Finding $\displaystyle555^{555^{555}} \bmod{1000}$

Question

I try to find $\displaystyle555^{555^{555}} \bmod{1000}$ and I find that $555^{555^{555}}\equiv375 \pmod{1000}$ but Wolframalpha gives me $555^{555^{555}}\equiv875 \pmod{1000}$. Are there any mistakes in my attempt?

My attempt

$\begin{align*}555^{555}={\left(500+55\right)}^{555}=\sum_{n=0}^{555}\binom{555}{n}{500}^{555-n}55^n&=\cdots+555\left(500\right){\left(55\right)}^{554}+{55}^{555}\\ &\equiv500+{\left(50+5\right)}^{555}\pmod{1000}\\&\equiv500+555\left(50\right){\left(5\right)}^{554}+5^{555}\pmod{1000}\\&\equiv500+750+125\pmod{1000}\\&\equiv375\pmod{1000}\end{align*}$ $\begin{align*}555^{375}={\left(500+55\right)}^{375}=\sum_{n=0}^{375}\binom{375}{n}{500}^{375-n}55^n&=\cdots+375\left(500\right){\left(55\right)}^{374}+{55}^{375}\\ &\equiv500+{\left(50+5\right)}^{375}\pmod{1000}\\&\equiv500+375\left(50\right){\left(5\right)}^{374}+5^{375}\pmod{1000}\\&\equiv500+750+125\pmod{1000}\\&\equiv375\pmod{1000}\end{align*}$

Does this answer your question? Modular arithmetic problem (mod $13$) — Parcly Taxel, Jul 28 '22 at 08:04
Does this answer your question? Modular exponentiation by hand ($a^b\bmod c$). Many techniques are explained there via examples. Also in many other threads including the one Parcly Taxel found. You need to combine some of the ideas. For example here $1000=5^3\cdot2^3$. Your number is obviously divisible by $5^3$, so by the Chinese remainder theorem it suffices to find it modulo $8$. But you also need to learn that then you won't calculate the exponent modulo $8$ but rather modulo something else :-) — Jyrki Lahtonen, Jul 28 '22 at 08:18
Without answering you question, I have verified (with PARI/GP) that $875$ is correct, $375$ incorrect. — Jeppe Stig Nielsen, Jul 28 '22 at 08:28
Does this answer your question? Modular exponentiation by hand ($a^b\bmod c$) — John Omielan, Jul 28 '22 at 08:37
Note that $555=5\cdot111$ so you can treat the two factors separately. The powers of $5$ modulo $1000$ are easily computed. For the other factor, which is coprime to $1000$ use Euler's theorem: https://en.wikipedia.org/wiki/Euler%27s_theorem — Andrea Mori, Jul 28 '22 at 08:37
You need to use the last three terms of $(50+5)^{555}$, because $50^2$ isn't $0\pmod{1000}$. — JMP, Jul 28 '22 at 08:52

Prem · Answer 1 · 2022-08-16T11:16:14.877

You asked : Are there any mistakes in my attempt?

There are many comments on various Procedures and theorems to use to find the Answer.
I will leave that to you to explore.

Here I am only going to answer where you went wrong.
Though there are other ways to go about this, your approach is not wrong.
With the correction to your mistake, we will get the correct answer.

$\begin{align*} 555^{555} & = {\left(500+55 \right)}^{555} = \sum_{n=0}^{555}\binom{555}{n}{500}^{555-n}55^n = \cdots+555\left(500\right){\left(55\right)}^{554}+{55}^{555}\\ & \equiv 500+{\left(50+5\right)}^{555}\pmod{1000}\\ & \equiv 500+\cdots+[[\frac{(555!)}{(2!553!)}\left(50\right)^{2}{\left(5\right)}^{553}]]+555\left(50\right){\left(5\right)}^{554}+5^{555}\pmod{1000}\\ & \equiv 500+X+750+125\pmod{1000}\\ & \equiv X+375\pmod{1000} \end{align*}$

The term in $[[ .... ]]$ was missing.
That missing term is $X \equiv [[\frac{(555!)}{(2!553!)}\left(50\right)^{2}{\left(5\right)}^{553}]] \equiv (555) (277) (50)^{2} (5)^{553} \equiv 500 \mod {1000}$

We then get $500+375 \equiv 875$ which is the Correct Answer.

score 2 · Accepted Answer · answered Jul 28 '22 at 09:07

First of all the value mod 1000 does not mean that its power cycle is also 1000 as a cycle, so you cannot mod 1000 to it's power. You need find it's cycle.

$555^0 \equiv 1 \pmod{1000}$

$555^1 \equiv 555 \pmod{1000}$

$555^2 \equiv 25 \pmod{1000}$

$555^3 \equiv 875 \pmod{1000}$

$555^4 \equiv 625 \pmod{1000}$

$555^5 \equiv 875 \pmod{1000}$

So the cycle is $2$, start from power $= 3$

$555^{555^{555}} \equiv 555^{3 + (555^{555} - 3 \pmod 2)} \pmod{1000}$

$\equiv 555^{3} \pmod{1000}$

$\equiv 875 \pmod{1000}$

In fact, we get that 2 is its cycle period, so 1000 can also be considered as a cycle

But your calculation process is not entirely correct, leaving out the secondary terms

$\begin{align*}555^{375}={\left(500+55\right)}^{375}&=\sum_{n=0}^{375}\binom{375}{n}{500}^{375-n}55^n \\ &=\cdots+375\left(500\right){\left(55\right)}^{374}+{55}^{375}\\ &\equiv500+{\left(50+5\right)}^{375}\pmod{1000}\\ &\equiv500+\binom{375}{2}50^2 5^{373}+375\left(50\right){\left(5\right)}^{374}+5^{375}\pmod{1000}\\ &\equiv500+500+750+125\pmod{1000}\\ &\equiv875\pmod{1000}\end{align*}$

score 0 · Answer 3 · answered Aug 16 '22 at 13:46

Another way

$555^2\equiv5^2\pmod{1000}$ and $(5^2)^n\equiv5^3\pmod{1000}$ for $n\gt1$. It follows $$555^{2\cdot277+1}=555^{2\cdot277}\cdot555\equiv5^3\cdot555\equiv875\pmod{1000}$$ Which is equivalent to$$555^{555}=875+1000m$$Now $$555^{875+1000m}\pmod{1000}=555^{2(437+500m)+1}$$ Then $$\displaystyle555^{555^{555}} \bmod{1000}\equiv625\cdot555\equiv875\pmod{1000}$$

Finding $\displaystyle555^{555^{555}} \bmod{1000}$

3 Answers3