Find the last two digits of $$ N=\left(1!+2!+3!+\cdots+101!\right)^{101}. $$

Solution

Finding the last two digits of \(N\) is equivalent to finding

$$ N \pmod{100}. $$

Since

$$ 10!=2^8\cdot3^4\cdot5^2\cdot7, $$

we have

$$ 10!\equiv0\pmod{100}. $$

Hence, every factorial \(n!\) with \(n\ge10\) is also divisible by \(100\). Therefore,

$$ N\equiv\left(1!+2!+3!+\cdots+9!\right)^{101}\pmod{100}. $$

Now,

$$ \begin{aligned} 1!+2!+\cdots+9! &=1+2+6+24+120+720+5040+40320+362880\\ &\equiv1+2+6+24+20+20+40+20+80\\ &\equiv13\pmod{100}. \end{aligned} $$

Thus,

$$ N\equiv13^{101}\pmod{100}. $$

Observe that

$$ \begin{aligned} 13^2&=169\equiv69\pmod{100},\\ 13^4&\equiv69^2=4761\equiv61\pmod{100},\\ 13^5&\equiv61\cdot13=793\equiv93\pmod{100}. \end{aligned} $$

Hence,

$$ 13^{20}=(13^5)^4\equiv93^4\equiv1\pmod{100}. $$

Therefore,

$$ \begin{aligned} N &\equiv13^{101}\\ &=13(13^{20})^5\\ &\equiv13\cdot1^5\\ &\equiv13\pmod{100}. \end{aligned} $$

Answer

$$ \boxed{13} $$

Therefore, the last two digits of $$ \left(1!+2!+\cdots+101!\right)^{101} $$ are 13.