Modular Arithmetic with higher powers

Question

Find an integer $0\leq n<1000$ such that the last three digits of $n^{267}$ are $321$.

I have tried this method:

$n^{267}\equiv 321(\text{ mod }1000)$

So, I break it up into two congruences and thought to apply the Chinese remainder theorem.

$n^{267}\equiv n^3\equiv321\equiv 1(\text{ mod }8)$
$n^{267}\equiv n^{67}\equiv321\equiv 71(\text{ mod }125)$

But it's very hard to solve the above congruences. Therefore, I do suspect that there is another method to solve the above congruence. Please help

@DietrichBurde IMO I don't think it does, unless you want OP to cycle through all values of $a$ coprime to 1000. — Calvin Lin, Mar 25 '20 at 16:10
I am sorry, I meant using it like your hint. Such questions have been posted here already. One need's to find the right one, of course. — Dietrich Burde, Mar 25 '20 at 16:13
Right, the hard part is "find the right one", esp if we don't know OP's ability (or willingness) to abstract from a slightly different situation. — Calvin Lin, Mar 25 '20 at 16:15
Since someone wants to delete this topic, I think we need to open it. — Michael Rozenberg, Apr 05 '20 at 10:44

Calvin Lin · Answer 1 · 2020-03-25T16:15:45.153

1

Hint: $267 \times 3 = 801 = 1 + \phi(1000)$

Can you take it from here?

If you want a step-by-step approach:

Step 1: Show that $ \gcd(n, 1000) = 1$.

$ $

Step 2: Calculate $ n^{801} \pmod{1000}$ from the given conditions using $ 267 \times 3 = 801$.

$ $

Step 3: Calculate $ n \pmod{1000}$ applying Euler's Theorem

edited Mar 25 '20 at 16:15

answered Mar 25 '20 at 16:08

Calvin Lin

score 1 · Answer 2 · answered Mar 25 '20 at 16:13

1

To solve $n^{67}\equiv71\bmod125$, note $(n^{67})^3\equiv n \equiv71^3\bmod 125$.

answered Mar 25 '20 at 16:13

J. W. Tanner

2 Answers2