How to get last digit of $7^{7^7}$

Question

I want to find the last digit of $7^{7^7}$. I found out already that $7^7$ (mod 10) last digit is 3. But how do I use that to get the last digit of the whole thing?

Thanks

Related : http://math.stackexchange.com/questions/390685/the-last-2-digits-of-7777 — lab bhattacharjee, Dec 11 '14 at 06:13

score 4 · Accepted Answer · answered Dec 11 '14 at 06:01

4

Note that $7^2 \equiv 49 \equiv -1 \bmod 10$, and so $7^4 \equiv 1 \bmod 10$. Hence if $a=4q+r$ then $$7^a \equiv 7^{4q+r} \equiv (7^4)^q \cdot 7^r \equiv 7^r \bmod 4$$ This reduces your problem to finding $7^7 \bmod 4$... but this should be easy since $7 \equiv -1 \bmod 4$.

answered Dec 11 '14 at 06:01

Clive Newstead

64,925

How does getting that $7^r$ mod 4 mean the problem reduces to $7^7$ mod 4? – omega Dec 11 '14 at 06:14
1

Because if $7^7 \equiv r \bmod 4$ then $7^{7^7} \equiv 7^r \bmod{10}$. – Clive Newstead Dec 11 '14 at 06:19

score 2 · Answer 2 · answered Dec 11 '14 at 06:03

Notice that the last digits of powers of $7$ run in the repeating sequence $7,9,3,1$. Thus, what you really need to know is what $7^7$ is congruent to modulo $4$, not modulo $10$, so as to tell where in the sequence $7^{7^7}$ falls. $7^k\bmod 4$ is easily computed from $k$; how?

How to get last digit of $7^{7^7}$

2 Answers2