7m+4=11n+6=13p+3 smallest such number will be say x req number =x+(7*11*13)l to find the smallest numberfrom the first two parts=39 from the last two parts=94 hence 39+77k=94+143f=809 req number=809+2002=2811 P(2811,17)=6
How many positive integers between 100 & 200 are coprime to 630?
But (x-1)^5=x^5-5x^4+10x^3-10x^2+5x-1= -x^2+5x-10-10x^2+5x-1= -11x^2+10x-11= -11(x-1)+10-11=-x, so (x-1)^5+1/(x-1)^5= -x+1/(-x)=-1 If x+1/x=-1 then result of (x+1)^5+1/(x+1)^5=1
How many nos are coprime to 7000 below it but are more than 1500? (7000-1400)(1/2)(4/5)(6/7) =1920. Now 1400 to 1500 sake as 1 to 100 Till 70 : 24 70 to 100 : 71, 73, 79, 81, 83, 87, 89, 93, 97, 99 So 34 Answer 1920 - 34 = 1886 how many 8 digit numbers are there having sum of of digits even? OA: 45*10^6 9*10^7 /2
how many 6 digit numbers have exactly 4 distinct digits
Parajumble #3 1. Such pious statements are meaningless when there is so much loose talk about exercising the nuclear option. 2. There is no need to exhibit any injured pride about the West showing concern about the ability of the two countries to handle nuclear weapons. 3. But there is no point in the strategists telling us how India was one of the first countries to be opposed to nuclear weapons, implying that India's weapons are meant for deterrence and not use. 4. It was perhaps a deliberate ploy to whip up fears of nuclear war in South Asia. 5. It does not help either when the Government says India would use nuclear weapons only in retaliation when there are enough politicians, strategists and defence experts egging the Government on to "finish" Pakistan. 24351
1. He must have a social conscience and unerring political antennae but cannot become a passionate advocate or plug a party’s political agenda. 2. He must deliver a daily illustrated epigram to reader’s breakfast tables saying just enough to raise a smile, give pause f or thought and raise the spirit, but nothing so biting that it has the opposite effect. 3. He must never transgress the fine line that divides cheerful caricature from brutal satire. 4. The political cartoonist’s life is a funny business. 5. He can banter but he must not lampoon 42531
Parajumble#1
1. This suggests that between the dialectic and the eristic elements themselves there operates a higher dialectic. 2. A survey of the prose of thought of the twentieth century reveals a certain rhythmic and even cyclic development. 3. Indeed without such a higher dialectic, language would constantly be moving towards one or the other extreme and thus failing as a means of communication. 4. The idealist philosophers, taking the Hegelian System as their point of departure, found themselves confronted with the New Realism. 5. In this development, we may detect, at each stage, a dialectic and an eristic aspect. 25413
Q. Sum of all the possible values of x in the equation |x-2|+|x+4|=5 is ? if x is even both |x-2| and |x+4| are even but 5 is odd. if x is odd both |x-2| and |x+4| are odd and their sun will be even, but 5 is odd. No solutions That is not the logic 5 is less than the minimum possible distance
If X and Y are positive real numbers and X + Y = 1, what is the least value of (X + 1/X)^2 + (Y+1/Y)^2. 12.5 At x = y = 1/2 by symmetry If x = (15^3+16^3+19^3+20^3),then x is divided by 70 leaves a remainder of ? 0 => a^n+b^n+c^n+d^n is divisible by a+b+c+d given that n is odd
907=kN+x 907=mN+x equating, N=2m/(k-m).N to be max k-m should be min.as they are integers.so min is. k-m=1,k=m+1,N=2m putting this in any of the above equation and arranging, 2m²+2m+(x-907)=0 as m is integer,so D is also an integer. D=4-4*2*(x-907)=3632-4x..to be it a perfect square x should be equal to 8. Find the missing number 2, 11, 36,?, 504, 2525 123 127 131 133
(2+9)*1=11 (11+7)*2=36 (36+5)*3=123 0 4 18 48 120 294 ? n^3-n^2 In a race of 100 m, A beats B by 20 meters and C by 25 meters. B beats C by how many meters? a/b= 100/80= 5/4 => a/c= 100/75= 4/3 => b/c= (4/3)/(5/4)= 16/15 So, B beats C by 100 -(100/16)*15= 50/8= 25/4= 6.25m
A Integral solutions for x^2+y^2=19^4*37^7 with solution
Agar 4k+3 type hoga and odd power to zero solutions, Agar 4k+1 and 4k+3 dono hai toh sirf factors of 4k+3 will give positive solutions, 4*positive = integral solutions.
