Is there a quicker way to get $ 207^{321} \mod 7 $

Question

For $207^{321} \pmod{7},$ I got $$ 207^{321} = 207^{6\cdot 53+3}$$ and $$207^{\Phi(7)} \equiv 207^6 \equiv 1 \pmod{7}$$ by Euler's Theorem. Then $$207^3 \equiv 4^3 \equiv 1 \pmod{7} $$

Is there any simpler way? I'm also not sure about the format of module symbol.Should there be only one (mod 7) written on the right of the equation so as to avoid redundancy ?

I have also seen equation like this 26 mod 5=1,rather than $26\equiv 1 \mod{5}$. What's the difference?

@Omnomnomnom: What's the difference between pmod and bmod, by the way? — Brian Tung, Oct 05 '15 at 17:17
@BrianTung \bmod is meant for use as a binary operator. \pmod puts parentheses around the second argument (and spaces things differently). \mod uses the spacing of \pmod but not the parentheses. — Ben Grossmann, Oct 05 '15 at 17:19
Also $207^{6^{53}}$ is not the same as $\left(207^6\right)^{53}$
Which has been already corrected, as I can see. — Kuba, Oct 05 '15 at 17:22
@Omnomnomnom: Excellent, thanks; I've added it to my own answer. — Brian Tung, Oct 05 '15 at 17:26
@I want to make games but they are congruent since any exponent goes beyond where $\phi^m(n)$ hits $1$, or the remainder hits $0$ won't matter to the result. — Roddy MacPhee, Sep 04 '21 at 00:31

Brian Tung · Answer 1 · 2015-10-05T17:38:40.030

2

I don't know, your line of reasoning seemed pretty quick and easy to me. I suppose you could observe that (modulo $7$)

$$ 207^{321} \equiv (207 \bmod 7)^{(321 \bmod 3)} \equiv 4^0 = 1 $$

Is that simpler, by your lights?

The difference in notation, incidentally, is that when you write $26 \bmod 5 = 1$, the mod is treated as an operation—essentially the remainder left when you divide $26$ by $1$. When you write $26 \equiv 1 \pmod 5$, you mean that $26$ and $1$ fall into the same equivalence class, modulo $5$. The two formulations are equivalent under ordinary circumstances.

edited Oct 05 '15 at 17:38

answered Oct 05 '15 at 17:22

Brian Tung

34,160

1

That should be $321 \bmod 6 = 3$ – Ben Grossmann Oct 05 '15 at 17:29
I had that first (well, I had that second, after I fixed an earlier mistake). But isn't it the case that powers of $4$ have periodicity $3$ modulo $7$? – Brian Tung Oct 05 '15 at 17:31
oh, you're right; I didn't notice. – Ben Grossmann Oct 05 '15 at 17:35
@BrianTung - good solution. (+1). how do we know that powers of $4$ have periodicity $3$ mod $7$? what about other powers and other mods, and how can this be shown? – Hypergeometricx Oct 05 '15 at 17:45
Thank you.But I didn't got your method.Why did you take mod 3 on the power? How can I use 'powers of 4 have periodicity 3 mod 7'? – Charlotte Gu Oct 05 '15 at 17:50
Charlotte, @hypergeometric: There's nothing clever about it, I'm afraid. Once we know that things are down to modulo $7$, all the numbers are small enough to try by hand. Since powers of $4$ are even powers of $2$, they are easy enough to remember: $4, 16, 64, 256, 1024, \ldots$ We test each of these modulo $7$, and see where we get a $1$. The successive results are $4, 2, 1, 4, 2, 1, 4, \ldots$ That's what I meant when I said the powers of $4$ have periodicity $3$ modulo $7$. – Brian Tung Oct 05 '15 at 18:11

score 0 · Answer 2 · answered Oct 05 '15 at 17:45

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Note that $207^{321}=(207^{6})^{53} \cdot 207^3=1\cdot 4^3=1$ (mod. $7$).

answered Oct 05 '15 at 17:45

Euler88 ...

2,090

score 0 · Answer 3 · answered Sep 04 '21 at 00:39

You could use $a^p\equiv a\pmod p$ a Fermat"s little theorem derivative,then use exponent rules to show only the digital root in prime base $p$ of any exponent matters to this calculation. So as $321_{10}=636_{7}$ which has digital root $3$ we can decrease the exponent to $3$ we then apply the fact that $(px+y)^n$ has only $1$ term that doesn't include a factor of $p$ so we can calculate $y^3$ in this case. Your way uses much of the same theory but is less verbose.

Is there a quicker way to get $ 207^{321} \mod 7 $

3 Answers3