<s>[INST] How many zeros are there in $1 \times 2^2 \times 3^3 \times \cdots \times 100^{100}$? How many zeros are there at the end of $1 \times 2^2 \times 3^3 \times \cdots \times 100^{100}$ ?
I tried it by grouping all the $2$'s and $5$'s and $5$'s and $6$'s but cant get my answer...
 [/INST] To get a zero you need to multiply a factor $5$ by a factor $2$. There are clearly more factors of $2$ than of $5$ in the product, so you need to find a systematic way of counting the factors of 5.
Well these come from the terms $5^5\cdot 10^{10} \cdot 15^{15} \dots 100^{100}$
All of these terms provide a factor $5$ so we have $5^5\cdot 5^{10} \cdot 5^{15} \dots 5^{100}$, with exponent $5+10+15+ \dots +100$.
However the multiples of $25$ contribute twice, and we've only counted them once so far, so we need to add $25+50+75+100$ to the exponent.
We have no multiples of $5^3=125$ which would have contributed three factors.
The exponent counts the number of factors of $5$ and hence the number of zeros.
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