Calculation of the last three digits of $132^{1601}$ (solving $x \equiv 132^{1601} \pmod {1000}$)

Question

I want to calculate the last three digits of $132^{1601}$. This is equivalent to find $x \equiv 132^{1601} \pmod {1000}$.

This is how I've solved it:

$\Phi(1000)=400,$

$132^{400} \equiv 1 \pmod {1000},$

So $x \equiv 132^{1601} \pmod {1000} \equiv (132^{400})^4132 \pmod {1000} \equiv 132 \pmod {1000}.$

Is this approach correct?

Thanks.

EDIT: one of my friends suggest that it must be split using the Chinese reminder theorem and that the solution is $632 \pmod {1000}$. How is that possible?

Answer 1

You can not apply Euler to this directly, since $132$ is not relatively prime to $1000$. Indeed, it is clear that $132^{400}\not \equiv 1 \pmod {1000}$ since this would imply that $2\,|\,1$.

To solve the problem, work mod $2^3$ and $5^3$ separately. Clearly $132^{1601}\equiv 0\pmod {2^3}$. Now, $\varphi(5^3)=100$ and Euler applies here (since $\gcd(132,5)=1$) so we do have $$132^{100}\equiv 1 \pmod {5^3}\implies 132^{1600}\equiv 1 \pmod {5^3}$$

Thus $$132^{1601}\equiv 132\equiv 7\pmod {5^3}$$

It follows that we want to find a class $n\pmod {1000}$ such that $$n\equiv 0 \pmod 8\quad \&\quad n\equiv 7 \pmod {125}$$ The Chinese Remainder Theorem guarantees a unique solution, which is easily found to be $$\boxed {132^{1601}\equiv 632\pmod {1000}}$$

Note: with numbers as small as these, the CRT can be solved by mental arithmetic (or, at least, by simple calculations). We start with $7$. Clearly that isn't divisible by $8$ so we add $125$ to get $132$. That's divisible by $4$, but not by $8$. Now, adding $125$ to this would give an odd number so add $250$. We now get $382$, still no good. Adding $250$ again gives $632$ and that one works, so we are done.

If you prefer to solve it algorithmically, write the solution as $n=7+125m$ We want to solve $$7+125m\equiv 0\pmod 8\implies 5m\equiv 1 \pmod 8\implies m\equiv 5 \pmod 8$$ In that way we get $n=7+5\times 125=632$.

Answer 2

Like Find the last two digits of $2^{2156789}$ and Last Two Digits Problem and How to find last two digits of $2^{2016}$,

let us find $P=132^{1601-2}\pmod{125}$

Now $132\equiv7\pmod{125},1601-2\equiv-1\pmod{\phi(125)}$

$\implies P\equiv7^{-1}\pmod{125}\equiv18$

$\implies132^2P\equiv18\cdot132^2\pmod{125\cdot132^2}$

$\equiv18(100+32)(100+32)\pmod{1000}$

$\equiv18(200\cdot32+32^2)$

$\equiv18(400+24)\equiv200+432$

Answer 3

$2\mid 132,1000\,$ so Euler $\phi$ doesn't apply. Use CRT, or simpler (a minute of mental calculation)

$ 4k^{\large 1+100N}\!\bmod 1000\, =\, 8 \overbrace{\left[ \dfrac{(4k)^{\large 1+\color{#c00}{100}N}}8\bmod \color{#c00}{125}\right]}^{\qquad \large \color{#c00}{100\ \ = \ \ \phi(125)} } \!$ $= 8\underbrace{\left[ \dfrac{k}2\bmod 125\right]} =\!\!\!\!\!\!\begin{align}\overbrace{4k\!+\!500}^{\ \ \large 632\ {\rm if}\ 4k\ =\ 132}\!\!\!& {\rm if}\ \ 2\nmid k \\ 4k\qquad & {\rm if}\ \ 2\mid k \\ \phantom{.} \end{align} $
by $\,\ ab\bmod ac\, =\, a(b\bmod c)\ $ [mod distributive law] $\ $ & $\ \ \dfrac{k}2\equiv \dfrac{k\!+\!125}2\,\pmod{\!\!125}\ \,$ if $\ 2\nmid k$

Answer 4

I would do it in a slightly different way: split $132$ as a factor of $1000$ times a factor coprime to $1000$: $$132=4\cdot 33.$$

On the other hand, $\;\varphi(1000)=\varphi(2^3)\,\varphi(5^3)=4\,(4\cdot 5^2)=400$, so by Euler's theorem $$33^{1601}\equiv 33^{1601\bmod400}=33^1.$$

As to $4$, we'll use the Chinese remainder theorem, in the form:

If $a$ and $b$ are coprime, the solutions of the system of congruences $\;\begin{cases}x\equiv\alpha\mod a,\\ x\equiv \beta\mod b,\end{cases}\;$ are given by $$x\equiv\beta ua+\alpha vb\mod ab.$$

