19-08-2021 Answers

1. A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two. If 99 is added to the number the digits are reversed. Find the last two digits of the number.

Answer: 53

Solution: Ans: 53 (as the number is 253).

Let the number be xyz.

x+z=y. Hence 2y=10, y=5.

As x+z=5, the x and z can be (0,5) (1,4), (2,3). The numbers can be 055, 154, 253.

Among these three numbers only 253 is reversed when 99 is added to it.

2. What is the unit digit of the expression 8^56 - 2^56 ?

Answer: None of these

Solution: Ans: None of the above (as answer is 0)

If n is even x^n - y^n is always divisible by (x+y).

Hence the expression is divisible by 8+2 = 10 which implies the unit digit is zero.

3. How many numbers are divisible by 3, 5, 7 between 11000 and 22000 inclusive ?

Answer: 105

Solution: To be divisible by 3, 5, 7, it should be divisible by their LCM 105

The first number divisible within the given range is 11025

The arithmetic progression is like 11025, 11130 .... 21945

where a = 11025, d = 105, An = 21945

21945 = 11025 + (n - 1)*105

By solving n = 105

4. In a division problem given in a school, the remainder is 0 as the student mistook the divisor as 12 instead of 21 and obtained 70 as quotient. What is the correct quotient?

Answer: 40

Solution: The number is = divisor * quotient + remainder = 12 * 70 + 0 = 840.

Correct Quotient = 840/21 = 40

5. A number when divided by 296 leaves 75 as remainder. When the same number is divided by 37, the remainder is

Answer: 1

Solution: Let the quotient be q and the number be x.

x = 296q + 75 = 37(8q) + 74 + 1 = 37(8q+2) + 1. 

Hence when x is divided by 37, the remainder is 1.

6. A 3-digit number 4a3 is added to another 3-digit number 984 to give a 4-digit number 13b7, which is divisible by 11. What is the value of (a + b) ?

Answer: 10

Solution: For a number to be divisible by 11, the recursive sum of digits in odd position and even position should be equal.

So, 1+b = 3+7, Solving b=9.

 

4a3+984 = 1397, 4a3 =1397-984 = 413. Thus a=1.

a+b = 9+1 = 10.

7. What is the remainder when the product 5435 × 22 × 17 is divided by 13?

Answer: 10

Solution: The remainders when 5435, 22, 17 are divided by 13 are 1, 9, and 4 respectively. Hence the final remainder is the remainder when the product 1*9*4 is divided by 13.

Hence the remainder when 36 is divided by 13 is 10.

8. The last digit of (2004) ^5 is

Answer: 4

Solution: As 2004 = 2000 + 4, the last digits of (2004) ^5 is same as 4^5.

The power cycle of 4 ends with 4 for odd powers and 6 for evenpowers.

As 5 is odd power, unit digit is 4.

9. 5^6 -1 is divisible by

Answer: 31

Solution: 5^6 -1 can be written as (5^3 + 1)(5^3 -1)

which is nothing but 124*126.

Among the given options only 31 is a factor of 124 or 126.

10. If the remainder is 75 when a number is divided by 85, what is the remainder when the same number is divided by 17?

Answer: 7

Solution: Let the number be x. Then x = 85k + 75 where k is the quotient.

Remainder of (85k+75)/17 = remainder of 85k/17 + remainder of 75/17

= 0 +7 = 7

11. Which is the largest digit that can replace M if the number 5139756MN is divisible by both 5 and 8?

Answer: 8

Solution: To be divisible by 5 the number should end with 0 or 5.

But a number ending with 5 is not divisible by 8. Hence N is 0.

To be divisible by 8, the last three digits should be divisible by 8.

Hence 6M0 should be divisible by 8.

Possible values are 600, 640, 680

The largest digit among 0,4,8 is 8.

12. Which of the following numbers is divisible by 45?

Answer: 4275

Solution: To be divisible by 45, a number should be divisible by both 5 and 9.

Applying the rules of divisibility we find 4275 as the answer.

13. What is the remainder when 9^1+9^2+9^3+....+9^9 is divided by 6?

Answer: 3

Solution: 9 is an odd multiple of 3 and 6 is an even multiple of 3. Hence always the remainder will be 3 when 9^N is divided by 6.

Hence the net remainder will be remainder of (3+3+....+3) (9 times) divided by 6

= remainder of 27 divided by 6 = 3

14. hat is the remainder when 4^31 is divided by 10?

Answer: 4

Solution: 4^1 = 4, 4^2=16, 4^3 = 64, 4^4 = 256.

Thus we see for odd powers the last digit is 4 and for even digit it is 6.

As 31 is odd, the remainder will be 4 when divided by 10.

15. What is the largest 3 digit number that leaves a remainder of 6 when divided by 15?

Answer: 996

Solution: The largest 3 digits number is 999.

On dividing 999 by 15, the quotient is 66 with a remainder of 9.

Hence the required number is 999-9+6 = 996