Number Theory, Number Systems : CAT Quant Since
CAT
2001,
this
is
an
oft-repeated
topic
in
CAT
Number Theory / Number systems is a topic from which you can expect anywhere between 4 to 10 marks worth of questions in the quant section of CAT. This topic will definitely feature amongst the A category topics if one were to do an ABC analysis of topics that figure in CAT Math. Ascent's Arithmetic Refresher covers this topic in depth. The chapter starts from absolute basic and covers advanced topics in Number Theory. The following areas are covered in this topic in the refresher material. 1.
Definition of different types of numbers from Natural Number to Real numbers, including irrational numbers, surds, binomial quadratic surds and their conjugates, complex numbers and their conjugates.
2.
Prime numbers and method to determine if a given number is a prime number. Co-prime numbers, applications and examples.
3. 4.
Tests of divisibility including shortcuts for divisibility of numbers such as '7' and '13'. Concepts and solved examples to elucidate the concept of LCM and HCF.
5. 6.
Fractions - Proper fractions, improper fractions, LCM and HCF of fractions. Factors and method to find the number of factors for a given integer. Method to find the sum of all positive factors of a natural number.
7.
Properties of remainders and quotients of division of large numbers for e.g., 371 by 8.
8.
Remainder theorem and its application with examples - an oft repeated question in CAT since CAT
9.
2001. Factorials explained with examples. Highest power of a number contained in a factorial.
10. Properties of unit digits of numbers raised to large powers. For example unit digit of 14328971179. 11. Conversion of numbers from one base to another and back to decimal notation. 12. Successive division of numbers and remainders of such successive division The chapter contains
•
Over 15 illustrative examples to explain the concepts
•
Around 35 solved examples (with shortcuts wherever applicable) to acquaint you with as many different questions as possible
•
30 exercise problems with answer key and explanatory answers to provide you with practice and
•
An objective type speed test with around 60 questions. Explanatory answers and answer key are provided for the speed test
Sample Question in Number Theory This question appeared as a 2-Marks question in CAT 2005 What is the righmost non-zero digit of the number 302720? (1) 1 (2) 3 (3) 7 (4) 9 Correct Answer: Choice (1).
Explanatory Answer As the given number is a multiple of 10, the last 2720 digits will be 0s. The right most non-zero digit of the number will be the unit digit of the number 32720. The unit digit of the powers of "3" follow a cyclicity of "4" i.e., The units digit of 31 is 3 The units digit of 32 is 9 The units digit of 33 is 7 The units digit of 34 is 1. Then the cyclicity sets in. The units digits of 35 is 3. The units digits of 36 is 9 and so on. In 32720 as 2720 is a multiple of "4", the number will have the same units digit as 34 which is "1". Hence, the rightmost non-zero digit of the number is "1". A collection of questions that typically appear in the Common Admission Test (CAT) from the topic Number Theory. These questions will guide you through your CAT and other MBA entrance exam preparation. 1>
If both 112 and 33 are factors of the number a * 43 * 62 * 1311, then what is the smallest possible value of a? 1.
121
2.
3267
3.
363
4.
33
Correct choice (3). Correct Answer - (363) Explanatory
Answers
112 is a factor of the given number. The number does not have a power or multiple of 11 as its factor. Hence, "a" should
include
112
33 is a factor of the given number. 62 is a part of the number. 62 has 32 in it. Therefore, if 33 has to be a factor of the
given
number
a
*
43
*
62
*
1311,
then
we
will
need
at
least
another
3.
Therefore, if "a" should be at least 112 * 3 = 363 if the given number has to have 112 and 33 as its factors.
2>
Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28.
(1)
99840
(2)
99900
(3)
99960
(4)
99990
Correct Choice is (3) and the correct answer is 99960 Explanatory The Hence,
number it
Answers should is
be
enough
exactly to
divisible check
by the
99960 is the only number which satisfies the given condition.
15
(3,
divisibility
5), for
21
(3, 3,
7), 4,
28 5
(4,
7).
and
7.
3>Question Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product
went
up
by
540.
What
is
the
new
product?
(1)
1050
(2)
540
(3)
1440
(4)
1590
Correct choice is (4) and Correct Answer is 1590 Explanatory Let
Answers
the
She
number
was
Instead,
that
Anita
expected she
wanted
to
find
found
to
multiply
the
the
be
value
value
of
of
'X'. 35X.
