Where are the numbers?

Number Theory Level 3

True or False:

Suppose there is a positive integer $a$ such that $a!$ has $m$ trailing number of zeros . There is another distinct positive integer $b$ such that $b!$ has $n$ trailing number of zeros. Then, $(a+b)!$ must have $m+n$ trailing number of zeros.

Notation : $!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

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False True

1 solution

展豪張
Jun 14, 2016

Number of trailing zeros $m$ of $a!$ is actually the maximum power of $5$ which divides $a!$ and have a formula:
$m=\lfloor\dfrac a5\rfloor+\lfloor\dfrac a{25}\rfloor+\lfloor\dfrac a{125}\rfloor+\cdots$
which is not a linear function.
In fact, $(a+b)!$ has $\large[$ $m+n+$ 'number of carries when $a$ is added to $b$ in base $5$ ' $\large]$ trailing zeroes. (Proof is left as an exercise for readers. ;))

Very nice explanation on how and why trailing zeroes increases at some points. I am impressed :)

Abhay Tiwari - 5 years ago

Thank you! =D

展豪張 - 5 years ago

Where are the numbers?

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