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Find the last three digits of the number:

$7^{7^{7^7...}}$

where there are 1001 sevens.

I know how to do it for when there are 4 and 5 sevens. I get an answer of 343. But how do I find it for such a large number of sevens?

Deepak
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Vince
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    HINT: http://math.stackexchange.com/questions/166083/last-few-digits-of-nnn-cdot-cdot-cdotn?rq=1 – rugi Mar 28 '16 at 12:10
  • Thank you for the hint. Even then I don't understand how the conclusion was made that "any tower taller will end in 343"? – Vince Mar 28 '16 at 20:09

1 Answers1

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Render $7^4=2401=2400+1$. Raise to the fifth power with the Binomial Theorem, thus:

$7^{20}=(2400+1)^{5}=(\text{ a multiple of }2400^2)+(5×2400)+1$

Both the "multiple of $2400^2$" and $5×2400=12000$ are obviously multiples of $1000$, so $7^{20}\equiv1\bmod 1000$ forcing $7^n\equiv343$ whenever $n$ is three more than a multiple of $20$.

Oscar Lanzi
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