Arun Sharma Based Level 1: Remainder & Divisibility - Free MCQ Practice

If P = 1! = 1
Then P + 2 = 3, when divided by 2! remainder will be 1.
If P = 1! + 2 × 2! = 5
Then, P + 2 = 7 when divided by 3! remainder is still 1.
Hence, P = 1! + (2 × 2!) + (3 × 3!)+ ……+ (10 × 10!)
Hence, when p + 2 is divided by 11!, the remainder is 1.
Alternative method:
P = 1 + 2 × 2! + 3 × 3! + …..10 × 10!
= (2 – 1)1! + (3 – 1)2! + (4 – 1)3! + …+ (11 – 1)10!
= 2! – 1! + 3! – 2! + …..+ 11! – 10!
= 11! – 1
Hence p + 2 = 11! + 1
Hence, when p + 2 is divided by 11!, the remainder is 1

aⁿ +bⁿ is always divisible by a + b when n is odd.
Therefore 15²³ + 23²³ is always divisible by 15 + 23 = 38.
As 38 is a multiple of 19, 15²³ + 23²³ is divisible by 19.
Therefore,the required remainder is 0.

First of all, we have to identify such 2 digit numbers.
Obviously, they are 10, 17, 24, ….94
The required sum = 10 + 17 … 94.
Now this is an A.P. with a = 10, n = 13 and d = 7
Hence, the sum is

In the successive division, the quotient of first division becomes the dividend of the second division and so on.
Let the last quotient be p, so the last dividend will be 7p + 4 which is the quotient of the second division.
So, the second dividend is (7p + 4) × 4 + 1.
Applying the same logic, the number = 3 {4(7p + 4) + 1} + 2 = 84p + 53
Hence, if the number is divided by 84, the remainder is 53.

We have N = 55³ + 17³ – 72³ = (54 + 1)³ + (18 – 1)³ – 72³
When N is divided by 3, we get remainders (1)³ + (- 1)³ – 0 = 0
Hence, the number N is divisible by 3.
Again N = (51 + 4)³ + 17³ – (68 + 4)³
When N is divided by 17, the remainder is (4)³ + 0 – (4)³ = 0
Hence, the number is divisible by 17.
Hence, the number is divisible by both 3 and 17.

As we know the last digit depends upon the unit digit of the multiplier numbers so the unit digit of 13⁴¹
is same as the last digit of 3⁴¹ and we know that the cyclicity of 3 is 4
On dividing the number 41 with 4 and we will get the remainder as 1 and the last digit will be 3¹ = 3

Here the last digit is depend upon the 5
The cyclicity of 5 is 1 so there is no need to divide the power with 1
because whatever is the power of 5 the last digit will remains 5 so the last
digit of 45⁴⁵ will be 5 .
5¹ = 5
5² = 25
5³ = 125

Last digit of 3⁴ = 1
Last digit of 4⁵ = 4
Last digit of 5⁶ = 5
So as we discussed earlier that the last digit depends upon the last digits
So 1 x 4 x 5 = 20 so the last digit is 0; also whenever 5 is multiplied by even
number then last digit will be 0

Last digit of 79⁴ is depends upon the the 9⁴ i.e 1
Last digit of 87⁵ is depends upon the the 7⁵i.e 7
So the last digit of the whole expression will be 1 +7 = 8
So the last digit will be 8

Last digit of 43⁵⁶ = 1 because cyclicity of 3 is 4 and 56 is completely
divisible by 4 so the last term of 3⁴ is 1.
Last digit of 5⁶⁷ is 5 and last digit of 45³⁴ is also 5
so that means the last digit of whole of the expression is 1 x 5 x 5 = 5 therefore the digit is 5.