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Q1. What is the remainder when five digit number 23abc is divided by 5 if 'c' is non-zero natural number not more than 5.

• c
• c - 1
• Zero
• Either c or 0

Solution

A number to be divisible by 5 should have 0 or 5 at one's place. Since c is non zero natural number not more than 5, it will be 0 c 5, therefore on dividing the remainder will be either c or 0.

Q2. A two digit number is 5 times the sum of its digits. What is the remainder when this number is divided by 9?

• Zero
• 1
• 4
• 5

Solution

10a + b = 5(a + b) 10a + b = 5a + 5b     ....(1) 5a = 4b Putting 5a = 4b in (1), we get 10a + b = 4b + 5b = 9b i.e. multiple of 9. Hence, dividing the number by 9 gives 0 as remainder.

Q3. A two digit number with distinct digits is divisible by 3. What is maximum possible value of the sum of its digits?

• 9
• 18
• 15
• 12

Solution

The biggest two distinct digit number divisible by 3 is 96 and the sum of its digits is 9 + 6 = 15.

Q4. A three digit number is 297 more than its reversed number. How many such numbers are possible?

• 6
• 27
• 60
• 36

Solution

(100a + 10b + c) - (100c + 10b + a) = 297 Thus, 99(a - c) = 297 a - c = 3From 1 to 9, there are 6 pairs which have the difference of 3. For each such pair, 10 three digit numbers can be formed. Hence, 60 such three digit numbers exist.

Q5. Replace * by the smallest number, so that the number 1*1 is divisible by 3.

• 1
• 2
• 3
• 4

Solution

Here, Sum the known digits of 1*1 = 1 + 1 = 2   Now the number next to 2, which is divisible by 3 is 3.   Thus if we replace the * by 1 the sum of digits will become 3 and the number becomes divisible by 3.

Q6. Sum of a two digit number and the number obtained after reversing the digits is always divisible by:

• 10
• 11
• 13
• 17

Solution

Divisible by 11.

Q7. Rita wrote a three digit number on the white board and then subtracted the sum of its digits from the number itself. Two of the digits of the resulting number are 3 and 4. What is the remaining third digit of resulting number?

• 1
• 2
• 3
• 4

Solution

For a three digit number, the sum of its digits a + b + c cannot be more than 9. Since the two digits are 4 and 3, the third digit is 2.

Q8. A two digit natural number leaves 1 and 3 as remainders when divided by 2 and 9 respectively. If the number is multiple of 5, what is the digit at its ten's place?

• 2
• 3
• 7
• 8

Solution

By applying divisibility test of 2, the digit at unit's place should be 0, 2, 4, 6, 8. But remainder is 1 and the number is multiple of 5. Hence, the digit at unit's place will be 6 - 1 = 5 For a number to be divisible by 9, the sum of the digits should be divisible by 9. It is given that when a number is divided by 9, the remainder is 3. Thus, 75 is the only number which satisfy the given conditions. Hence, 7 is the digit at ten's place.

Q9. 123 + 231 + 312 is divisible by:

• 31
• 34
• 35
• 37

Solution

123 + 231 + 312 = 666 which is divisible by 37.

Q10. If an even number leaves 3 as remainder when divided by 5, what is the digit at one's place?

• 4
• 2
• 8
• 6

Solution

For any number to be divisible by 5, the digit at one's place should be 0 or 5. Number being even and remainder to be 3, the digit at one's place should be 8.

Q11. Difference of a two digit number and the number obtained after reversing the digits is always divisible by:

• 7
• 11
• 9
• 13

Solution

Divisible by 9.

Q12. 68 + 86 is divisible by:

• 11
• 9
• 5
• 13

Solution

68 + 86 = 154 is divisible by 11.

Q13. Which of the following number is divisible by 5 and 10 both?

• 45
• 405
• 450
• 412

Solution

If the unit place of a number is zero then it is divisible by both 5 and 10.

Q14. 718531 is divisible by:

• 7
• 5
• 3
• 11

Solution

Sum of digits at odd places = 7 + 8 + 3 = 18   Sum of digits at even places = 1 + 5 + 1 = 7   Difference = 18 - 7 = 11 Since, 11 is divisible by 11. Hence, 718531 is divisible by 11.

Q15. There is a two digit number ab. Its reversed number is obtained by interchanging the digits as ba. Sum of ab and ba is 77 and one's place of the number is 3. What is the digit at ten's place of the number?

• 3
• 7
• 4
• 2

Solution

(10a + b) + (10b + a) = 77b = 3 Thus, 10a + 3 + 30 + a = 77 11a + 33 = 77 a = 4

Q16. A number is divisible by 6 if it is divisible by:

• 2, 4
• 3, 4
• 2, 3
• 3, 5

Solution

A number is divisible by 6 only if it is divisible by 2 and 3.

Q17. Expanded form of 432 is:

• 4 x 100 + 3 x 1 + 2 x 10
• 4 x 100 + 3 x 10 + 2
• 4 x 1000 + 3 x 100 + 2
• 4 x 100 + 3 x 10 + 1

Solution

432 = 4 x 100 + 3 x 10 + 2.

Q18. A two digit number is 27 more than its reversed number. What is the difference between its digits?

• 2
• 3
• 4
• 7

Solution

ab - ba = 27(10a + b) - (10b + a) = 279a - 9b = 27a - b = 3

Q19. Expanded form of 412 = 

• 4 x 100 + 1 x 10 + 2 x 1
• 4 x 100 - 1 x 10 + 2 x 1
• 4 x 1000 + 1 x 100 + 2 x 10
• 4 x 10 + 1 x 1 + 2 x 0

Solution

4 x 100 + 1 x 10 + 2 x 1

Q20. If the number 6a34b4 is divisible by 3. How many values of a + b are possible?

• 1
• 2
• 3
• 4