Back

Divisibility

ALL MODULES

CAT 2025 Lesson : Divisibility - Past Questions

5. Past Questions

Question 1

How many even integers

n

, where

100 \le n \le 200

, are divisible neither by seven nor by nine?
[CAT 2003 L]

		40
		37
		39
		38

Observation/Strategy

1

) We are dealing with even integers only. So, we look for numbers divisible by

14

and

18

instead of

7

and

9

.
2) We find the number of

14

18

multiples and subtract them from the total.
3) Multiples of LCM(

14

18

)

= \space 126

will be double counted as they are multiples of both

14

and

18

100 \le n \le 200

, where

n

is even.
Number in the range are from

2 \times 50

2 \times 100

Total numbers in range

= 100 - 50 + 1 = \bm{51}

14

multiples in the range are from

14 \times 8

14 \times 14

Number of $\bm{14}$ multiples in the range

= 14 - 8 + 1 = 7

18

multiples in the range are from

18 \times 6

18 \times 11

Number of $\bm{18}$ multiples in the range

= 11 - 6 + 1 = \bm{6}

126

multiples in the range

= \bm{1}

(which is

126

itself)

Number of

14

18

multiples in range

= 7 + 6 - 1 = \bm{12}

Numbers that are not multiples of

14

18

in the range

= 51 - 12 = \bm{39}

Answer: (3)

39

Question 2

For two positive integers

a

and

b

, if

(a + b)^{(a + b)}

is divisible by

500

, then the least possible value of

a \times b

is:
[XAT 2016]

		8
		9
		10
		12
		None of the above

Observation/Strategy

1

)

500 = 2^2 \times 5^3

2

and

5

are it's only prime factors.

2

) The base of

(a + b)

should be the smallest multiple of

2

and

5

, which is

10

3

) Smallest common multiple of

2

and

5

10

10^{10}

is divisible by

500

.

∴

a + b = 10

Least possible value of

a \times b = 1 \times 9 = 9

Answer: (2)

9

Question 3

The unit digit in the product of

(8267)^{153} \times (341)^{72}

is
[IIFT 2012]

		1
		2
		7
		9

Observation/Strategy

1

) The two terms are being multiplied.

2

) We can ignore

341^{72}

as it's units digit will be

1

3

)

7

has a cyclicity of

4

. The power of

153

has to be written in

4k

format.

U\left((8267)^{153} \times (341)^{72} \right) = U\left((8267)^{153} \right)

= U(7^{153})

153

when divided by

4

yields a remainder of

1

U(7^{153}) = U(7^{1}) = 7

Answer: (3)

7

Loading Video....