None of these
is
= 3 = – 1 = 7 = – 1 = 3.7 – 2.3 = 15 = – – 1 = 3.15 – 2.7 = 31 = – 1 =
– 1 = 511.
9.
A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1’s, then six consecutive 2’s, and likewise six 3’s, six 4’s and six 7’s in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible div isible by 11. Find the number of possible values of the three-digit number. (a) 82 (b) 92 (c) 81 (d) 91
9. c
The original number is 111111222222333333444444777777. It is divisible by 11. Let the three-digit number that is inserted in the number be ‘abc’. = 111111222222333333444444abc777777 For the resulting number to be divisible by 11, abc must also be divisible by 11.
∴
Hence, the number of possible values of abc is the number of three-digit numbers that are divisible by 11 = 81
UNPROC 5 17.
How many two-digit natural numbers satisfy integer less than or equal to x? (a) 7 (b) 10 (c) 9 (d) None of these
17.d
*+ * +
, where [x] denotes the greatest
*+ * +
Since the value of the right hand side of the given equation is 1 for all two-digit natural numbers, therefore, the values of x that will satisfy the equation are 10, 11, 12, 13, ..., 17. Hence, the number of possible solutions = 8.
26.
Let Z = {1, 11, 111, 1111, ..., 11111 up to 55 times} and and T be a subset of Z such that the sum of no two elements of T is divisible by 3. The maximum possible number of elements in T is (a) 20 (b) 18 (c) 19 (d) 37
26. a
Z = {1, 11, 111, 1111, ..., 11111 up up to to 55 times} The number of elements in Z that leave remainders 1, 2 and 0 when divided by 3 are 19, 18 and 18 respectively. T will have maximum number of elements if it contains all the 19 elements of Z which leave remainder 1 when divided by 3 and one more element which leaves remainder 0 when divided by 3. Hence, T can have a maximum of 19 + 1 = 20 elements.
UNPROC 6 5.
Find the last two digits of (a) 34 (b) 76
. (c) 24
(d) None of these
5. c
The last two digits of = 24. Any number with the last two digits as 24 when raised to any odd power gives the last two digits as 24 and when raised to any even power gives the last two digits as 76. = = 1024 raised to any odd power will have the last two digits as 24. Hence, will have the last two digits as 24 .
15.
A natural number consists of onl y 0’s and 1’s. If the number is divisible by 375, then what is the least possible number of 0’s and 1’s in the number respectively? (a) 6 and 3 (b) 3 and 6 (c) 6 and 9 (d) None of these
15. d
375 = 3
If a number is divisible by 125, then the last three digits of the number should be divisible by 125. Since the number consists only 0’s and 1’s, the last three digits of the number will be 000. As the number is divisible by 3, the sum of digits of the number should be divisible by 3. Therefore, there should be at least three 1’s in the number. Hence, the smallest such number = 111000.
18.
A shopkeeper has 2721 apples, 3081 bananas and 3501 peaches. He makes baskets of these fruits such that the number of fruits of each type in a basket is either ‘x’ or 0. In the end the shopkeeper is left with p (< x) fruits of each type. Find the maximum possible value of x. (a) 1 (b) 10 (c) 15 (d) 60
18. d
As the the shopkeeper shopkeeper is left with with the same number of fruits of each type, 2721, 3081 and 3501 when divided by x should leave the same remainder i.e. p. The maximum possible value of x = HCF (3081 – 2721, 3501 – 3081) = HCF (360, 420) = 60.
24.
A society of 380 people organized a tournament comprising three different games. The number of people who participated in at least two games was 42% more than those who participated in exactly one game. At least one person participated in exactly n games, where n = 1, 2, 3. If the number of people who did not participate p articipate in any of the three games was minimum possible, then what was the maximum possible number of people who participated in exactly two games? (a) 149 (b) 150 (c) 212 (d) 213
24. c
Let the number of people people who participated participated in exactly exactly 0, 1, 2 and 3 games be N, I, II and III respectively. Therefore, II + III = 1.42 I …(i) and N + I + II + III = 380. N + 2.42 I = 380 [Using (i)] I= Since N is minimum possible, (380 – N) must be the largest possible multiple of 121. 380 – N = 363 N = 17 I = 50 × 3 = 150 From (i), II + III = 1.42 × 150 = 213 Hence, the maximum possible value of II = 213 – 1 = 212.
⇒⇒ ⇒⇒ ⇒
27.
Rohan and five of his friends contributed equal money and hired a car to Goa. At least one of his friends pulled out of the trip at the last moment and so they decided to divide the expenses between the rest of the people. During the journey j ourney the car required some minor repair for which they paid Rs250 extra. The total travelling expenses, including the repair charges, amounted to Rs410 per person. If each of the six friends had contributed an integral amount more than Rs100 initially, then what was their total initial contribution? (a) Rs1,800 (b) Rs570 (c) Rs2,460 (d) Cannot be determined
27. a
Let the initial amount (in !) contributed by each of the six friends be ‘p’ (> 100) and number of people (including Rohan) who went to Goa be ‘n’.
Total travelling expenses (in Rs) including the repair amount = 410 × n Initial amount (in Rs) contributed by the six friends = 6p. As per the given information, 410n – 6p = 250
⇒ p =
Since p > 100 and 0 < n < 6, n = 5 and p = 300. Hence, the initial amount contributed by the six friends = 6 × 300 = Rs1,800.
Unproc 7 7.
Which of the following are factors of (i) 16 (ii) 349 (iii) 2443 (a) Only (iii)
7.b
(b) (ii) and (iii)
?
(c) (i) and (iii)
(d) All of them
As 37 is an odd number, the given expression will be divisible by ( + ) i.e. 2443. It will also be divisible by 349 since it is a factor of 2443, but not by 16 as the value of the given expression is odd.