Now $4^k\equiv 0\mod 8$ for all $k>1$, and as $4$ is coprime to $125$, $\;4^{1601}\equiv 4^{1601\bmod \varphi(125)}= 4^1 \mod 125 $, so that a Bézout's relation between $8$ and $125$: $$47\cdot 8-3\cdot 125=1$$ (obtained with the extended Euclidean algorithm) yields the congruence $$4^{1601}\equiv 4\cdot47\cdot8=1504\equiv 504\mod 1000, $$ and ultimately $$132^{1601}=4^{1601}33^{1601}\equiv 504\cdot 33 =500\cdot 32+500+4\cdot 33=632\mod 1000.$$

Answer 5

Euler's theorem $a^{\phi n} \equiv 1 \pmod n$ only works if $\gcd(a,n) = 1$. Which is not the case with $132, 1000$

However $1000 = 8*125$

And CRT theorem does guarantee that if we can solve $132^{1601}\pmod 8$ and $132^{1601} \pmod {125}$ those two solutions will provide a unique solution to $132^{1601} \pmod {1000}$

....

We can't use Euler's Theorem for $132^{1601} \pmod 8$, of course, but $132 = 33*4$ so $132^{1601} = 33^{1601}*4^{1601}$ and $8|4^k$ for all $k \ge 2$ so $132^{1601} \equiv 0 \pmod 8$.

And for $132^{1601} \pmod {125}$ we CAN use Euler's theorem.

As $125|1000$ then $\phi{125}|\phi{1000}$ so $132^{1601}\equiv 132 \pmod {125}$. (in fact $\phi(125) = 20$ but... why redo work you already did.)

So we need to find the unique solution $x \equiv 0 \pmod 8$ and $x \equiv 132 \pmod {125}$. That is $x = 8m = 132 + 125k$ where $0 \le m < 125$ and $0 \le k < 8$.

As $8\not \mid 132$ we can't have $8\mid k$ but as $4|132$ we must have $4|k$.

In other words $k =4$ and $x \equiv 132 + 500\equiv 632 \pmod {1000}$ is the unique solution.

.....

If we want to verify this:

$132^{1601} = 4^{1601}*33^{1601}$ And $33^{1601} \equiv 33\pmod{1000}$ so

$4^{1601}\equiv 4\pmod {125}$ so $4^{1601} \equiv 4,129,254,379,504,629,754,$ or $879 \pmod {1000}$. But as $8|4^{1601}$ then $4^{1601}\equiv 504\equiv 500 + 4 \pmod{1000}$

So $132^{1601} = 4^{1601}33^{1601} \equiv (500 + 4)33 \pmod{1000}$

$\equiv 500 + 132\equiv {1000}$

==========

In general. If you have $a$ and $n$ and $\gcd(a,n) = d$ then we can set up $n = n'D$ where $\gcd(n',D) = 1$ and $d|D$.

Then we can solve $a^k \equiv x\pmod n$ by solving $a^k \pmod{n'}$ and $a^k \equiv \pmod D$.

$a^k \pmod{n'}$ can be solved by Euler's Theorem.

$a$ can also be written as $a = a'\delta$ where $\gcd(a',n) = \gcd(a', d) = 1$ and $d|\delta$ (note that either $\delta$ or $D$ equals $d$). And so we can solve $a^k\pmod D$ by solving $a^k = a'^k*d^k*(\frac {\delta}d)^k = MD = Md*\frac Dd\implies$

$a'^k d^{k-1}(\frac {\delta}d)^k = M\frac Dd$ . If $D= d$ then will mean $a^k\equiv 0 \pmod D$. Other wise this means $a'^k d^{k-1} = M\frac Dd$. Now $\frac Dd$ has the same prime factors of $d$ so this will usually mean $a^k \equiv 0\pmod D$ but might not if the powers of the prime factors of $\frac Dd$ are higher than the prime factors of $d^{k-1}$. But if that is the case we can reduce and and solve by Euler's theorem.

So Euler's theorem in combination with CRT will always allow us to solve these.

Answer 6

$\overbrace{132^{\large 1+\color{#c00}{100}N}}^{\large X}\!\!\equiv 132\,\ \overbrace{{\rm holds} \bmod \color{#c00}{125}}^{\large\color{#c00}{100\ =\ \phi(125)}}\,$ & $\overbrace{\!\bmod 4}^{\large 0^K \equiv\ 0}\,$ so mod $500,\,$ so it's $\overbrace{ 132\ \ {\rm or} \underbrace{132\!+\!500}_{\large \rm must\ be \ this }\!\pmod{\!1000}}^{\large 132\ \not\equiv\ X\ \ {\rm by}\ \ N>1\ \,{\Large \Rightarrow}\,\ 8\ \mid\ 132^{\LARGE 2}\, \mid\ X\!\! } $

Answer 7

We have $132 = 2^2 \cdot 3 \cdot 11$.

If we want to apply the exponential laws we'll need to find a reducing exponential relation for the $2$ factors. Since

$$ 2^{10} \equiv 2^3 \cdot 3 \pmod{1000}$$

we have all we need to solve the problem.

Since

$$ (2^{2})^{1601} \equiv 2^3×3^{457} \pmod{1000}$$

we write

$$ 132^{1601} \equiv 2^3 \cdot 3^{2058} \cdot 11^{1601} \pmod{1000}$$

and

$$ 132^{1601} \equiv 8 \cdot 689 \cdot 11 \equiv 632 \pmod{1000}$$