Divisible by this year??? v2015 (Part 3: A Jinx from the future!!!)

Number Theory Level 5

Find the minimum value of $n$ where $n$ is an integer such that the number of trailing zeroes of $(n+20)! \times (n + 16)!$ is divisible by $2015$

( $201500000$ has $5$ trailing zeroes while $2000015000$ has $3$ )

1 solution

Rindell Mabunga
Jan 3, 2015

Let $f(n) = (n + 20)! \times (n + 16)!$

The domain of this function is $[-16, + \infty )$

So first let's try $f(-16)$ .

$(n + 20)! \times (n + 16)!$

$= ( - 16 + 20)! \times ( - 16 + 16)!$

$= 4! \times 0!$

$= 24$

Since there are $0$ trailing zeroes in $24$ and $0$ is divisible by $2015$ , the answer must be $-16$

very tricky but fortunately, I realize it when submiting my 3rd answer :)

Hunter Killer - 6 years, 5 months ago

Great... :P very tricky,

Terza Reyhan - 5 years, 11 months ago