10.
If (6) n1 (7)n2 = (33)n2, where n1 and n2 are bases such that n 1 can n1 assume? (a) 7 (b) 6 (c) 8 (d) 13
n2, how many different values
10. a
(6 × n 10) × (7 × n 20) = 3 × n 2 + 3 × n 20 n2 = 13 The value of n1 must be greater than 6 as the digit 6 has been used in base n 1. Therefore, the values which n1 can assume are 7, 8, 9, …, 13. Hence, the answer = 7.
16.
A bag contains 35 tokens numbered 1 to 35. Five of them are picked randomly and are found to have five consecutive numbers. If the product of the five numbers is P, how many distinct values that P can assume are multiples of 16? (a) 23 (b) 15 (c) 21 (d) 31
16. a
Any sequence of five consecutive natural numbers contains either two or three even numbers.
⇒
Case (i): The sequence contains three even numbers. In this case the product of the five numbers will always be a multiple of 16 as the product of any three consecutive even numbers is a multiple of 16. The number of such sequences possible = The number of even numbers less than 31 = 15. Case (ii): The sequence contains two even numbers.
As the H.C.F. of two consecutive even numbers is 2, their product will be a multiple of 16 if and only if one of the two numbers is a multiple of 8. Since there are four multiples (i.e. 8, 16, 24 and 32) of 8 which are less than 35 and the number of required sequences in which each e ach of these multiples can appear is 2, the number of the possible required sequences will be 8. Hence, the answer = 15 + 8 = 23. 17.
If a girl has coins of denominations Rs1, Rs2 and Rs5, in how many ways can she make a payment of exactly Rs11? (a) 12 (b) 9 (c) 10 (d) 11
17. d
Let the number of coins of denominations Rs1, Rs 2 and Rs 5 used to make the payment be x, y and z respectively. x + 2y + 5z = 11 Now the possible values of z are 0, 1 or 2.
∴ ∴
Case (i): z = 0 x + 2y = 11 The ordered pairs (x, y) that satisfy the above equation are (1, 5); (3, 4); (5, 3); (7, 2); (9, 1) and (11, 0). Therefore, the number of ways in this case = 6.
∴
Case (ii): z = 1 x + 2y = 6 The ordered pairs (x, y) that satisfy the above equatio n are (0, 3); (2, 2); (4, 1) and (6, 0). Therefore, the number of ways in this case = 4.
∴
Case (iii): z = 2 x + 2y = 1 The only ordered pair (x, y) that satisfies the above equation is (1, 0). Hence, the total number of ways = 6 + 4 + 1 = 11.
29.
If B f represents the number of bijective functions from S 1 = {a, b, c, d, e} to S 2 = {p, g, r, s, t} such that f(c) t and f(e) p, what is the value of B f? (a) 120 (b) 84 (c) 114 (d) 78
29. d
As a bijective function function is a one-to-one one-to-one onto onto function, the n number umber of bijections from from S 1 to S2 is nothing but one-to-one associations of each element of o f S 1 to a unique element of S 2 and viceversa under the given constraints. The number of one-to-one associations from S 1 to S2 = 5! = 120. The number of one-to-one associations from S 1 to S2 when ‘c’ is associated with ‘t’, but ‘e’ is not associated with ‘p’ = 3 × 3 × 2 × 1 = 18. The number of one-to-one associations from S 1 to S2 when ‘e’ is associated with ‘p’, but ‘c’ is not associated with ‘t’ = 3 × 3 × 2 × 1 = 18. The number of one-to-one associations from S 1 to S2 when ‘c’ is associated with ‘t’ and ‘e’ is associated with ‘p’ = 3 × 2 × 1 = 6.
Hence, the answer = 120 – (18 +18 + 6) = 78.
Unp 8
6.
Which of the following could be a possible value of ‘x’ for which each of the fractions
is in its simplest form, where [x] stands for the greatest integer less than or equal to ‘x’? (a) 45.45 (b) 49.49 (c) 51.51 (d) 53.53
6. c
A fraction is said to to be be in its simplest form when the numerator and the denominator are co-prime. If we observe the fractions carefully, we find that in each term a remainder of 2 is left when the integer part of the numerator is divided by b y the denominator. E.g. 2 by 10, 13 by 11, 26 by 12, 41 by 13, and so on. The fractions can be written as:
,1+
, 2+
,3+
and so on…
Thus ‘x’ needs to be such that [x] + 2 is co -prime with 10, 11, 12, …49 and 50. Among the options, the only such value is 51.51.
7.
The sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17,…. has one odd number followed by the next two even numbers, then the next three odd numbers followed by b y the next four even numbers and so rd on. What is the 2003 term of the sequence? (a) 3953 (b) 3943 (c) 3940 (d) 3950
7. b
We have the following alternate sequences of odd and even terms: Number of Terms Terms 1 1 2 2,4 3 5,7,9 4 10,12,14,16 5 17,19,21,23,25 6 26,28,30,32,34,36 7 37,39,41,43,45,47,49 8 50,52,54,56,58,60,62,64 If we observe the sequences carefully the last term in any sequence is the square of the number of terms, i.e. when n = 3, last term = 9; when n = 4, last term = 16; when n = 8, last term = 64 and so on… Also, the total number of terms in the sequence is the sum of the number of terms in the alternate sequences of even and odd terms. Since
rd
=1953, we can say that the 2003 term will lie in a sequence of odd terms and will
be the 50th term in that sequence. The last term in the sequence of even terms with n = 62 will be 2 th 62 = 3844. Hence, the next odd sequence begins at 3845. The 50 term in this sequence will be 3845 + 49 × 2 = 3943.
11.
