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What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$

My approach:

Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.

So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.

Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.

Any help is appreciated!

J. W. Tanner
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Chirag Maheshwari
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5 Answers5

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By geometric sum formula we have $$\frac{7^{2005}-1}{6} = \frac{7^{2005}-1}{7-1} = 1+7+7^2+\dots+7^{2004}$$

The sequence $1,7,7^2,\dots$ has period $20 $ modulo $1000$ (since $7^{20} \equiv 1 \pmod{1000}$).

$$1+7+7^2+\dots+7^{2004} \equiv 100(1+\dots+7^{19})+1+7+7^2+7^3+7^4 \equiv 100\cdot 0+801\equiv 801 \pmod{1000} $$

hgmath
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    I'm not sure how you know $7^{20}\equiv1\bmod1000$, but I know $7^{100}\equiv1\bmod1000$ by Carmichael – J. W. Tanner Feb 28 '21 at 16:34
  • @J.W.Tanner it's a small exponent so we could just calculate by squaring $7^2\equiv 49$, $7^4\equiv 401$, $7^8\equiv 801$, $7^{16}\equiv 601$ and now $7^{20}\equiv 601\cdot 401\equiv 1$ modulo $1000$ – hgmath Feb 28 '21 at 16:39
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    Alternatively, $7^{20}=(1-50)^{10}\equiv1-10\cdot50+{10\choose2}50^2$ mod $1000$. – Barry Cipra Feb 28 '21 at 16:41
  • @JyrkiLahtonen oops, you are right! I'll correct it :) – hgmath Feb 28 '21 at 16:49
  • $7^4\equiv401\pmod{2000}$ implies easily that $7^{20}\equiv1\pmod{2000}$ @J.W.Tanner. You are right in that it is better to do that calculation modulo $2000$ so that we can cancel the extra $2$. – Jyrki Lahtonen Feb 28 '21 at 17:29
  • That's a lot of unneeded calculation vs. using Euler totient or similar e.g. see my answer. – Bill Dubuque Feb 28 '21 at 18:38
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We must multiply the modulus by $\,6\,$ to balance the division by $6$, i.e.

$\qquad 6\mid a \,\Rightarrow\, a/6 \bmod 1000 = (a \bmod 6000)/6\ $ by the mod Distributive Law

$6000\!=\! 2^4\cdot 3\cdot 5^3$ whose totients $2^3,2,100\mid 2000\,$ so $\,\color{#c00}{7^{2000}\!\equiv 1}\bmod 2^4,3,5^3$ so also mod $6000,\,$ so $\bmod 6000\!:\ a\equiv 7^5 \color{#c00}{7^{2000}}-1 \equiv 7^5-1\equiv4806,\,$ so $\bmod 1000\!:\ a/6\equiv 4806/6 \equiv 801$.

Bill Dubuque
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You can distinguish between the two possibilities you found by starting modulo $2000$:

because the Carmichael function of $2000$ is $100$, $7^{2000}\equiv(7^{100})^{20}\equiv1\bmod2000$,

so $7^{2005}\equiv7^5\equiv807\bmod 2000$.

Therefore, $7^{2005}-1\equiv806\bmod2000$.

Therefore, $\dfrac{7^{2005}-1}2\equiv403\bmod1000$.

Therefore, $\dfrac{7^{2005}-1}6\equiv403\times3^{-1}\bmod{1000}.$

In general, if $n=3k-1$ then $3^{-1}\equiv k\pmod n$,

and if $n=3k+1$ then $3^{-1}\equiv -k\pmod n$,

so $3^{-1}\equiv-333\equiv667\bmod{1000}$.

Therefore, $\dfrac{7^{2005}-1}6\equiv403\times667\equiv801\bmod1000$.

J. W. Tanner
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  • You don't say how you calculated $,3^{-1}\equiv 667\pmod{!1000},$ but we can instead do $,403/3\equiv -597/3\equiv -199\equiv 801$ by twiddling. We don't need to calculate any inverses if instead we work mod $6000,,$ e.g. as in my answer. – Bill Dubuque Feb 28 '21 at 19:10
  • $3\times333\equiv-1\bmod1000$, so $3^{-1}\equiv-333\bmod1000$ – J. W. Tanner Feb 28 '21 at 19:20
  • You should say what method (algorithm) you used since that is likely not clear to beginners (it can be done by various methods in the link I gave). – Bill Dubuque Mar 01 '21 at 01:54
  • Your edit still doesn't reveal the method. How does it work to compute modular inverses in general? If I can't figure out precisely what you did how can a beginner? – Bill Dubuque Mar 01 '21 at 09:55
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Just to give a different approach, calculating mod $2000$ and using the fact that

$$3^{2000}=(1-10)^{1000}=1-1000\cdot10+\cdots\equiv1\mod2000$$

we have

$$\begin{align} 7^{2005}-1&=-1-(3-10)^{2005}\\ &\equiv-1-3^{2005}+2005\cdot3^{2004}\cdot10-{2005\choose2}3^{2003}\cdot10^2+{2005\choose3}3^{2002}\cdot10^3\\ &\equiv-1-3^5+5\cdot3^4\cdot10-{5\choose2}3^3\cdot100+{5\choose3}3^2\cdot1000\\ &\equiv-1-243+4050-27000+90000\\ &\equiv-244+50-1000\\ &=-1194\mod2000 \end{align}$$

and thus

$${7^{2005}-1\over6}\equiv{-1194\over6}=-199\equiv801\mod1000$$

Barry Cipra
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If $k$ and $n$ are not co-prime, you have to be very careful when dividing by $k\bmod n$ $-$ there may be more than one answer, as you saw. So $806/6\bmod 1000$ is not well-defined, because both $6\cdot 301$ and $6\cdot 801$ are equal to $806\bmod 1000$.

But in this case the problem is well-defined, because the expression $\dfrac{7^{2005}-1}{6}$ is an integer. So we can calculate it before reducing $\bmod 1000$. hgmath's answer shows one way to do this (and Raffaele's answer doesn't).

TonyK
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  • But we don't need to calculate it as an integer (which could be an arbitrarily huge integer in bigger problems). Rather we only need to calculate it modulo $6$ times the modulus as I explain in my answer. – Bill Dubuque Feb 28 '21 at 19:14
  • @BillDubuque: Of course you are right. Which means that the ideal answer for the OP would probably consist of my answer followed by yours. – TonyK Feb 28 '21 at 20:48