Test: Divisibility - 2 - UPSC MCQ

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

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.

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

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.

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

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.

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.

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

The answer is E

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.

We first expand (2n + 2)²
(2n + 2)² = 4n² + 8n + 4
Factor 4 out.
= 4(n² + 2n + 1)
(2n + 2)² is divisible by 4 and the remainder is equal to 0. The answer is A.