In how many ways can 1000 be written as a sum of ‘n’ consecutive natural numbers, where ‘n’ is greater than 1? (a) 0 (b) 1 (c) 2 (d) 3
11. d
Case I: The number of terms ‘n’ is odd. The middle term will be the average of all the terms. Let the average be k. k n 1000 (Both k and n are natural numbers.) n umbers.) Odd factors of 1000 are 1, 5, 25 and 125. (i) When n = 5, k = 200, the consecutive numbers are from 198 to 202. (ii) When n = 25, k = 40, the consecutive numbers are 28 to 52. (iii) When n = 125, k = 8, the consecutive numbers will not be natural numbers.
∴
Case II: The number of terms ‘n’ is even. The average of the middle two terms will be the average of all the terms. Let the average be k. k n 1000 (n is a natural number and k is a positive rational number whose decimal part is 0.5.) (i) When n = 16, k = 62.5, the consecutive numbers are from 55 to 70. (ii) When n = 80, k = 12.5, the consecutive numbers will not be natural numbers. (iii) When n = 400, k = 2.5, the consecutive numbers will not be natural numbers. So the total number of possible ways is 3.
∴
Unproc 9
1.
How many positive integer values of ‘a’ are possible such that a (a) 12
1.
a a 220
(b) 13
=
a 4 216
1
(c) 15
a 220
is an integer? a4 (d) 16
216
a4 a4 a4 Therefore, (a + 4) must be a factor of 216. The number of factors of 216 = 16 But (a + 4) cannot be equal to 1, 2, 3 and 4 as ‘a’ has to be a positive integer. Total possible values = 16 – 4 – 4 = 12
4.
4.
In a test consisting of 15 questions, 3 marks are awarded for a correct answer, 1 mark is deducted for an incorrect answer and no mark is awarded for an unattempted question. If a student attempts at least one question in the paper, what is the number of distinct scores that he can get? (a) 57 (b) 58 (c) 59 (d) None of these b
A student student can get any integer integer score from from –15 –15 to 45, except 44, 43 and 40. So the answer is 61 – 3 = 58.
Unp 9 7.
What is the remainder obtained obtained when the sum of the squares of any thirty consecutive natural numbers is divided by 12? (a) 0 (b) 3 (c) 11 (d) Cannot be determined
7.
c Let a, a + 1, a + 2, …, a + 29 be thirty consecutive natural numbers and N be the sum of their squares. N = a2 + (a + 1) 2 + (a + 2)2 +…+ (a + 29) 2 N = 30a 2 + 2a(1 + 2 + ….+ 29) + (1 2 + 22 + …+ 292) 29 30 29 30 59 2 N 30a 2a 6 2 Remainder when
6 29 30
Now 30a 30a 2a 2
29 30 59
is divided by 12 = 11
2 30(a 29a)
2 As (a + 29a) is always even, remainder obtained when 30(a 2 + 29a) is divided by 12 = 0. So the remainder obtained when the sum of the squares of any thirty consecutive natural numbers is divided by 12 = 11. 2
10.
If a 2 + b2 = 1, c 2 + d2 = 1 and ac + bd = 0, where a, b, c and d are real numbers, then what is the value of a 2 + c2? (a) 2 (b) 1 (c) 0 (d) Cannot be determined determined
10.
b ac + bd = 0 ac = – = –bd bd 2 2 2 2 ac =b d 2 2 2 2 a (1 – (1 – d d ) = (1 – (1 – a a )d 2 2 a = d 2 2 2 2 Since c + d = 1, c + a = 1.
18.
The sum of the digits of a four-digit four-digi t number is 31. What fraction of such numbers are are divisible by 11? (a) ¼ (b) 1/5 (c) 1/6 (d) None of these
18.
d Let the four-digit number be abcd. For the number to be divisible by 11, (a + c) – (b – (b + d) = 0 or +11k, where k is a natural number. Let us assume that a + c = x and b + d = y. x + y = 31 and x – y – y = 11
(For x and y to be integers, x – y – y can neither be 0 nor an even multiple of 11.) Solving the above equations, we get x = 21 and y = 10. As x is the sum of two two single digit digit numbers, numbers, the maximum maximum possible value of x is 18. Therefore, no such number is possible.
23.
a and b are natural numbers such that a > b > 1. If 8! is divisible by a 2 × b2, then how many such sets (a, b) are possible? (a) 5 (b) 6 (c) 7 (d) 8
23.
c There are four prime numbers less than 8 i.e. 2, 3, 5 and 7. The highest powers of 2, 3, 5 and 7 in 8! are 7, 2, 1 and 1 respectively. The possible sets of values of (a 2, b2) are: (32, 22) (42, 22) (42, 32) (62, 22) (82, 32) (62, 42) (122, 22)
Unp 9
27.
27.
w, x, y and z are natural numbers such that: 2 5 (i) log x (ii) log w y z 3 4 What is the value of ‘x – w’? (a) 81 (b) 117
(iii) y – z – z = 9 (c) 93
c
24
Since both x and w are natural numbers, the above is true only 2 1 when x 3 5 and w 5 = 4 Hence, x = 125, w = 32 and x – w – w = 93.
30.
What is value of 10.11.12.13 + 11.12.13.14 + … + 96.97.98.99?
(d) 109
(a) 1806869592 30.
(b) 1806869594
(c) 1806869596
(d) 1806869598
a
96.97.98.99 =
1 5
[100.99.98.97.96 – [100.99.98.97.96 – 99.98.97.96.95] 99.98.97.96.95]
Adding the the terms given above, above, we get 10.11.12.13 + 11.12.13.14 + … + 96.97.98.99 1 = [100.99.98.97.96 – 9.10.11.12.13] – 9.10.11.12.13] 5 1806869592 Note: We need not calculate the exact value; just checking the last digit of the answer will suffice in this case. 1 [100.99.98.97.96 – 9.10.11.12.13] – 9.10.11.12.13] 5 = 20.99.98.97.96 – 20.99.98.97.96 – 9.2.11.12.13 9.2.11.12.13 [ends with 0] [ends with 8] Hence, the unit digit of the answer must be 10 – 8 – 8 = 2 and there is just one option with unit digit 2.
