2015 Countdown Problem #6: Summing Up The Years - Revisited

For the term $i=0$ , $i^{2005}=0$ , so that term can be ignored. Every term other than the term we can ignore is divisible by 5.

$i^{2015}$ is odd if and only if $i$ is odd, and multiplying by an odd number doesn't change its parity. Therefore, every odd term is odd and all other terms are even. There are $\frac{2015+1}{2}=1008$ odd terms, so their sum is even, so the entire sum is even.

Therefore the entire sum is divisible by 5 and even, so it is divisible by 10, so its last digit is 0.

nice logic

correct answer = 0 .

The original problem is here for those who haven't seen it yet

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Solution:

Note that for any integer $i>0$ , the last digit of ${(2015)}^i$ = the last digit of $5^i$ =5. Hence the last digit of ${(2015)}^i$ =5, simplifying the sun to $5\sum_{i=1}^{2015} {i^{2015}}$ . It can be further simplified to $5\sum_{i=1}^{2015} {(last-digit-of-i)}^{2015}$ .

For $0≤i≤9$ , the last digits of $i^k$ where $k$ is an integer are tabulated below (note the pattern):

|| $x^5$ || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 ||

Note that 2015=3 (mod 4)=1 (mod 2)=5 (mod 10). Hence,

$0^{2015}$ =0 (mod 10),

$1^{2015}$ =1 (mod 10),

$2^{2015}$ =2^{3 (mod 4)}=8 (mod 10),

$3^{2015}$ =3^{3 (mod 4)}=7 (mod 10),

$4^{2015}$ =4^{1 (mod 2)}=4 (mod 10),

$5^{2015}$ =5 (mod 10),

$6^{2015}$ =6 (mod 10),

$7^{2015}$ =7^{3 (mod 4)}=3 (mod 10),

$8^{2015}$ =8^{3 (mod 4)}=2 (mod 10), and

$9^{2015}$ =9^{1 (mod 2)}=9 (mod 10).

Therefore, the last digit

$=5(\frac{2010}{10})(0+1+8+7+4+5+6+3+2+9)$ $+5(0+1+8+7+4+5)$

$=45350$ = 0 (mod 10).