## [Soln] Fall 2014 USA Online Math Open Qn 9

Fall 2014 USA OMO Qn 9. Let $N = 2014! + 2015! + 2016! + \dots + 9999!$. How many zeros are at the end of the decimal representation of $N$?

To count the number of zeros at the end of $N$, we need to find out the highest power of 2 that divides $N$ and the highest power of 5 that divides $N$, then the number of zeros that the decimal representation of $N$ has is the lower of the two. Concretely, if $N = 2^\alpha 5^\beta M$ for some integer $M$ such that $\gcd (M, 10) = 1$, then the number of zeros in $N$‘s decimal representation is $\min (\alpha, \beta)$.

The short explanation above suggests that it would be a good idea to find the prime factorisation of $N$. The key to the solution is realising that each succeeding term is a multiple of the previous one, hence all succeeding terms are a multiple of the first one. That is, we can write $N$ as

$N = 2014! [ 1 + 2015 + (2015)(2016) + \dots + (2015)(2016)\dots(9999) ].$

At this point, you might notice that for the second factor (the one in square brackets), all its terms except the 1 in front are divisible by 5. This means that the second factor will not be “contributing” any factors of 5 to the prime factorisation of $N$. To simply the line above, we can write

$N = 2014! [1 + 5k]$

for some integer $k$. While $[1+5k]$ factor might contribute some factors of 2 to the prime factorisation of $N$, it won’t contribute any more factors of 5. It should also be clear that the $2014!$ factor has more factors of 2 than factors of 5. Hence,

$\text{number of '0's at end of }N = \text{number of '0's at end of } 2014!$.

There is a well-known formula that the highest power of a prime $p$ which divides $n!$ is

$\displaystyle\sum_{i = 1}^\infty \left\lfloor \frac{n}{p^i} \right\rfloor = \left\lfloor \displaystyle\frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \dots$

Hence, number of ‘0’s at the end of $N$ is

\begin{aligned} \left\lfloor \frac{2014}{5} \right\rfloor + \left\lfloor \frac{2014}{25} \right\rfloor + \left\lfloor \frac{2014}{125} \right\rfloor + \left\lfloor \frac{2014}{625} \right\rfloor &= 402 + 80 + 16 + 3 \\ &= \boldsymbol{501}. \end{aligned}