National Olympiad of Argentina

Let $a,b,c\in \mathbb{N}$, $a+b+c=1810$, and let $abc$ end in $N$ zeros. The following triples $(a,b,c)$ show that $N$ can be $0,1,\dots ,7$: $(1808,1,1)$, $(1799,10,1)$, $(1709,100,1)$, $(1765,40,5)$, $(1200,605,5)$, $(1700,100,10)$, $(1250,520,40)$, $(1250,400,160)$.

We prove that $N\le 7$ always holds. Clearly $N$ is at most the total number of 5's in the prime factorizations of $a,b,c$. Let $5^k,5^l,5^m$ be the highest powers of 5 dividing $a,b,c$ respectively. Assume

$k\ge l\ge m$ by symmetry. We have $k\le 4$ as $5^5>1810$. If $m=0$ (i. e. $5\nmid c$) then $l=0$ because $5\mid 1810$, so that $N\le k\le 4$. And if $m\ge 1$ then $m=1$ since $1810$ is exactly divisible by $5$. Hence $N\le 4+4+1=9$. However $N=8$ and $N=9$ cannot occur. Indeed $N\ge 8$ only if $k=4$ and $m=1$, hence $k+l+m\ge N\ge 8$ yields $l\ge 3$. Thus $(k,l,m)=(4,3,1)$ or $(k,l,m)=(4,4,1)$. In the first case $a$ is divisible by $5^4=625$, in the second case so are $a$ and $b$. Note that some of $a,b,c$ can be a multiple of $5^4=625$ only if it is even. Indeed having an odd summand in $a+b+c=1810$ means having exactly two odd ones as $1810$ is even. Since $N$ is at most the total number of 2's in the prime factorizations of $a,b,c$, $N\ge 8$ implies that the third number must be divisible by $5\cdot 2^8=1280$. This cannot hold as $625+1280>1810$. So $a\ge 2\cdot 625=1250$ for $(k,l,m)=(4,3,1)$ and $a,b\ge 1250$ for $(k,l,m)=(4,4,1)$. Thus the latter is rejected together with $N=9$ ($1250+1250>1810$). There remains $N=8$ with $(k,l,m)=(4,3,1)$ and $a=1250$ ($a$ is an even multiple of 625 and also $4\cdot 625>1810$). Here $b+c=560$ and $5^3|b$; moreover it is clear that $b$ is even. Thus $b=2\cdot 125=250$ or $b=4\cdot 125=500$, but neither one gives a product $abc$ divisible by $10^8$. The desired $N\le 7$ is established.