1,3 3,1 Sarah intended to multiply a two digit number and a three digit number but she left out the multiplication sign and simply placed the 2 digit number to the left of 3 digit number thereby forming a 5 digit number. The number is exactly nine times the product Sarah should have obtained. What is the sum of the 2 digit number and the 3 digit number ?
1). 101^1 = 101 The last two digits are always = 01 untill 100^9. 100^9 -> 901 2).
101^10 =10 01 The last 3 digits are 001 untill 100^99. 3). 101^100 = 100 01 The last five digits are 100 01 If we subtract 1 It gives u 10001-1=10000
:)
#iQuanta #D411 3/a+5/b = 1/24 No. of integral and natural no. solutions Natural number 56 Integral 111 natural number should be 55 i think we should subtract the 0/0 case 2(4*2*7)-1= 111
Please explain the divisibility rule for 99 and 101. Not by the coprime numbers multiplication method. Please explain the other method that is there..using 100-1 for 99. I do not understand that
finding the divisibility of a number by 7,13,17,19.the method is known as osculator method.first we have to know the concept of osculator.the osculator of 7 is 2 ( 7* 3 = 21 = 20+1 here we are adding 1 so we need to consider the osculator as negative osculator ) similarly for 13 ,13 * 3 = 39 = 40-1 so the osculator is 40 and is one more osculator and the value is 4 and for 17 ,17 * 3 = 51 = 50+1 so again the osculator is negative and it is 5 and for 19 it is one more osculator 19 = 20-1 ,so the osculator is 2 . if you didn't understand the concept of osculator just remember these for 7 osculator = 2 , sign = ' - ' for 13 osculator = 4 , sign = ' + ' for 17 osculator = 5 , sign = ' - ' for 19 osculator = 2 , sign = ' + ' lets proceed with an example 55277838 is considered and to check it's divisibility by 7. 5527783 8 : 5527783 - 8 *2 =5527767 ( here we have taken the units digit and multiplied it by osculator 2 and subtracted it from the remaining part as the sign is negative as suggested above we have subtracted ) the next step is the following is repeated for the resulting value in the a bove step 552776 7 : 552776 - 7 *2 = 552762 55276 2 : 55276 - 2 * 2 = 55272 5527 2 : 5527 - 2 *2 = 5523 552 3 : 552 - 3 *2 = 546 54 6 : 54 - 6*2 = 42 as 42 is divisible by 7 so the number is divisible by 7.
x + y + z = 13 ( x -3) + (y-2) + (z-1) = 13-3-2-1= 7 So keep all the terms equal = 7/3 Product : (7/3)^3 = 343/27 The sides of the quadrilateral are 20 16 12 10.Find the maximum area of quadrilateral using this sidess? The largest area of a quadrilateral with given sides is that of a cyclic quadrilateral. Hence apply the formula √(s -a)(s-b)(s-c)(s-d) where s = 58/2. So 194.4
In how many ways can 4 married couples seat themselves around circular table a) if there are no restrictions? b) if each husband must sit beside his wife? c) if at least one husband sits beside his wife? d) if no wife sits next to her husband? e) if men and women alternate? f) if each husband must sit beside his wife and men and women alternate? g) if spouses sit opposite each other
a)
If there are no restrictions?
Number of ways in which 4 married couples can be arranged around the circular table = (8-1)!