Unp 10
5.
There are two signals which keep changing their colours continuously according to a pattern. The first signal S 1 remains Red for 5 seconds, then
turns Green for the next 7 seconds, then turns Blue for the next 6 seconds, then Red again for 5 seconds and so on. The second signal S 2 remains Red for 7 seconds, then turns Green for the next 11 seconds, then turns Blue for the next 8 seconds, then Pied again for 7 seconds and so on. If S 1 and S2 turn Red simultaneously, then after how much time (in seconds) will they turn Blue simultaneously simultaneously for the first time? (a) 140 (b) 66 (c) 72 (d) 78 1. b
One cycle of S1 takes 18 seconds and that of S 2 takes 24 seconds. Therefore, the two signals turn Red R ed simultaneously after every 72 seconds (L.C.M of 18 and 24). Also, exactly 6 seconds before they turn Red, R ed, they would have turned Blue simultaneously. Hence, the two signals will turn Blue simultaneously simultaneously for the first time exactly 66 seconds after the start.
10.
Find the number of positive integer solutions of the equation + (b) 1
(a) 0
10.b We have +
(c)2
(d)3
= 5
= 5
=> 2y + 15x = 5xy => 5xy – 5xy – 2y 2y -15x = 0 => (y – (y – 3) 3) (5x – (5x – 2) 2) = 6 Now, 6 can be written as as 2 x 3, -2 x -3, 1 x 6 Or -1 x -6. The only possible case is 5x – 5x – 2 2 = 3 and y – y – 3 3 = 2. Therefore, x=1 and y=5
19.
N = 7777......................7777, where the digit 7 repeats itself 429 times. What is the remainder left when N is divided by 1144? (c) 777 (a) 913 (b) 1129 (d) None of these
19.c
A natural number formed by using the same digit written 6 times (eg. 777777) is always divisible by 1001 i.e. 7 11 13. Also, N can be written as:
7777...................7777 1000 + 777 ------- 426 times ------times -------→ .’. 7777...............7777 1000 is always divisible ------- 426 times ------times -------→ by 1144 i.e. 11 13 8.
Required remainder =777
28.
How many factors of 20! Are there whose unit digit is 5? (a) 2068 (b) 1868 (c) 1728 (d) None of these
28.c The prime factorized form of 20! 18 8 4 2 1 1 1 1 =2 3 5 7 11 13 17 19 8 2 1 1 1 1 1 2 When any factor of 3 7 11 13 17 19 is multiplied by 5 , 5 , 3 4 5 or 5 , the unit digit of the resulting number will be 5. 8 2 1 1 1 1 The number of factors of 3 7 11 13 17 19 = 9 3 2 2 2 2 = 432. The number of factors of 20! Whose unit digit is 5 = 432 4 = 1728.
Unp 11
3. If N = 1! – 2! + 3! – 4! +…..+ 47! – 48! + 49!, then what is the unit digit of NN? (a) 0 (b) 9 (c) 7 (d) 1 3. d 3. d The unit digit of every term from 5! to 49! is 0. Also, 1! – 2! + 3! – 4! = 1 – 2 + 6 – 24 = –19. Hence, the unit digit of N will be 10 – 9 = 1. The unit digit of NN will also be 1.
18.
4 ≤ p ≤ 9, –0.25 ≤ q ≤ 0.36, –0.49 ≤ r ≤ –0.01 and
r
2 s
p
2
, where p, q, r and s
q
are real numbers. What is the difference between the maximum and the minimum possible values of s? (a) 0.0875 (b) 0.105 (c) 0.210 (d) None of these
18.
d r
2 s
p
2
q
= qr = p 2q2 But as –0.49 < r < –0.01, we must have q < 0 or –0.25 < q < 0. 2 qr s p
2
Maximum possible value of s 2
(0.25)(0.49) 42
Maximum and minimum possible values of s are 0.0875 and –0.0875 respectively. Hence, the answer = 2 × 0.0875 = 0.175.
Unp 11 22. Two different two-digit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers? (a) 70 (b) 71 (c) 72 (d) 73
22.
d As the larger larger number number is written written on the the left, the the larger number number is either either 54 or 55. 55. Let the smaller number be x. Case I: The larger number is 54. 5400 + x = 5481 + 54 – x 2x = 5535 – 5400 = 135 (In this case x will not be a natural number.) Case II: The larger number is 55. 5500 + x = 5481 + 55 – x 2x = 5536 – 5500 = 36 x = 18 Hence, the required sum = 73.
27. A = 2812, B = 188 and C = 216. How many natural numbers are there by which at least one among A, B and C is divisible? (a) 499 (b) 501 (c) 504 (d) 505
27.
d A = 2812 = 224 × 712 B = 188 = 28 × 316 C = 21 6 = 36 × 76 The total number of factors of A will be (24 + 1)(12 + 1) = 325. Similarly, the total number of factors of B and C will be 153 and 49 respectively. Case I: Factors common to A, B and C. HCF of (A, B and C) = 1. The only factor common to A, B and C is 1. Case II: Factors common to exactly two among A, B and C. Factors common to A and B = 2 0, 21,…, 27, 28 Factors common to B and C = 3 0, 31,…, 35, 36 Factors common to A and C = 7 0, 71,…, 75, 76 Hence, the answer = (325 + 153 + 49) – (9 + 7 + 7) + 1 = 505.
Unp 12 2.
