remainder

a^+b^n will be divisible by a+b when 'n' is odd.

here, n is 21 and a+b+c+d = 60.

So, 17^21 +18^21 +19^21 + 6^21 will be divisible by 60. C IMO

powers of 7
1/7
2/9
3/3
4/1
5/7
17^21 :: 21/4=5*4 + 1, unit's digit 7

powers of 8
1/8
2/4
3/2
4/6
5/8
18^21 :: 21/4=5*4 + 1, unit's digit 8

the same applied for 19 and 6
unit's digit for 19 is 9 and the unit's digit for 6 is 6 (always)
7+8+9+6=30, unit's digit is 0

among the answer represented we have only one which unit's digit is 0, this is 0 itself (answer 3, C)

given we don't know the ten's digit and cannot check the full remainder we are directed by the unit's digit, and this cannot be anything except for one with unit's digit of 0 (we cannot even reduce the denominator here like 40/60 2/3 as we are asked about the remainder of number divided by 60!)

vishal chugh wrote:what is the remainder when 17^21 +18^21 +19^21 + 6^21 is divided by 60?
1. 12
2. 3
3. 0
4. 45
5. 56