(Summing up the Years) which has been reshared on Brilliant's official Facebook page, and this year I've decided to introduce a twist to this deceptively simple problem:
Last year I've posted the following problemFind the last digit of i = 0 ∑ 2 0 1 5 ( ( ( 2 0 1 5 i ) i ) ( i 2 0 1 5 ) )
This problem is part of the set 2015 Countdown Problems .
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The original problem is here for those who haven't seen it yet
Solution:
Note that for any integer i > 0 , the last digit of ( 2 0 1 5 ) i = the last digit of 5 i =5. Hence the last digit of ( 2 0 1 5 ) i =5, simplifying the sun to 5 ∑ i = 1 2 0 1 5 i 2 0 1 5 . It can be further simplified to 5 ∑ i = 1 2 0 1 5 ( l a s t − d i g i t − o f − i ) 2 0 1 5 .
For 0 ≤ i ≤ 9 , the last digits of i k where k is an integer are tabulated below (note the pattern):
x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
x 2 | 1 | 4 | 9 | 6 | 5 | 6 | 9 | 4 | 1 |
x 3 | 1 | 8 | 7 | 4 | 5 | 6 | 3 | 2 | 9 |
x 4 | 1 | 6 | 1 | 6 | 5 | 6 | 1 | 6 | 1 |
|| 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 2 0 1 5 =0 (mod 10),
1 2 0 1 5 =1 (mod 10),
2 2 0 1 5 =2^{3 (mod 4)}=8 (mod 10),
3 2 0 1 5 =3^{3 (mod 4)}=7 (mod 10),
4 2 0 1 5 =4^{1 (mod 2)}=4 (mod 10),
5 2 0 1 5 =5 (mod 10),
6 2 0 1 5 =6 (mod 10),
7 2 0 1 5 =7^{3 (mod 4)}=3 (mod 10),
8 2 0 1 5 =8^{3 (mod 4)}=2 (mod 10), and
9 2 0 1 5 =9^{1 (mod 2)}=9 (mod 10).
Therefore, the last digit
= 5 ( 1 0 2 0 1 0 ) ( 0 + 1 + 8 + 7 + 4 + 5 + 6 + 3 + 2 + 9 ) + 5 ( 0 + 1 + 8 + 7 + 4 + 5 )
= 4 5 3 5 0 = 0 (mod 10).
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For the term i = 0 , i 2 0 0 5 = 0 , so that term can be ignored. Every term other than the term we can ignore is divisible by 5.
i 2 0 1 5 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 2 2 0 1 5 + 1 = 1 0 0 8 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.