What is the number of common terms in the two sequences given below? S1 = 4, 10, 16, 22, …., 562. S2 = 3, 8, 13, 18, …., 573. (a) 20 (b) 40 (c) 36 (d) 18
2.
d The first common term of the two sequences is 28. As the common difference of the two sequences are 6 and 5 respectively, we can say that any number of the form ‘30n – 2’, where n is a natural number, will be common to both the sequences. The maximum possible value of such common term will be 30 × 18 – 2 – 2 = 540 – 540 – 2 2 = 538. Hence, the number of common terms in S 1 and S2 is 18.
6.
For how many positive integer values of ‘x’ is ||||x – 1| – 1| – – 2| 2| – – 3| 3| – – 4| 4| < 5? (a) 10 (b) 14 (c) 4 (d) 15
6.
b For the highest possible value of integer ‘x’ which satisfies the given inequality, say xmax, all the modulus brackets will open with a positive sign i.e. |x – 1| – 1| > 0, ||x – ||x – 1| 1| – – 2| 2| > 0, |||x – |||x – 1| 1| – – 2| 2| – – 3| 3| > 0 etc. When all the brackets open with a positive sign, x – 1 – 1 – – 2 2 – – 3 3 – – 4 4 < 5 or x < 15 or x max = 14.
15.
How many numbers are there between 0 and 500 which when when divided by 3, 4, 6 and 8 leave remainders 1, 2, 4 and 6 respectively? (a) 10 (b) 20 (c) 21 (d) None of these
15.
b L.C.M. of 3, 4, 6 and 8 is 24. Each remainder left on division is 2 less than the divisor. So if k is a divisor then the dividend must be of the type nk – 2 – 2 where n is a natural number. To satisfy the condition for all the four divisors we have to consider L.C.M. of 3, 4, 6 and 8 in place of nk. So the smallest such number = 24 – 2 – 2 = 22. The next number will be 22 + L.C.M. (2, 4, 6 and 8) = 22 + 24 = 46. A general general expression for any number N satisfying satisfying the the given condition condition is N = 24k – 2. – 2. If 0 < N < 500, then N can be: 22, 46, 70,... 478. As 478 = 24 × 20 – 20 – 2, 2, the answer is 20.
Unp 12
20.
How many four-digit four-digi t even numbers are are there which have 0 and 1 as two of the digits and the remaining two are distinct digits among 2, 4 and 6? (a) 42 (b) 36 (c) 38 (d) 30
20.
a Suppose we do not select digit 6 and the four digits are 2, 4, 0 and 1. When digit 1 is at the thousands place: The digit 0 can be placed at any position. Total such numbers = 1 × 3 × 2 × 1 = 6.
CASE I.
When digit 1 is not at the thousands place: Possible positions for the digit 1 = 2. Possible positions for the digit 0, once the digit 1 has been fixed = 2. Now digits 2 and 4 can be placed at the remaining two positions in 2 ways. Total such numbers = 2 × 2 × 2 = 8. Adding the the above two cases, cases, there are 14 such numbers. numbers. Similarly, we will get 28 more such numbers when instead of 2 and 4 the digits selected are either 2 and 6 or 4 and 6. Hence, the answer = 14 × 3 = 42 CASE II.
23.
Amit wrote down the squares of of first ‘n’ natural natural numbers numbers in his notebook. notebook. He He then erased erased one of the numbers written by him and found that the sum of the numbers remaining on
the notebook was 1432. The square of which of the following numbers was erased by Amit? (a) 1 (b) 4 (c) 6 (d) 8 23.
d Sum of the squares of first n natural numbers =
n(n 1)(2n 1) 6
By substitution we can see that 1 + 22 + 33 + 44 + … + 16 2 = 1496. As 1496 – 1496 – 64 64 = 1432, the number whose square was erased must be 8. 24.
The H.C.F. of a, b and c is 8. If a – b – b = b – b – c c = 8 and the L.C.M. of a, b and c is a fourdigit number, then what is the maximum possible value of c? (a) 80 (b) 88 (c) 96 (d) 100
24.
c Let c = 8k. Hence, b = 8(k + 1) and a = 8(k + 2), where k is a natural number. As the H.C.F. H.C.F. of ‘k’, ‘k ‘k + 1’ and ‘k ‘k + 2’ is always always 1, we can we can say that the L.C.M. of ‘k’, ‘k + 1’ and ‘k + 2’ will be kk1k2, when k is odd, k(k 1)(k 2) and , when k is even. 2 The L.C.M. of a, b and c is either 8k k1k2 or 4k k1k2. L.C.M < 10000 (given) For highest value of k we’ll take 4k(k+1)(k+2) < 10000, where k is even. Hence, k(k + 1)(k + 2) < 2500. The maximum possible value of k = 12. The maximum possible value of c = 96.
Unp 13 2.
2.
Which of the following number(s) is/are not prime? (i) 25001 + 1 (ii) 25002 + 1 (iii) 25003 + 1 (a) (i) and (ii) (b) (i) and (iii) (c) (ii) and (iii) d
an bn
4.
Remainder 0, if n is odd. ab 25001 + 1 and 2 5003 +1 are divisible by 3. Now, 25002 + 1 can be written as 4 2501 +1. 42501+1is divisible by 5. Hence, none of the given numbers is a prime number. How many positive integer pairs (x, y) satisfy x+ y= 2003 ? (a) 0 (b) 1 (c) 2 (d) 3
(d) (i), (ii) and (iii)
4.
a
x 2003 y Squaring both the sides, we get x 2003 y - 2 2003 y For x to be an integer, y must be of the form 2003a 2, where ‘a’ is a natural number. Similarly, we can say that x must be of the form 2003b2, where ‘b’ is a natural number. 2003a 2003b 2003 2003a 2 2003b 2 2003 a b 1 Hence, no solution is possible. 2
2
9.