(∵ there are 8 persons)
= 7! = 5040
(Note : If 8 persons are arranged in a linear way, there are 8! ways to do so. But when it come to a circular arrangement, the arrangements ABCDEFGH, BCDEFGHA, CDEFGHAB, DEFGHABC, EFGHABCD, FGHABCDE, GHABCDEF, HABCDEFG are same (due to rotation). i.e., for each of the 8! arrangements, there are 8 same arrangements. Hence we need to divide 8! by 8. i.e., total number of arrangements = 8!/8 = 7!.
Another way to approach this : Arrange one person at first. Since it is a circular table, it does not make any difference where he sits. Other seats are then distinguishable relative to the occupied seat. Hence the remaining 7 can be arranged in 7! ways. Total number of ways = 7!
In general, n objects can be arranged in (n-1)! ways around a circular table.
Another aspect is when observations can be made from both sides like in the case of necklace. see the first question)
b) If each husband must sit beside his wife?
Group each couple. These 4 groups can be arranged in (4-1)! ways. The two persons in each group can be arranged themselves in 2! ways. Required number of ways = (4-1)! × (2!)4 = 96
for n married couples, required number of ways = (n-1)! × 2n
c) If at least one husband sits beside his wife?
We need to use inclusion–exclusion principle for this. Let's start by finding out the number of ways in which at least one couple are sitting in nearby seats. Group any one couple, say couple1. therefore we can consider there are total 7 objects(the grouped couple and other 6 persons). These 7 objects can be arranged around the circular table in (7-1)! × 2 ways (we multiplied by 2 because husband and wife can interchange their seats within the group). We will get similar arrangements for each of the 4C1 groups. So the estimated ways = 4C1×(7-1)! × 2
But we overcounted the arrangements where two couples are both together(say couple1 and couple2). Let's find out the number of arrangements where couple1 and couple2 are both together. Consider couple1 as one group and couple2 as another group. there are total 6 objects (couple1, couple2 and other 4 persons). These 6 objects can be arranged in (6-1)! ways × 22 ways (we multiplied by 22 because husband and wife can interchange their seats within any of the two groups). Similar overcounting happened with every two particular groups (i.e., couple1 and couple2,
couple1 and couple3, couple1 and couple4, couple2 and couple3, couple2 and couple4, couple3 and couple4) Total overcouting due to this = 4C2 × (6-1)! × 22 therefore, estimated ways = 4C1×(7-1)! × 2 - 4C2×(6-1)! × 2 2
With similar reasoning, we can say overcounted arrangements by considering each three couples together = 4C3 ×(5-1)! × 2 3
Similarly, total overcounted arrangements with all the four couples together = 4C4 ×(4-1)! × 24
Adjusting the overcounts using the principle of inclusion–exclusion, required number of ways = 4C1×(7-1)! × 2 - 4C2 × (6-1)! × 22 + 4C3×(5-1)! ×2 3 - 4C4 (4-1)! × 2 4 = 3552
We can generalize this. For n married couples, required number of ways = n∑i=1(−1)(i−1) nCi×2i(2n−1−i)!∑i=1n(−1)(i−1) nCi×2i(2n−1−i)!
d) If no wife sits next to her husband
This is known as relaxed ménage problem.
Required number of ways = Number of ways in which 8 persons can be seated around a circular table - Number of arrangements where at least one husband sits beside his wife (as we have seen in the previous case) = 7! - 3552 = 1488
e) if men and women alternate?
4 men can be seated around the circular table in (4-1)! ways. 4 women can then be seated in the 4 seats in 4! ways Required number of ways = (4-1)! × 4! = 144
for n married couples, required number of ways = (n-1)! × n!
f) If each husband must sit beside his wife and men and women alternate?
Group each couple. The 4 groups can be arranged around the circular table in (4-1)! ways
in each of these arrangements, we can change the order of all the couples (like AaBbCcDd to aAbBcCdD) to get a new arrangement. Hence, required number of ways = (4-1)!×2 = 12
Alternatively, we can approach this problem in a different way.
