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The number obtained from the last two nonzero digits of $90!$ is equal to $n$. What is $n$?

Link to the sample solutions for the problem. Please look at Solution 1. (Solution 2 is even less clear, I feel) I understand that we first get rid of the trailing zeroes by setting $N = \frac{90!}{10^{21}}$. Then we can reduce this mod 25.

My issue is with the clarity of the following statements:

  1. If instead we find $N\pmod{25}$, we know that $N\pmod{100}$, what we are looking for, could be $N\pmod{25}$, $N\pmod{25}+25$, $N\pmod{25}+50$, or $N\pmod{25}+75$.

    How do we reduce our search to these numbers specifically?

  2. If we divide $N$ by $5^{21}$ by taking out all the factors of $5$ in $N$, we can write $N$ as $\frac M{2^{21}}$ where $$M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,$$

    Ok, this makes no sense. I thought all factors of 5 were divided out when we got rid of the trailing zeros? And where did the $2^{21}$ come from? After that, what's with $M$? I don't understand this at all...

I'd really appreciate it if someone who understands the solution could explain it to me.

1 Answers 1

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Well lets say for some $p$ $$N\equiv p\pmod{25}$$ That implies that for some $k$ $$N=p+25k$$ Then looking $25k\pmod{100}$ we see that it can be $25,50,75,0$ from this we see why the solution $\mod{100}$ can be $N\pmod{100},N\pmod{100}+25,N\pmod{100}+50,N\pmod{100}+75$.

For the second statement we have that $$M=\frac{90!}{5^{21}}$$ And $$N=\frac{90!}{2^{21}\cdot 5^{21}}=\frac{90!}{5^{21}}\cdot \frac{1}{2^{21}}=M\cdot \frac{1}{2^{21}}=\frac{M}{2^{21}}$$ Also $$M=1\cdot 2\cdot 3\cdot 4\cdot \color{red}{1}\cdot 6\cdot 7\cdot 8\cdot 9\cdot \color{red}2 \cdots \cdot 89\cdot \color{red}{18}$$ Now the parts in the red are the remainders of the $5$'s that got divided,and the other ones are groups of four consecutive integers not divisible by $5$'s.Now the red numbers go $\color{red}{1,2,3,4,1,6,7,8,9,2\cdots 16,17,18}$ so every fifth number is divided by $5$(since $M$ isn't divisible by $5$'s) those are numbers that were divisible by $25$ in $90!$ and he just grouped the terms like that.