The sum of thirty-two thirty- two consecutive natural numbers is a perfect square. What is the least possible sum of the smallest and the largest of the thirty-two numbers? (a) 81 (b) 36 (c) 49 (d) 64
9.
c Let the numbers be a, a + 1, a + 2, ……, a + 31. 31 32 Sum of these numbers = 32a = 16(2a + 31) 2 As 16 is a perfect square, square, the the least possible possible value value of 2a + 31 31 = 49. Therefore, a = 9 and a + 31 = 40. The least possible sum = 49.
12.
In a three-digit number, the unit digit is twice the tens digit and the tens digit is twice the hundreds digit. The same number is written as 1XY and 1YX in base 8 and base 9 respectively. Find the sum of X and Y in the decimal system. (a) 15 (b) 7 (c) 11 (d) Cannot be determined
12.
c Let the hundreds digit be n. The tens digit will be 2n. The unit digit will be 4n. The possible values of ‘n’ are 1 and 2 and hence the possible numbers are 124 and 248 respectively. On converting 248 in base 8 and base 9, the given condition gets violated. On converting 124 in base 8 and base 9, we get (174) 8 = (147) 9. Required sum = 4 + 7 = 11.
24.
What are the last two digits of 948367483? (a) 21 (b) 87 (c) 27
(d) None of these
24.
b If N is a natural number such that N = a pbqcr ... where a, b, c etc are prime numbers, then 1 1 1 Euler’s Totient function is function is given by (N) = N1 1 1 a b b If P is some other natural number which is prime to N, then the remainder when P (N) is divided by N is 1. To find the last two digits we need to find the remainder when 9483 67483 is divided by 100. 1 1 1 100 = 100 1 1 1 = 40 2 5 b 3
Required remainder = = 83 67483 = 8340k+ 3 = 83 = 87.
26.
26.
Two sets A and B are given below. A = {20, 21, 22, 23, 24} B = {30, 31, 32, 33, 34} How many different proper fractions can be made by picking the numerator from one of the sets and the denominator from the other set? (a) 24 (b) 20 (c) 12 (d) None of these a A = {1, 2, 4, 8, 16} 16} B = {1, 3, 9, 27, 81} The number of possible proper fractions when denominator is equal to 2, 4, 8 and 16 are 1, 2, 2 and 3 respectively. The number of possible proper fractions when denominator is equal to 3, 9, 27 and 81 are 2, 4, 5 and 5 respectively. Since no two of these fractions can be equal, the answer = 24. Alternate method: All method: All the combination combinations s can result result in a proper proper fraction except except when when 1 is chosen from both the sets. So the answer = 5 × 5 – 1 – 1 = 25 – 25 – 1 1 = 24.
28.
The question given below is followed by two statements, A and B. Mark the answer using the following instructions: Mark (a) if the question can be answered by using one of the statements alone, but cannot be answered by using the other statement alone. Mark (b) if the question can be answered by using either statement alone. Mark (c) if the question cannot be answered even by using both the statements together. Mark (d) if the question can be answered by using both the statements together, but cannot be answered by using either statement alone.
Q.
If p and q are natural numbers, then what is the remainder when (p + q) is divided by 7? A. 6q – 6q – p p = 5 B. 8q + p = 45
28.
b
From Statement A: 6q – p = 5 p = 6q – 5 p + q = 7q – 5 (7q – 5) when divided by 7 leaves remainder 2. This statement alone is sufficient to answer the question. From Statement B: 8q + p = 45 p = 45 – 8q p + q = 45 – 7q (45 – 7q) when divided by 7 leaves remainder 3. This statement alone is sufficient to answer the question.
Unp 14 2.
In a class of 57 students, each student plays at least one of the three games – H o c k e y , C r i c k e t a n d Footba Football. ll. If 27 27 play play Hock Hockey, ey, 28 play play Cricke Cricket, t, 32 32 play play Foot Footba ballll and and 8 play play all all the three games, wh what at is the number of students who who play exactly two games ? (a) 13 (b) 14 (c) 15 (d) 16
2.
b Let x, y and z be the number number of students who who play exactly one, one, two and three games respectively. Then, x + y + z = 57 x + y = 49 ……………. (i) (Given, z = 8) Also Al so,, x + 2y + 3z = 27 + 28 + 32 x + 2y = 87 – 2 – 2 4 x + 2y = 63 ………….. (ii) Solving (i) and (ii), we get y = 14. Hence, the number of students in the class who who play exacly two games = 14
Unp 14 6.
If ‘C’ is a positive integer, how many three -digit numbers ‘abc’ are there such that both ‘abc’ and ‘cba’ are divisible by 4? (a) 40 (b) 30 (c) 34 (d) None of these
6.
a For a number to be divisible by4, the last two digits should be divisible by 4. Therefore, ‘bc’ and ‘ba’ should be divisible by 4 sim ultaneously. If b is even, then both a and ca can take two value – 4 – 4 or 8. If b is odd, then both a and c can take two values -2 or 6.
Hence, for any value of b, there will be 4 numbers. As th ere er e are ar e 10 ways wa ys (0 -9 ) to ch oose oo se b, the th e to ta l such su ch numb nu mb er s = 4 10 = 40.
7.
The question given below is followed by two statements, A and B. Mark the answer using the following instructions: Mark (a) if the question can be answered by using statement B alone, but not by using statement A alone. Mark (b) if the question can be answered by using statement A alone, but not by using statement B alone. Mark (c) if the question can be answered by using both the statements together, but cannot be answered by using either statement alone. Mark (d) if the question cannot be answered even by using both the statements together.
