17-08-2021 Answers

 1. Find the remainder when 3^126 is divided by 8.

Answer: 1

Solution: 3^126 = 9^63.

Hence remainder = remainder of (9/8) ^ 63 = 1^63 = 1

2. What is the remainder when 4^82 is divided by 6?

Answer: 4

Solution: Remainder of 4^n mod 6 is 4 irrespective of the value of n, where n > 0.

3. Find the last two digits in 4761^25

Answer: None of these

Solution: The unit digit will be 1.

The tens digit will be the unit digit in the product of 6*5(the unit digit in power and the tens digit of the number)

= unit digit in 30 = 0

Hence last two digits will be 01

4. If you wrote all of the numbers from 300 to 400 on a piece of paper, how many times would you have written the number 3?

Answer: 120

Solution: Numbers from 300 to 399 will have 100 3's in hundredth position.

They will have 10 3's in tenth position (from 330 to 339)

and 10 3's in unit position (for 303,313,....,393).

Hence overall 120.

5. Which of the following divides the difference between cubes of two consecutive positive even integers without leaving a remainder?

Answer: 8

Solution: Let the numbers be a,a+2.

Difference = a^3 + 6a^2 + 12a + 8 - a^3

= 6a^2 + 12a + 8

As a is even (has 2 as a factor), all the terms have 8 as common factor and hence is divisible by 8.

6. What is the remainder when 516^3 -759^3 + 1159^3 -116^3 is divided by 4?

Answer: 0

Solution: The expression can be rearranged as  516^3 -116^3 + 1159^3 -759^3 and now formula for (a-b)^3 can be used.

The common factor is a-b which is 516-116 = 400 = 1159-759.

Hence the expression is divisible by 400 and hence 4 (4 is factor of 400).

So remainder is zero.

7. How many odd numbers are there between 7000 and 10000 containing atleast one 8?

Answer: 690

Solution: From 8001 to 8999 there are 500 ODD numbers containing atleast one 8.

Now from 7001 to 7999:

When 8 is in unit digit, the number is even and hence count =0,

When 8 is in ten's digit, count = 10*1*5 = 50 (unit digit can contain only 1,3,5,7,9)

When 8 is in hundredth digit, count = 1*10*5 = 50

Here 7881,7883,7885,7887,7889 will be repeated twice in the counting.

Hence total count between 7001 and 7999 = 95.

In a similar fashion, count of odd numbers between 9001 and 1000 is also 95.

Hence total numbers = 500+95+95 = 690

8. If the number 109236345978x  is divisible by 13, what is the value of x?

Answer: 5

Solution: The shorcut to find divisibility by 13 is divide the number into groups of 3 digits from right.

1, 092, 363, 459, 78x

Now add alternate groups and find their sums.

1+363+78x=364+78x  and 092+459= 551.

Now their difference = 78x+364-551 should be either zero or divisible by 13.

Hence 78x - 187 should be divisible by 13.

When x=0, 780-187 = 593 which leaves a remainder 8.

Hence adding 5 to 780 we get 785 as the required last group.

Hence x=5.

9. What is the remainder when -4x^3 + 8x^2 + 12x +16 is divided by x+2 ?

Answer: 56

Solution: Using remainder theorem, here a=-2.

Hence f(a) = -4 (-2^3) +8 (-2^2) +12*-2  +16 

= 32 + 32 -24 +16 = 56

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

Answer: 2223

Solution: (4)^1111 * (5)^2222 = (2^2)^1111 * (5)^2222

=(2*5)^2222 = 10^2222

So the given expression will contain 1 and 2222 zeros,

 Hence answer is 1+2222 = 2223 digits.

11. Find the HCF of 20^14  -1 and 20^12  -1

Answer: 399

Solution: To find HCF of x^a -1 and x^b -1, the shortcut is the HCF of x^( HCF of (a,b)) -1.

Here HCF of 14 and 12 = 2.

Hence required answer = 20^2 -1 = 399

12. The product of a two digit number by a number consisting of same digits written in reverse order is 1300. Find the smaller number.

Answer: 25

Solution: Reducing it to prime numbers,

1300 = 13 * 5 * 5 * 2 * 2.

As the unit digit in the product is 0, the possible combination for x,y such that xy < 99 (remember its a two digit number) are

65,20  and 52,25.

But 65,20 is not meeting the rules as 20 is not reverse of 65.

Hence the pair 52,25 is valid and 25 is the smaller number.

13. The numbers from 1 to 29 are written continuosly like 1234567891011...272829 and if the big number formed thus is divided by 9, what is the remainder?