53X.
The difference between the value that she got (53X) and what she was expected to get (35X), according to the question,
is
i.e.,
53X
or
540.
-
(53
-
35X 35)
=
*
X
X
540 =
540
=
30
Therefore, new product = 53 * 30 = 1590
4.>
Question
Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which one of the following
statements
(1)
cannot
(x-z)2
be
y
true?
is
even
(2)
(x-z)y2
is
odd
(3)
(x-z)y
is
odd
(4)
(x-y)2z
is
even
Correct Choice is (1) and Correct Answer is (x-z) 2 y is even Explanatory x
Answer
and
y
(x x
are
-
odd
and
z)2 z
positive
y is
and
z
is odd
is
even
even and
and
cannot y
is
positive true odd
Therefore, (x - z)2 will be odd and (x - z)2 y will be odd
5>Question When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided
by
12?
(1)
10
(2)
7
(3)
192
(4)
None
of
these
Correct Choice is (2) and Correct Answer is 7 Explanatory
Answer
Let When
the 'a'
is
divided
by
number 36,
i.e.,
let
the
quotient
and
or
=
a
'q'
and
we
know
remainder
a
when
be
be
is
36q divided
by
'a'. the
remainder
is
19
is
19
+
19
12,
we
get
or 36q
is
perfectly
divided
by
12
Therefore, remainder = 7
6>Question The
sum
of
(1)
the
first
100
2,
(2)
numbers,
1
to
100
4
is
divisible
and
2
8
and
(3)
4
2
(4)
by
only
None
of
these
Correct Choice is (3) and Correct Answer is 2 only Explanatory
Answer
The sum of the first 100 natural numbers is given by (n(n + 1))/2 = (100(101))/2 = 50(101). 101 is an odd number and 50 is divisible by 2. Hence, 50*101 will be divisible by 2.
7>Question How
many
different
factors
are
there
for
the
number
48,
excluding
1
and
48?
(1)
12
(2)
4
(3)
8
(4)
None
of
these
Correct Choice is (3) and the correct answer is 8 Explanatory
Answers
To find the number of factors of a given number, express the number as a product of powers of prime numbers. In
this
case,
Now,
increment
48
the
can
power
of
be each
written of
the
as
prime
16
*
numbers
by
3 1
=
and
(2 4
multiply
* the
3) result.
In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1) Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 - 2 = 8 factors.
8Question: 1025 - 7 is divisible by 1.
2
2.
3
3.
9
4.
Both (2) and (3)
Correct Answer - 3. Correct choice is (2) Explanatory
Answers
1025 The
-
7
number
Therefore,
1025
(1025
-
-
= 1
1)
= -
6
(1025 99.....9 =
(24
(25
digits)
nines
and
1) is
divisible
unit
digit
by is
3 3)
6 and
4.
99.......93.
This number is only divisible by 3 (from the given choices).
9>Question Find
the
G.C.D
of
12x2y3z2,
18x3y2z4,
and
24xy4z3
(1)
6xy2z2
(2)
6x3y4z3
(3)
24xy2z2
(4)
18x2y2z3
Correct Choice is (1) and Correct Answer is 6xy 2 z 2 Explanatory G.C.D
Answer of
12,
18
and
24
is
6.
The common factors are x, y, z and their highest powers common to all are 1, 2 and 2 respectively. Therefore, G.C.D = 6xy2z2
10>Question What is the value of M and N respectively? If M39048458N is divisible by 8 and 11; Where M and N are single digit integers? (1)
7,
8
(2)
8,
6
(3)
6,
4
(4)
5,
4
Correct Choice is (3) and correct answer is 6, 4 Explanatory
Answers
If the last three digits of a number is divisible by 8, then the number is divisible by 8 (test of divisibility by 8). Here,
last
three
digits
58N
is
divisible
by
8
if
N
=
4.
(Since
584
is
divisible
by
8.)
For divisibility by 11. If the digits at odd and even places of a given number are equal or differ by a number divisible
by
11,
then
the
given
number
is
divisible
Therefore, (M+9+4+4+8)-(3+0+8+5+N)=(M+5) should be divisible by 11 => when M = 6.
by
11.