Q. If ‘n’ is a natural number less than 10, is the number N = nnnn. ….nnnn divisible by 11? A. A . N i s o f t h e f o r m ( 6 k + 5 ) . B . N c o n s i s t s o f ( 2 p - 6) digits gits,, where here p is a natu natura rall numbe mber gre greater ter than than 3. 7.
a From Statement A: We cannot determine if N is divisible by 11 or not. Therefore, Statement A alone is not sufficient. From Statement B: any number with identical digits is divisible by 11 if the number of digits is even. The number of digits in N is (2p – (2p – 6), 6), which is even. Therefore, N is divisible by 11. Hence, Statement B alone is sufficient to answer the question.
8.
If x, y and z are natural numbers such that x < y < z and How many values can ‘y’ take? (a) 0 (b) 1
8.
(c) 2
1 x
1
1
1
=
1
1
1
x y z 3 4 5 Therefore, x = 2 1 1 1
1 x
1 y
1
>1. z
<1.
x z 2 Now, y has to be 3 since if y 3, then the maximum value of 1 1 1 1 1
y z 4 5 2 Therefore, z = 6 Hence, only one value of ‘y’ is possible
y
1
=1,
(d) None of these
b It is obvious that x 1 as
1
z
Unp 14 20.
For how many of the first 300 300 even natural numbers is the total number of factors even ? (a) 88 (b) 276 (c) 288 (d) 90
20. c All the naturals that have an odd numbers of factors are perfect square and vice versa. Out of the f irst 300 even natural numbers, there are 12 perfect square i.e. 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484 and 576. Hence, there are 300 – 1 – 1 2 = 288 numbers among the first 300 even natural numbers that have an even number of factors.
x
23.
A number N = 5 y is the product of twenty twenty consecutive three-digit numbers. If x and y are positive integers, what is the maximum possible value value of x?
23.
b Since 625 contains the highest power of 5 among all the three-digit numbers, 625 must be one of the twenty numbers. In any gr oup of twenty consecutive number including 625, there will be three other multiples of 5. Hence, the maximum possible value of ‘X’ = 4 + 3 = 7.
Unp 14 26.
What is the sum of all the five-digit five-digi t numbers that can be formed by arranging the digits 2, 2, 3, 4, 5 and6, and6, using each each digit digit once and only only once? once? (a) 5333280 (b) 3333390 (c)6999390 (d) 5666901
26.
a There will be 5! = 120 such numbers. In these 120 numbers, each of 2, 3, 4, 5 and 6 appear at units place exactly 20 times. The sum of all digits at units place = 24 2 + 24 3 + 24 4 + 24 5 + 24 6 = 24 20 = 480. Same will be true for tens digit, hundreds digit, thousands digit and ten thousands digit as well. The sum of all such numbers = ( 480 10000 + 480 100 + 480 10 + 480 1) = 5333280 Note: We could have used the formula (given below) for finding the sum of all the numbers that can be formed by using ‘n’ distinct digits without repetition. (n – (n – 1)! 1)! (111…n times) sum of n digits.
Unp 15
10. If (24) is a factor of 5! X 10! X 15! , what is the largest possible integer integer value of k?
(a) 5
(b) 7
(c) 11
(d) 4
10. b In order to find the maximum possible value of k, we have have to calculate the highest power of 24 ( =
Highest power of
=[
]=[
x 3)in 5! x 10! x 15!
in 5! x 10! x 15!
] = 7
Where [] denotes the greatest integer function. Highest power of 3 in 5! X 10! X 15! 1+4+6 =11 Therefore the highest power power of 24 in 5! X 10! X 15! Is7. Hence the maximum possible value of k is 7.
11. The product of three natural numbers – numbers – H.C.F. How many ordered triplets ( (a) 12
(b) 15
11.d Let the values of
(c) 18
,
,
,
- is twelve times their
) are possible?
(d
21
be xh , yh , and zh respectively , where h is
the H.C.F. of the three numbers.
∴
Xh x yh x zh = 12 x h
→
x x x y x z = 12
Since h is a natural number, there are two possible cases. Case (1): h = 1 x x y x z = 12 = 1 x 1 x 12 = 1 x 2 x 6 =1 x 3 x4 = 2 x 2 x 3 Number of ordered triplets of (x, y, z) i.e.
,
= 3 + 6 + 6 + 3 = 18
Case (2) h = 2 Xxyxz=3 =1x1x3 Number of ordered triplets of ( x , y, z ) i.e. ( Hence, the answer = 18 + 3 = 21
,
) = 3
14. The mean, median and mode temperature temperatur e of the seven days of a particular week in a city were 5.429 temperature (i
(approximately),
respectively. If the
) on each of the seven days was a positive integer, than which
of the following could not be the temperature (i (The data set contains only one mode) (a) 1
and
(b) 3
(c) 2
14.b Let the temperature (i d, 9 and 9.
) on any of the seven days?
(d) None of these
) on the seven days in ascending order be a, b, c, 7,
a + b + c + d = 13 As d ≥ 7 and only mode as 9, the only possible solution is a = 1, b = 1, c = 2 and d = 9 Hence,
wasn’t the temperature on any of the seven days.
Unp 15
24. Which of the following numbers cannot be expressed as the product of two prime numbers that lie between 30 and 45? (a) 1147
(b) 1333
(c) 1779
(d) 1591
24. c The prime numbers between 30 and 45 are 31, 37, 41 and 43. Therefore, all the possible pairs of the numbers are (31, 37); (31, 41); (31, 41); (37, 43) and (41, 43). Thus, thus possible products can be 1147, 1271, 1333, 1517, 1591, and 1763. Hence, the answer is 1779. Alternate solution: It could have been observed that the last digit of the product of any two prime numbers between 30 and 45 cannot be 9.
28. if f(n) represents the sum of the digit digit (s) of n for n = 1,2 , 3, 4,….., find find the remainder when f(1) f(2) f(3) f(4) …. F(100) is divided by 90. (a) 1
(b) 11
(c) 46
(d) 0
28.a while writing the the first 99 natural numbers, numbers, each of the nine digits ( 1, 2, 3, ……, 9) is used exactly 20 times.