Answer: 3

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

Digital sum from 1 to 9 = 45

From 10 to 19 = 10+45 = 55

From 20 to 29 = 20+45 = 65

Overall total = 165, reducing further the digital sum is 12.

Hence required remainder is the remainder when 12 is divided by 9 which is nothing but 3.

14. The sum of the odd numbers between 1 and n is 11025, where n is an even number. What is the value of n?

Answer: 210

Solution: Sum of odd numbers from 1 to n (where n is even) is given by the formula (n/2)^2.

Square root of 11025  is 105.

Hence 105 = n/2.

So n=210

15. If p is a prime number greater than 3, then which of the following numbers will always divide P^2 - 1?

Answer: 24

Solution: As beyond 3, all prime numbers are odd. P^2 -1 can be written as (p-1)*(p+1) which both are even numbers (as p is odd).

Among these one is divisible by 4 (as both are consecutive even numbers) and the other by 2. Also one among them is divisible by 3 as p-1,p,p+1 are consecutive.

Hence one of the factor must always be 4*2*3 = 24.

16. What is the value of x if 9^x = 9 ÷ 3^x  ?

Answer: 2/3

Solution: 9^x = 9/(3^x),

27^x = 9,

3^3x = 3^2,

Comparing powers 3x=2,

x=2/3

17. A four-digit number is a multiple of 9 and can have a digit repeated exactly three times consecutively. How many such numbers are possible?

Answer: 20

Solution: As all four digits cannot be same, the maximum sum of digits should either be 9 or 18 or 27.

Let the number if in the format mmmn or nmmm.

Case 3m+n = 9:

Possible numbers are 3330, 2223, 3222, 1116, 6111, 9000

Case 3m+n = 18:

Possible numbers are 6660, 5553, 3555, 4446, 6444, 9333, 3999

Case 3m+n = 27:

Possible numbers are 9990, 8883, 3888, 7776, 6777, 6669, 9666

Hence total possible numbers = 6+7+7 = 20

18. If the sum of digits of an even number is divisible by 9, then that number is always divisible by

Answer: 18

Solution: Expressing 9 in terms of prime factors, 9 = 3*3.

As an even number, the number is also divisible by 2.

Hence the number will be divisible by 3*3*2 = 18

19. Which of the following option does not divide 5^6 - 1 completely (that is remainder is zero)?

Answer: 27

Solution: 5^6 - 1 = (5^3 + 1)*(5^3 -1)

= 126*124

= (2*3*3*7)*(2*2*31)

Among the options 27 is not a factor.

20. Which number when added to 5/4 gives the same result as when it is multiplied 5/4?

Answer: 5

Solution: Let the number be x.

5/4 + x = 5x/4,

5/4 = x/4, x=5.

21. How many positive numbers less that 10000 exist so that they are perfect squares but not perfect cubes?

Answer: None of these

Solution: Numbers which are perfect square and cubes will have a number raised to the power 6 (LCM of 2 and 3).

Such numbers are 1^6, 2^6, 3^6, 4^6  (5^6 is 15625 which is more than 10000).

Total numbers less than 10000 and are perfect squares = 99 (As 100^2 is 10000).

Hence required count = 99-4 = 95

22. A number when divided by 765 leaves a remainder 42. What will be the remainder if the number is divided by 17?

Answer: 8

Solution: Number = 765k+42 (k is the quotient when divided).

When divided by 17, it is

(765k+42)/17 = (45k*17 + 34 + 8)/17

Hence remainder is 8 (As other numerator terms are divisible by 17)

23. What highest power of 8 will divide 26! exactly?

Answer: 7

Solution: 8 represented using primes is 2*2*2

Number of 2's in 26! multiplication = 26/2 + 26/4 + 26/8 + 26/16 = 13 + 6 + 3 + 1 = 23.

As 8 contains 3 twos, 23/3 = 7.

Hence 8 to the power 7 will exactly divide 26!

24. How many numbers are there between 7000 and 8000 containing atleast one 9?

Answer: 271

Solution: When 9 is in unit digit, possible numbers = 10 * 10 * 1 = 100

When 9 is in ten's digit, possible numbers = 10 * 1 * 10 = 100

When 9 is in hundred's digit, possible numbers = 1 * 10 * 10 = 100

But 7099 to 7899 , 7909 to 7989, 7990 to 7998 will be common and repeated  twice. Hence 27

7999 will be common and repeated thrice. Hence 2 is to be subtracted.

Total = 100+100+100 - (27+2) = 271

25. What is the remainder when (55^55 + 55) is divided by 56?

Answer: 54

Solution: (55^55 + 55) mod 56

= ((-1)^55 + -1) mod 56

= -2 mod 56

Hence the remainder = 56 - 2 = 54