Solution: |
You'll need an easy lemma before you can solve this:
Lemma: For any integer n>=1, 4^(2n-1) ends in a 4 and 4^(2n) ends in a 6. (That is, the units digits are 4 and 6 respectively).
This can be proven by induction, but the idea is that multiplying by four flips the last digit between 4 and 6. This can be seen
by doing the multiplication by hand. For example, 4^1 = 4, 4^2 = 16, 4^3 = 16*4 = 10*4 + 6*4 = 64, 4^4 = 64*4 = 60*4 + 4*4 = 256, and so on.
So, 4^1997 ends in a 4
Also:
| 1997^4 = |
(1995+2)^4 |
| = |
1995^4 + 4*(1995^3)*2 + 6*(1995^2)*(2^2)
+ 4*1995*(2^3) + 2^4 |
| |
(by the Binomial Theorem) |
Now, the first three terms in this last expression are clearly divisible by 5. So 1997^4 + 4^1997 is divisible by 5 if 2^4 + 4^1997 is
divisible by 5. Since 2^4 = 16 and 4^1997 ends in 4, 2^4 + 4^1997 will end in a 0, so it will be divisible by 5. Thus, 1997^4 + 4^1997 is
divisible by 5.
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