∴
f (1) (1) f(2) f(3) f(4) …… f(100) = 20 x (1 2 3 …..9) 1 = 901
Hence, the remainder = Rem [
] =1.
Direction for question 29 and 30: Answer the question on the basis of the information given below.
Unp 16 11.
N = 2.xxyxxyxxy…, where at most one of the digits x and y is 0. Which of the following numbers, when multiplied by N, necessarily results in an integer? (a) 198 (b) 666 (c) 990 (d) 1998
11.
d N 2.xxyxxyxxy... 1000N2xxy.xxyxxy... Therefore, 999 × N = 2xxy – 2. – 2.
2xxy 2
999 Hence, N must be multiplied by 1998, as 1998 is a multiple of 999. 1421
15.
What is the unit digit of + 7 (a) 3 (b) 5
15.
25 15 +4 ? (c) 7
(d) 9
b We know that the cyclicity of the unit digit of 7 is 4 and that of 4 is 2. 4k p 1421 We can express 7 as 7 . 21 Now, 4k + p = 14 . 1421 (2) 21 (4)10 2 = Rem 0 = Rem So the value of p = Rem = 4 4 4
Therefore, the unit digit of 7 Similarly, the unit digit of 4
1421
= 7
4k
will be 1.
15 25
will be 4. 25 1421 15 4 Hence, the unit digit of (7 ) will be (1 + 4) = 5.
Unp 16 20.
How many natural numbers are factors of at least one of 6 14, 1012 and 15 20? (a) 835 (b) 834 (c) 794 (d) 795
20.
d 614 = 214 × 314; 1012 = 212 × 512; 1520 = 320 × 520 The number of factors of 6 14 = (14 + 1)(14 + 1) = 225 The number of factors of 10 12 = (12 + 1)(12 + 1) = 169 The number of factors of 15 20 = (20 + 1)(20 + 1) = 441 Among the the three, the lowest lowest power of of 2 is 12, that of 3 is 14 and that of 5 is 12. 212 has 13, 3 14 has 15 and 5 12 has 13 factors and they have 1 as their common divisor. Hence, the total number of factors = 225 + 169 + 441 – 13 – 13 – – 15 15 – – 13 13 + 1 = 795.
21.
Avnish counted the digits used to number the pages of a book and found that the total number of digits used was 3441. Find the number of pages in the book. (a) 1128 (b) 1137 (c) 1413 (d) None of these
21.
b From pages 1 to 9, 9 digits were used. From pages 10 to 99, a total of 2 × 90 = 180 digits were used.
From pages 100 to 999, a total of 3 × 900 = 2700 digits were used. There were 3441 – 3441 – (9 (9 + 180 + 2700) = 552 digits remaining, 552 which would have = 138 pages. 4 Hence, the total number of pages = 9 + 90 + 900 + 138 = 1137.
Unp 16
25.
4
Find the value of
1 1
2 5 10 1
(a) −4 5 25.
(c) 5+ 5
(b) 8
4 1 1
(d) None of these
1 1
1
52 52
10 4 5
(2 5 )(2 5 ) 2
1
=
2 5 2 1 5 22
4
=
2 5 10
4
2 5 10
1
4
=
4
=
2 5 10 1
1
52
b =
=
52
2 5 4 5 10 4 5
4 1
1
4
=
1
2 5 2 2 5
= 8.
2
26.
A table of of ‘n’ rows and and ‘n’ columns columns is created created such that the value value of the cell in the i th row and the j th column is given by (i + 1) + j. Some numbers are selected from the table. If it is found that exactly one number has been selected from each row and each column, then the sum of the selected numbers will be equal to n(n 1) (a) n2 + 1 (b) (c) n(n + 2) (d) n(n + 1) 2
26.
c One number is selected from each row. Let us say in row one, the number is picked from –1 and so on. column j n, in row two the number is picked from column j n –1
–1 + 1, 3 + j n –2 –2 + 1,…, i + jn –i+1 –i+1 +1,…, So all the numbers in the set will be 1 + j n + 1, 2 + j n –1 n + j1 + 1. Sum of all the elements of any such set
Alternate solution: Consider a 1 × 1 table. The only value in the table is (1 + 1) + 1 = 3. Putting n = 1 in all the options, only n(n + 2) gives the value as 3.
27.
From the first ‘n’ consecutive natural numbers written on a blackboard, one of the numbers is removed. If the average of the remaining numbers is 32 number removed. (a) 50
27.
(b) 64
(c) 46
2 9
, then find the
(d) None of these
a Let a number ‘x’ be removed from the first n consecutive natural numbers written on the blackboard.
We know that the maximum decrease in the average of the first ‘n’ natural numbers can be caused by removing ‘n’ and minimum decrease (or maximum increase) in the same average can be caused by removing 1.
as 1< x < n
n < 64.44 < (n+2) n = 63 or 64
will be an integer, (n – 1) – 1) should be a multiple of 9. Therefore, n = 64.
n(n 1) 64 65 x x 2 2 2 290 290 32 (n 1) 9 (64 (64 1) 9 x = 32 × 65 – 65 – 290 290 × 7 = 50.
Alternate solution: Maximum change in the average can be of
1 2
or
1 2
, therefore ‘n’ should be either 63
or 64. It can also be seen that the average of the first ‘n’ natural numbers is close to
will be an integer, (n – 1) – 1) should be a multiple of 9. Therefore, n = 64.
64 65 n(n 1) x x 2 290 290 2 2 32 (n 1) 9 (64 (64 1) 9 x = 32 × 65 – 65 – 290 290 × 7 = 50.
n
2
