18-08-2k21 Answers

 1. How many numbers are there between 500 and 600 containing atleast one 9?

Answer: 19

Solution: When 9 is in unit digit, possible numbers = 10 * 1 = 10

When 9 is in tens digit, possible numbers = 1 * 10

But 599 is common and hence repeated twice.

Total = 10+10-1 = 19

2. Find the number of prime factors of 6^10 * 7^17 * 55^27.

Answer: 91

Solution: 6^10 * 7^17 * 55^27 = 2^10 * 3^10 * 7^17 * 5^27 * 11^27

So prime factors = 10+10+17+27+27 = 91 

3. If a positive integer N is divisible by 44 it is also divisible by

Answer: 11

Solution: 44 = 2*2*11.

Hence 1,2,4,11,22,44 are it's factors.

Among the given options, 11 is the answer.

4. Find the unit digit of 1544^56781298

Answer: 6

Solution: For a number ending with 4, even power gives six as unit digit and odd power gives 4 as unit digit.

As 56781298 is even, the unit digit is 6.

5. Find the remainder when 43^23 + 37^23 is divided by 40.

Answer: 0

Solution: 43^23 mod 40 = 3^23 mod 40

37^23 mod 40 = (-3)^23 mod 40 (Using remainder theorem)

So adding these two we get the remainder as 0 mod 40 and hence the remainder is 0

6. In how many ways can 1200 be expressed as a product of two co primes?

Answer: 4

Solution: 1200 = 2^4 * 3 * 5^2

As there are 3 distinct prime numbers, required ways = 3+1 = 4

(We add 1 as 1 and 1200 are co-primes)

7. A number is divided strictly into two unequal parts such that the difference of the squares of the two parts equals 50 times the difference between the two parts. What is the number?

Answer: 50

Solution: Let the number be N and x,y are the parts.

x^2 - y^2 = 50(x-y).

(x+y)(x-y)=50(x-y)

x+y = 50 which is N

8. What is the value of N if A39048458N is divisible by 8? 

Note: A and N represents a single digit from 0 to 9.

Answer: 4

Solution: To be divisible by 8, the last three digits must be divisible by 8.

Hence 58N is divisible by 8.

So N must be 4.

9. What is the least value that must be assigned to * so that the number 197*5462 is divisible by 9?

Answer: 2

Solution: To be divisible by 9, the digital sum must be 9.

Digital sum of 197*5462 = 1+9+7+5+4+6+2+* = 7+*.

Hence least possible value for * is 9-7=2

10. Find the total number of digits in the product of (4)^1111 × (5)^2222.

Answer: 2223

Solution: 4^1111 * 5^2222 = 2^2222 * 5^2222 = 10^2222 which is nothing but 1 followed by 2222 zeroes.

So total number of digits = 1+2222 = 2223

11. Find the number of zeroes in the end in 1^1 * 2^2 * 3^3 ............ * 49^49.

Answer: 250

Solution: 5 is present in 5,10,15,20,25,...45.

So number of fives = 5 + 10 + 15 + 20 + 50 (in 25, we have 5^2) + 30 + 35 + 40 + 45

= 250

Hence we have 250 zeroes in the end.

12. Find the highest power of 2 in 17!

Answer: 15

Solution: Highest power of 2 in 17! = 17/2^1 + 17/2^2 + 17/2^3 + 17/2^4 + 17/2^5 + ...

= 8 + 4 + 2 + 1 + 0 + 0 + ...

= 15

13. What is the highest power of 21 in 342! ?

Answer: 54

Solution: 21 = 3*7 and hence 7 is the largest prime number as factor.

Power of 7 in 342! = 342/7 + 342/49 = 48+6 = 54

14. Consider the set of first 20 positive integers. The sum of any 18 numbers from it cannot be less than

Answer: 170

Solution: Lowest possible sum is the sum of 1 to 18 which is 18*19/2 = 171.

Hence the sum of any 18 numbers cannot be less than 170.

15. A positive number when decreased by 2 is equal to 15 times the reciprocal of the number. The number is

Answer: 5

Solution: Let the number be x.

x-2 = 15 * 1/x,

x^2 -2x-15 = 0,

Solving x=5 or x=-3

As x is positive, x=5

16. What is the remainder when 28! is divided by 29?

Answer: 28

Solution: Using Wilson's theorem, 28! + 1 is divisible by 29.

Hence 28! mod 29 = 28.

17. Find the highest power of 5 which can divide 22! + 17000!

Answer: 4

Solution: 22 is smaller.

Hence power of 5 in 22 is = 22/5 = 4

18. Find the smallest perfect square number divisible by 9!

Answer: 25401600

Solution: 9! = 9*8*7*...*2*1

= 7*5*3^4*2^7

For a perfect square the number should contain even powers.

Thus the smallest perfect square number is = 7^2 * 5^2 * 3^4 * 2^8

= 49 * 81 * 64 * 100 = 25401600

19. Which of the following number cannot be expressed as a sum of two prime numbers?

Answer: 71

Solution: All prime numbers except 2 are odd. So odd+odd = even.

But 71 is odd and hence it is possible only if 2,69 are the numbers and both must be prime. As 69 is divisible by 3 it is not prime number.

Hence 71 cannot be expressed as a sum of two primes.

20. The highest power of 9 dividing 99! completely is

Answer: 12

Solution: 99! = 99 * 98 * ... * 2 * 1

11 multiples of 9 + one more 9 in 81 (as 81 = 9*9).

Hence the highest power of 9 dividing 99! is 11+1 = 12