A = {179, 180, 181, … , 360}. B is a subset of A such that sum of no two elements of B is divisible by 9. The number of elements in B cannot exceed

a. 102

b. 81

c. 82

d. 101

9x – 21 elements

9x + 1 – 20 elements

9x + 2 – 20 elements

9x + 3 – 20 elements

9x + 4 – 20 elements

9x + 5 – 20 elements

9x + 6 – 20 elements

9x + 7 – 20 elements

9x + 8 – 20 elements

9x – 21 elements

All elements of the form of 9x + 1, 9x + 2, 9x + 3 and 9x + 4 will be taken and one of 9x can be taken. So, total of 82 elements.

How many four-digit numbers, having distinct digits, using the digits 1, 2, 3, 4 and 5 can be formed such that the numbers formed are divisible by each of the digits used in the number?

If 5 is chosen, it cannot be divisible by 2 or 4. So, 5 cannot be chosen. If you choose 1, 2, 3 and 4, the number won’t be divisible by 3. So, we cannot make any numbers.

In a community of 80 members, 30 could speak French and 40 Urdu. 20 new members joined this community, out of which 9 could speak both Urdu and French and 11 could speak exactly one of these two languages. As a result, the fraction of the members of the community who could speak both Urdu and French doubled. The number of members who could speak exactly one of these two languages, finally, is

Let the initial number be x.

2 * x/80 = (x + 9)/100

x = 6. Hence, (30 – x) + (40 – x) + 11 = 69

V is a 56 digit number. All the digits except the 32nd from the right are the same. If V is divisible by 13, then which of the following can never be the unit’s digit of V?

a. 4

b. 7

c. 1

d. Both 4 and 7

32nd digit from the right will be the 25th digit from the left. So, (aaa …24 times)ba(aaa…30 times). Any digit written consecutively six times is divisible by 7, 11, 13, 3, and 111. So, the first and the last part will be divisible by 13. So, the only part that remains is ba. For ba to be divisible by 13, the values could be 13, 26, 39, 52, 65, 78, and 91. So, a can never be 4 or 7.

Find the sum of all the digits in the integers from 1 to 100000.

00000 to 99999 and 100000 is what is required. As sum of digits has been asked, we do not need place values. So, 100000/10 will be the occurrence of each digit at each place. So, 45 * 10000 * 5 = 2250000. Also, we have to add 1 that comes because of 100000. So, total of 2250001.

P is a prime number greater than 30. When P is divided by 30, the remainder is x. How many different values of x are possible?

a. 8

b. 9

c. 10

d. 11

P = 30k + x

x cannot be a multiple of 2, 3 or 5. Also, x has to be less than 30. So, the possible remainders are 1, 7, 11, 13, 17, 19, 23, and 29. Total of 8 values.

Raman took classes for 20 consecutive days. On nth day, there were (n + 1) students present in his class, where n is a natural number. Each day Raman distributed Rs.1100 equally among the students present in the class. Student ‘x’, who attended all the 20 classes of Raman, donated 1/(m + 1) th part of the amount received by him (from Raman on that particular day) to a charity at the end of each day, where ‘m’ is the number of students attending the class on that day. What is the total amount donated by student ‘x’ to the charity?

a. Rs. 300

b. Rs. 1000

c. Rs. 666

d. Rs. 500

On the first day, the student got Rs. 1100/2 and gave away 1/3 of what he got. So, amount donated = 1100/6

On the second day, the student got Rs. 1100/3 and gave away ¼ of what he got. So, amount donated = 1100/12

On the third day, he would have donated 1100/20, on the fourth day, 1100/30 and so on.

So, total amount donated will be

1100 (1/6 + 1/12 + 1/20 + … + 1/462)

1100 (1/2 – 1/3 + 1/3 – ¼ + ¼ - 1/5 + … + 1/21 – 1/22)

1100 * 10/22

Rs. 500.

Let M be a three-digit number denoted by ‘ABC’ where A, B and C are numerals from 0 to 9. Let N be a number formed by reversing the digits of M. It is known that M – N + (396 × C) is equal to 990. How many possible values of M are there which are greater than 300?

a. 10

b. 18

c. 30

d. 20

99A – 99C + 396C = 990

99A + 297C = 990

A + 3C = 10

C = 1, 2, 3 and A = 7, 4, 1. For each value of A and C, B can take 10 values. Also, A has to be greater than or equal to 3. So, 4B2 and 7B1 will be the numbers. B can take any value from 0 to 9. So, 20 values in total.

The product of three positive integers is 6 times their sum. One of these integers is the sum of the other two integers. If the product of these three numbers is denoted by P, then find the sum of all distinct possible values of P.

a. 432

b. 252

c. 144

d. 336

Let a = b + c

(b + c)bc = 6(2b + 2c)

bc = 12.

The values of a, b and c could be 13, 12, 1 or 8, 6, 2 or 7, 4, 3. So, the distinct products will be 156 + 96 + 84 = 336.

The number of factors common to 30^11 and 20^13 is

The common term is 2^11 * 5^11. So, 12 * 12 = 144 factors in total.

N is a nine-digit number, all of whose digits are the same. N is definitely divisible by

i. 111

ii. 27

iii. 11

iv. 9

v. 37

a. i, ii and v

b. i, iv and v

c. i, ii, iv and v

d. i and iv

It will definitely be divisible by 111 and subsequently 37. Also, the number will be divisible by 9 as the sum of the digits will be in the form of 9x. For 27, you can take 111,111,111 = 1001001 * 111. This number will be divisible by 9 but not by 27 as both the individual parts are divisible by 3 at maximum.

P1, P2 and P3 are three consecutive prime numbers and P1 × P2 × P3 = 190747. What is the value of P1 + P2 + P3?

Can start by understanding that the numbers will be close to the cube root of 190740 which should lie between 50 and 60 (between 125000 and 216000). Can start with the smallest prime numbers and check. It will be satisfied for 53 * 59 * 61. Total of 53 + 59 + 61 = 173.

googletag.cmd.push(function() { googletag.display('div-gpt-ad-1505009385004-0'); });A 4-digit number of the form aabb is a perfect square. What is the value of a + b?

The perfect squares that have the last two digits in the form of bb will end in either 00 or 44. So, we need to consider 38^2, 62^2 and 88^2. Only 88^2 will satisfy the condition and will be equal to 7744 (remember! if you don’t already). So, a + b = 11.

The number 44 is written as a product of 5 distinct integers. If ‘n’ is the sum of these five integers then what is the sum of all the possible values of n?

a. 11

b. 23

c. 26

d. 32

44 = 1 * 2 * 2 * 11. This should give you a clue that only integers have been asked and so, they could be negative. Hence, 44 = (- 1) * (1) * (- 2) * (2) * (11). There is only one way in which you can write these and so, the sum will be 11.

(500! + 505! + 510! + 515!) is completely divisible by 5^n, where n is a whole number. How many distinct values of n are possible?

500! (1 + something that ends in 0)

So, we have to find the highest power of 5 in 500! which will be 100 + 20 + 4 = 124. Hence, n can take values from 0 to 124 (remember that n is a whole number and the expression will be divisible by 5^0 (i.e. 1) - TRAP!). 125 values in total.