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This mock test of Test: Divisibility - 2 for UPSC helps you for every UPSC entrance exam.
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QUESTION: 1

How many three digit numbers are divisible by 5 or 9?

Solution:

Three digit numbers divisible by 5 or 9 = three digit numbers divisible by 5 + three digit numbers divisible by 9 – three digit numbers divisible by 5 and 9.

The three digit numbers divisible by 5 = 100, 105, 110….995

The sequence given is in A.P with common difference 5. Let 995 be the nth term of the A.P, then

995 = 100 + (n – 1)5 = 100 + 5n – 5

Thus, n = 180 – (1)

The three digit numbers divisible by 9 = 108, 118, … 999

The sequence given is in A.P with common difference 9. Let 999 be the pth term of the A.P, then

999 = 108 + (p – 1)9 = 108 + 9p – 9

Thus, p = 100 – (2)

The three digit numbers divisible by 45 = 135, 180, …990

The sequence given is in A.P with common difference 45. Let 990 be the qth term of the A.P, then

990 = 135 + (q – 1)45 = 135 + 45q – 45

Thus, q = 20 – (3)

Thus, from (1), (2) and (3) the three digit numbers divisible by 5 or 9 = 180 + 100 – 20 = 260

QUESTION: 2

If 8A5146B is divisible by 88, then what is the value of AxB?

Solution:

Since, the given number is divisible by 8, the last three digits should also be divisible by 8. Only when B = 4, 46B is a multiple of 8. Thus, B = 4.

As the given number is divisible by 11, the difference between the sum of its odd digits and even digits must be a multiple of 11.

Thus, (8 + 5 + 4 + 4) – (A + 1 + 6) = 14 – A should be divisible by 11. Only when A = 3, 14-A is divisible by 11.

Thus, the value of AxB = 4×3 = 12.

QUESTION: 3

What is the number of even factors of 36000 which are divisible by 9 but not by 36?

Solution:

36000=2^5 * 3^2 * 5^3

Since we are talking of even factors, there must be at least one 2 in the required factors.

Since the number is divisible by 9, we must have both the threes.

We cannot have more than 1 two as it will make the number divisible by 36.

So we have 1 way of choosing 2, 1 way of choosing 3, 4 ways of choosing 5.

Thus the required number of factors are

1*1*4 = 4

QUESTION: 4

The number A39K2 is completely divisible by both 8 and 11. Here both A and K are single digit natura. Which of the following is a possible value of A+K?

Solution:

The number is divisible by 11, so the difference between the sum of the digits at the odd places and the digits at the even places is either 0 or a multiple of 11.

Let the difference be a 0, so

11 + A = 3 + K

=> K – A = 8, the only possible value is 9,1

Now we have to check if it satisfies the divisibility by 8 test.

K= 9 makes the last 3 digits 992. This is divisible by 8.

Let’s check for other cases when the difference is 11

11 + A – 3 – K = 11 => A – K = 3

The possible values in this case are (9,6), ( 8,5), (7,4), (6,3), (5,2), (4,1)

Among these cases only (8,5) and (4,1) will be divisible by 8. So the possible values of sum are

13, 5 and 10

Now difference between the sum of odd and even places cannot be 22

11 + A – 3 – K = 22 => A – K = 14

Since both A and K are single digit natural numbers, this is not possible.

Thus the only possible values of sum are 5, 10 and 13.

In the given options only 10 is there. So it is the correct choice.

*Answer can only contain numeric values

QUESTION: 5

A positive number is divided by 100 to get a remainder thrice as the quotient. If the number is divisible by 11, then how many such numbers are possible that are less than 100000?

Solution:

Let the number be N and the quotient when divided by 100 be k. Then remainder is 3k. 3k < 100

N = 100k + 3k = 103k,

Also, N is divisible by 11.

=> k = 11p, where p is an integer.

=> N = 103*11p = 1133p...

As N < 100000, it implies that p can range from 1 to [100000/1133] i.e between 1 and 88

=> So, p can can range from 1 to 88

Also 3k < 100 => 3 * 11p < 100 => p < 4 Hence, p can take values 1,2,3

QUESTION: 6

If a positive integer n is divided by 5, the remainder is 3. Which of the numbers below yields a remainder of 0 when it is divided by 5?

Solution:

n divided by 5 yields a remainder equal to 3 is written as follows

n = 5 k + 3 , where k is an integer.

add 2 to both sides of the above equation to obtain

n + 2 = 5 k + 5 = 5(k + 1)

The above suggests that n + 2 divided by 5 yields a remainder equal to zero. The answer is B.

QUESTION: 7

If an integer n is divisible by 3, 5 and 12, what is the next larger integer divisible by all these numbers?

Solution:

If n is divisible by 3, 5 and 12 it must a multiple of the lcm of 3, 5 and 12 which is 60.

n = 60 k

n + 60 is also divisible by 60 since

n + 60 = 60 k + 60 = 60(k + 1)

The answer is D.

QUESTION: 8

What is the smallest integer that is multiple of 5, 7 and 20?

Solution:

It is the lcm of 5, 7 and 20 which is 140.

The answer is E

QUESTION: 9

When the integer n is divided by 8, the remainder is 3. What is the remainder if 6n is divided by 8?

Solution:

When n is divided by 8, the remainder is 3 may be written as

n = 8 k + 3

multiply all terms by 6

6 n = 6(8 k + 3) = 8(6k) + 18

Write 18 as 16 + 2 since 16 = 8 * 2.

= 8(6k) + 16 + 2

Factor 8 out.

= 8(6k + 2) + 2

The above indicates that if 6n is divided by 8, the remainder is 2. The answer is C.

QUESTION: 10

If n is an integer, when (2n + 2)^{2} is divided by 4 the remainder is

Solution:

We first expand (2n + 2)^{2}

(2n + 2)^{2} = 4n^{2} + 8n + 4

Factor 4 out.

= 4(n^{2} + 2n + 1)

(2n + 2)^{2} is divisible by 4 and the remainder is equal to 0. The answer is A.

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