Question 594344: Prove that for all integers n, that the last digit of n^5 is the same as the last digit of n.
Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! Just check ten cases, where each case covers a residue class (mod 10). In other words, we can simply check n = 0, 1, ..., 9 and we are done.
0^5 = 0
1^5 = 1
2^5 = 32 ≡ 2
3^5 = 243 ≡ 3
.
.
.
9^5 = 59049 ≡ 9
You can fill in the remaining exponents if you wish. Point is, n^5 is congruent to n (mod 10), so they must have the same last digit.
|
|
|