# Last 2 numbers of the product of divisors

Let $N$ be the product of all divisors of $2013^{2013}$. What are the last 2 numbers of $N$ in its decimal notation?

I don't know where to start in this exercise, would like to get a hint or two :) Thanks in advance!

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First note that the product of all divisors of an integer $n$ is $\sqrt{n^{\tau(n)}}$, where $\tau(n)$ denotes the number of divisors of $n$. Since $2013=3\cdot11\cdot61$, we see that

$$\tau(2013^{2013})=\tau(3^{2013})\tau(11^{2013})\tau(61^{2013})=2014^3$$

Moreover, as $\textbf{Ivan Loh}$ pointed out, Euler's totient theorem tells us that $a^{\varphi(100)}=a^{40}\equiv_{100}1$ for $\gcd(a,100)=1$. It follows that

\begin{align*}N&=\sqrt{(2013^{2013})^{\tau\left(2013^{2013}\right)}}=2013^{2013(1007)(2014)^2}\equiv_{100}13^{13(7)(14)^2}=\\&=13^{91(196)}\equiv_{100}13^{11(-4)}\equiv_{100}13^{-4}\equiv_{100}61^{-1}=\left\{a^{-1}=b(ab)^{-1}\right\}=\\&=(1−60)((1+60)(1−60))^{−1}\equiv_{100}-59\equiv_{100}41\end{align*}

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why did you divide 8169178744 by 2 and how can I tell the mod100 number without calculating the previous one? – darenn May 9 '13 at 17:50
Because $\sqrt{n^{2m}}=n^m$. I will elaborate on the calculations shortly. – Librecoin May 9 '13 at 17:52
You can show the product of all divisors formula by pairing a divisor $d$ with $\frac n d$ - that picks up each divisor once unless $n$ is a square, and it is easy to adjust for this case. So each pair of complementary divisors contributes $n$ to the product, with an adjustment if necessary for the square root. – Mark Bennet May 9 '13 at 18:09
Great thanks for your help! Just wondering if there's a way to do this without a calculator (you had to find those divisors somehow :) ) – darenn May 9 '13 at 18:15
You can avoid calculations as follows: We have $\tau(2013^{2013})=2014^3$. Also by Euler's totient theorem (since $\phi(100)=40$), $a^{40} \equiv 1 \pmod{100}$ for $\gcd(a, 100)=1$. Since $2013(1007)(2014)^2 \equiv 13(7)(14)^2 \equiv 91(196) \equiv 11(-4) \equiv -4 \pmod{40}$, so $$N=\sqrt{(2013^{2013})^{\tau(2013^{2013})}}=2013^{2013(1007)(2014)^2} \equiv 13^{-4} \equiv 61^{-1} \equiv 41 \pmod{100}$$ For the last step, you could do $61^{-1} \equiv (1-60)^{-1}((1+60)(1-60))^{-1} \equiv 41 \pmod{100}$. – Ivan Loh May 10 '13 at 1:56

Hint: The product of all divisors of N is a multiple of N.

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