Seat first woman in any of the 8 indistinguishable seats. Once we do this, all other seats will be distinguishable relative to the occupied seat. Her spouse can be seated next to her in clockwise direction (only 1 way). Any of the 3 remaining woman can be seated in the next clockwise seat (3 ways) and then her spouse(1 way). Any of the 2 remaining woman in the next clockwise seat (2 ways) and then her spouse(1 way). 1 remaining woman can be seated in the next clockwise seat(1 way) and then her spouse (1 way). Total number of ways = 3×2 = 6
This can be done in anticlockwise direction also which gives another 6 arrangements.
Therefore we can again see that required number of arrangements = 6+6=12
In a similar way, for n married couples, required number of ways = (n-1)!×2
g) If spouses sit opposite each other
Assume the first woman takes one of the eight indistinguishable seats. Her spouse must take the opposite seat.
Now there are 6 seats and these 6 seats are distinguishable with respect to the first woman. The second woman can then be seated in any of these 6 seats. Her spouse takes the opposite seat.
Now there are 4 distinguishable seats and third woman can take any of these 4 seats. Her spouse takes the opposite seat.
Now there are 2 distinguishable seats for the fourth woman and the opposite seat for her spouse.
Hence, total number of arrangements = 6 × 4 × 2 = 48
Similarly we can see that, for n married couples, required number of ways = (2n-2)×(2n-4)(2n-6)....2 = 2(n-1)×(n-1)!
Let use now see one more approach for handling the same problem with n married couples
Let the first woman take one of 2n in distinguishable seats and this seat can now be used as reference.Her spouse can be seated in the opposite seat. There are (n-1) seats between the seats occupied by the first woman and her spouse on one side and another (n-1) seats in the other side. Therefore, we have (n-1)! ways to arrange the remaining (n-1) couples. Position of each of these (n-1) couples can be interchanged. (we cannot interchange the position of the first woman and her spouse because such arrangements can be obtained by rotation of already obtained arrangements). So total number of ways = (n-1)!×2(n-1)
There is a more difficult problem, known as ménage problem, where we need to find out the number of ways in which n married couples can be seated at a circular table such that men and women alternate and each woman is not adjacent to her husband.
What is the length of the longest diagonal in a regular octagon each of whose sides is 10 cm?
Each internal angle is (n-2) pi / n = 6 pi / 8 = 3/4 pi Longest diagonal will be from two opposite ends.
It is also the diameter of the circumcircle of the octagon.
rt (2) *10 rt(2+rt(2))
A number has 288 factors. find the minimum number of composite factors of N
288=2^5*3^2 so take any 5 prime nos raised to powet 1 nd 2 othr prime no raised to power 2 so 288-7(prime no) =281
A natural number has 4 factors. sum of the factors excluding 1 and itself is 30. how many natural numbers are possible?
a+a^2=30 a=5 b+c=30- 3 sets of prime numbers total 4 A month is selected at random from a non-leap year. It is f ound that it has 5 Sundays. What is the probability that it has 5 Mondays? (4/11) * (1/2) + (7/11) * (2/3) = 2/11 + 14/33 = 6/33 + 14/33 = 20/33
How many numbers between 1 and 500 is not divisible by 5 or 6 or 8 ? (1+2+3)-(12+23+31)+(123)=200 ans=300 For how many integral values of x will f (x) = lX - 1! I +lX - 2! I + IX - 3!l+ lX - 4!l assume a minimum value? 23456=5values Remove 7 as it cant be present. Now remove 6 as it has to be present in every case. There is no choice for 6. Now remaining 5 elements in sets. Total 2^5 sets=32 in which 6 will always be present. How many subsets of the set (1,2,3,4,5,6,7) exist where 6 is the largest number? q. How many five digit numbers with distinct digits exist such that first 3 digits are in ascending order and last 3 are in descending order ? a. middle =4 5 6 7 8 9
resp case=3c2*2c2 4c2*3c2 5c2*4c2 .... 8c2*7c2=1134
Q)How many natural numbers 'n' upto 1000 exist such that (n-1)! is NOT divisible by n ?
