`Find the remainder when 1⁵+2⁵+3⁵+....+14⁵ is divided by 7`
Answers
Answer:
Let’s start by splitting the positive integers less than or equal to 100 into 50 pairs of the form (x,x+2): (1,3); (2,4); (5,7); (6,8); (9,11); (10,12); (13,15); (14,16); …; all the way through (97, 99) and (98, 100).
The modulo operator is distributive over addition - that is, (a+b)%c = ((a%c)+(b%c))%c. This is an important property for this problem.
Next, we need to prove that this holds for integer n:
(n⁵ + (n+2)⁵)%4 = 0.
Expanding (n+2)⁵ gives:
(n⁵ + (n⁵+10n⁴+40n³+80n²+80n+32))%4 = 0.
We already know that (40n³+80n²+80n+32)%4=0, given that n is an integer, so we can simplify the equation to:
(2n⁵ + 10n⁴)%4 = 0.
(2n⁵ + 10n⁴) will be a multiple of four if and only if half that quantity is a multiple of 2:
(n⁵ + 5n⁴)%2 = 0.
We know that for all postive integers (a,b), aᵇ will be even if and only if a is even, and a*b will be even if and only if either a or b is even, and since 5 is not even, 5n is even if and only if n is even. Thus, n⁵ + 5n⁴ is either the sum of two odd numbers or the sum of two even numbers, which is even in either case. While there’s probably an easier way to do so, we have just proven that (n⁵ + (n+2)⁵)%4 = 0 for all positive integers n. Therefore:
(1⁵ + 3⁵)%4 = 0.
(2⁵ + 4⁵)%4 = 0.
(5⁵ + 7⁵)%4 = 0.
(6⁵ + 8⁵)%4 = 0.
(9⁵ + 11⁵)%4 = 0.
(97⁵ + 99⁵)%4 = 0.
(98⁵ + 100⁵)%4 = 0.
Thus, adding up the first 100 5th powers will give us a multiple of 4 (Javascript confirms this, telling me that this number is 83,291,672,500).