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.

Download Number System for CAT Set-2 PDF

**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

**Â 1**

B.

B.

**4**

C.Â

C.Â

**3**

D.Â

D.Â

**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:**

**1)**Â Answer (C)

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.

**2)**Â Answer (C)

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

XAT Previous Papers with Solutions

**3) **Answer (B)

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

**4)**Â Answer (C)

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)

**5)**Â Answer (A)

let’s say k+4 = 7m

k = 7m-4

Now for k+2n or 7m+(2n-4) is also Â multiple of 7.

or 2n-4 should be a multiple of 7

So 2n-4 = 7p

or 2n = 7p+4

For p=2; n=9 (p cannot be 1 as n is an integer )