How many zeros are there in $1 \times 2^2 \times 3^3 \times \cdots \times 100^{100}$?

Question

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...

Answer 1

Hint: The number of 5's is $(5 + 10 + 15 + \ldots + 95 + 100) + (25 + 50 + 75 + 100)$.

Answer 2

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.

Answer 3

I checked the product using "PARI GP " program and found both the previous answers to be correct. The total number of zeroes =(5+10+15+...+95+100) +(25+50+75+100) =5*10*21+250=1300