Trailing Zeros in the Divisor Product of Product

Number Theory Level 5

A positive integer $m$ has 30 positive divisors in total. It is divisible by $2^4,$ but not divisible by $2^5.$ Another positive integer $n$ has 80 positive divisors in total. It is divisible by $5^7,$ but not divisible by $5^8.$ If $m$ and $n$ are coprime and $P$ is the product of all positive divisors of $mn,$ find the number of trailing zeroes in $P.$

The answer is 4800.

1 solution

Patrick Corn
Mar 6, 2018

The product of positive divisors of $x$ is $x^{d(x)/2},$ where $d(x)$ is the number of positive divisors of $x.$ In this case, $P = (mn)^{d(mn)/2} = (mn)^{d(m)d(n)/2} = (mn)^{1200}$ (note that $d(mn) = d(m)d(n)$ because $m$ and $n$ are coprime).

Now $mn = 2^4 5^7 b$ where $b$ is coprime to $10,$ so $P = 10^{4800} 5^{3600} b^{1200}$ has exactly $\fbox{4800}$ trailing zeros.

Trailing Zeros in the Divisor Product of Product

The answer is 4800.

1 solution

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