# Number System Questions for CAT With Solutions Test-2

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Number System is of the important topic for CAT. The number of questions asked from this topic is high in the exam.

Number System Questions for CAT:

You can download Number System questions or you can go through the details below.

Question 1:

A number is formed by writing first 54 natural numbers next to each other as 12345678910111213 … Find the remainderÂ when this number is divided by 8.

A.Â 1

B.Â 7

C.Â 2

D.Â 0

Question 2:

How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the number is divisible by 125?

A.Â
0

B.
Â 1

C.Â
4

D.Â
3

Question 3:

How many five digit numbers can be formed from 1, 2, 3, 4,Â  5, without repetition, when the digit at the unitâ€™s place must be greater than that in the tenâ€™s place?

A.Â 54

B.Â 60

C.Â 17

D.Â 2 x 4!

Question 4:

If x is a positive integer such that 2x +12 is perfectly divisible by x, then the number of possible values of x is

A.Â 2

B.Â 5

C.Â 6

D.Â 12

Question 5:

Let k be a positive integer such that k+4 is divisible by 7.Â  Then the smallest positive integer n, greater than 2, such that k+2n is divisible by 7 equals

A.Â
9

B.Â 7

C.Â 5

D.Â 3

Solutions:

For a number to be divisible by 8, last 3 digits must be
divisible by 8.
Last 3 digits of this number are 354.
354 mod 8 = 2
Hence, 2 is the remainder.

As we know for a number to be divisible by 125, its last three digits should be divisible by 125 So for a five digit number, with digits 2,3,8,7,5 its last threeÂ  digits should be 875 and 375 Hence only 4 numbers are possible with its three digits asÂ  875 and 375 I.e. 23875, 32875, 28375, 82375

Possible numbers with unit’s place as 5 = $$4 \times 3 \times 2 \times 1 = 24$$ Possible numbers with unit’s place as 4 and ten’s place 3,2,1 = $$3 \times 3 \times 2 \times 1 = 18$$
Possible numbers with unit’s place as 3 and ten’s place 2,1 = $$2 \times 3 \times 2 \times 1 = 12$$
Possible numbers with unit’s place as 3 and ten’s place 1 = $$1 \times 3 \times 2 \times 1 = 6$$
Total possible values = 24+18+12+6 = 60

If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
Hence, there are six possible values of x : (1,2,3,4,6,12)