VVIP concept
f(1) = 4f(4) = 7f(7) = 10f(10) = 13f(13) = 40 => 2f(2) = 5(f(5) = 8f(8) = 11f(11) = 14f(14) => 3f(3) = 6f(6) = 9f(9) = 12f(12) = 6 x 180 => 12f(12) + 13f(13) + 14f(14) = 0 => 14f(14) = -40 - 1080 = -1120 so, f(14) = -80
In how many ways can 4 people be selected from 10 people sitting in a row such that no 2 people are sitting together? Shortcut : (n-r+1)Cr r non consecutive out of n
There are 5 copies of math book, 4 copies of physics book and 3 copies of chemistry book. Find the number of ways in which one or more book can be given away. Cbd until unless mentioned books are identical or distinct If identical : (5+1)(4+1)(3+1)-1= 6x5x4-1= 119 If distinct : 2^(5+4+3) -1= 2^12-1
2^100 - 2^50 = 2^50(2^50 -1
10^h=g^x=10^100k . So k =h/100 10^h = g^h/100>g^g>g!. . 10h=10*10^g=10^g+1 . 10h
If Kush works alone, he will 20 more hours t o complete a task than if he works with Saptarshi. If Saptarshi works alone, he w ill take 5 more hours to complete than if he works with Kush. Ratio of time taken by Kush with Saptarshi? Let us assume that they complete the work together in x hours. Now, if Kush works alone, he will take 20 hours more to complete the task i.e. x+20 hrs, so in 1 hr, he completes 1/(x+20) work Similarly, if Saptarshi works alone, he will take 5 hours more to complete the task i.e. x+5 hrs, so in 1 hr, he completes 1/(x+5) work Now, if they work together, they will complete the task in x hours So, [1/(x+20) + 1/(x+5)] * x = 1 Solving, we get x= 10 hours So, Kush will take 30 hrs and Saptarshi will take 15 hrs. So, ratio is 2:1
when 98 is added to a perfect square, another perfect square is obtained How many such pairs of perfect squares exist? Acc. to question... x^2 + 98 = y^2 => y^2 - x^2 = 98 => (y-x)*(y+x) = 98 => No. of solutions = 0 As 98 is of the form 4k+2 How many numbers are coprime to 2304 and lie between 1000 and 2000? 2304 = consists of 2 and 3. So select range in 6k range . Upto from 1 to 1002, 1002 to 2004 each contains 1002(1/2)(2/3) =334 So including 1001 and excluding 2003 we get 334
If N is a natural number less than 100, then for how many values of N are the numbers 6N + 1 and 15N + 2 relatively prime? If 6N + 1 and 15N + 2 are divisible by x, then their difference i.e. 9N + 1 will always be divisible by x. Similarly (9N + 1) � (6N + 1) i.e. 3N will also be divisible by x. If x divides 3N then it c an also divide 6N. So, x divides 6N and 6N + 1 both i.e. two consecu tive numbers. Hence x cannot be anything but 1. So for all the values of N, the given two numbers will be co-prime.
Rohan has a certain number of (less than 10000) sweets with him. he would be left with 1 sweet, if he distributes them equally among 12,16,18 children. if he distributes them equally among 17 children, he would be left with no sweets.find the number of possibilities for the sweets he has? N = LCM(12,16,18)*x + 1 = 144x + 1 => Possible values = 1,145,289... Now, N = 17a => Possible value of N = 289 And the next number will occur after LCM(144,17) = 2448 numbers... => 289, 2737, 5185, 7633. => Possible numbers = 4
How many natural no. in the set of 1000 numbers can be expressed as the difference of square of 2 natural numbers in atleast one way? x^2 - y^2 = n for solutions, The value of 'n' needs to be any number except the form of 4k+2... => In 1000 numbers, No of numbers of the form 4k+2 = (998-2)/4 + 1 = 250 => the rest of the numbers can be expressed as the difference of square of 2 natural numbers in at least one way = 1000 - 250 = 750
Ans is 168. i dnt know the approach
