SOLUTION: The last digit of 7^1992 is a 1 b 2 c 6 d 7 e 9

Question 283081: The last digit of 7^1992 is
a 1 b 2 c 6 d 7 e 9

Found 2 solutions by nabla, Greenfinch:
Answer by nabla(475) (Show Source): You can put this solution on YOUR website!
You can figure this out a couple of ways, but I will show you how I would do it using modular arithmetic and congruences.
First, 7^2=49 which is congruent to (-1) modulo 10.
Second, 7^1992=(7^2)^996 which is congruent to (-1)^996 modulo 10.
Finally, (-1)^996=1. We conclude that the unit digit of 7^1992 is 1.
Without going through an entire explanation and lesson in modular congruences, I will explain this as follows:
Consider:
abcdefx where a,b,c,d,e,f,x are natural numbers that form the number abcdefx (we are not talking about multiplication here). Note that this is the same as abcdef0+x. Now, 10 divides a multiple of 10. So 10|(abcdef0). We are left with x mod 10 (as 10|abcdef0 implies abcdef0 is congruent to 0 mod 10). So when we take mod 10 of any number we can think of that as asking what the last digit will be.
Hope this helps.
Answer by Greenfinch(383) (Show Source): You can put this solution on YOUR website!
Since 7 is odd, powers will be odd. Therefore 2 and 6 will not be possible.
The last digit of powers of 7 go in sequence
7,9,3,1
7,9,3,1
1992 is divisible by 4, so I think the answer is 1 or answer (a)
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