Showing that a number ends in 007

Question

Find a positive integer x such that the last three digits of $7^{7^x} $ are 007 (do not use trial and error) i mean clearly when $x=1$ the number is divisible by 7 i believe that every $x\in \mathbb{N}$ is divisible by 7 as mod 7 it always 0 but ending by 007 and being divisible by 7 are different things.

Possible duplicate of Find the remainder when $7^{7^{7}}$ is divided by 1000 — Dietrich Burde, Nov 03 '17 at 16:27
Do you have a reason to believe that there is a method here besides trial and error? Discrete logarithms are notoriously difficult to compute. — Ben Grossmann, Nov 03 '17 at 16:27
How much would it help you if I told you that ending with 007 is the same thing as being equal to 7 modulo 1000? — Jack M, Nov 03 '17 at 16:28
You probably need to use some trial and error, but if you know the Chinese remainder theorem and Fermat's little theorem, things become a lot easier. — Arthur, Nov 03 '17 at 16:31
you may try binomial here...and 007 ...reminds one of james bond right:) — Pole_Star, Nov 03 '17 at 16:38
$7^{7^4} \equiv 7 \mod 1000$ therefore it ends by $007$. Please notice that $7^{7^4}$ is a number composed by $2,030$ digits. $7^{7^8}$ ends by $007$, too (about 5 million digits) and $7^{7^{12}}$ about ten billion — Raffaele, Nov 03 '17 at 17:03

lhf · Accepted Answer · 2017-11-03T17:01:35.600

4

A number $N$ ends with $007$ iff $N \equiv 7 \bmod 1000$.

We have $7^n \equiv 7 \bmod 1000$ iff $7^{n-1} \equiv 1 \bmod 1000$.

Now the order of $7 \bmod 1000$ is $20$. Therefore, $7^n \equiv 7 \bmod 1000$ iff $20$ divides $n-1$.

For $n=7^x$, we want to know when $7^x \equiv 1 \bmod 20$.

Now the order of $7 \bmod 20$ is $4$. Therefore, $7^x \equiv 1 \bmod 20$ iff $4$ divides $x$.

edited Nov 03 '17 at 17:01

answered Nov 03 '17 at 16:52

lhf

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Showing that a number ends in 007

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