# Show that at most half of the positive divisors of a positive integer end in 3.

If we assume that there does exist such an integer $n$, then it has to be 9 mod 10. Is this useful at all? I have no idea where to go from that.

By the way, I saw this problem in some old math test (I think from MOSP). But I cannot find the solution. I would appreciate any hints/ideas on this problem. Thanks!

If it has a prime factor $2$ at least half the factors will be even and can't end in $3$. Similarly if it has a prime factor $5$, so all the prime factors will end in $1,3,7,9$. You can think of each factor as pulling from a multiset of $1's, 3's, 7's$ and $9's$. If there is one $7$ you can match up each factor without a $7$ with one that has it. At most one ends in $3$. Similarly if there are more $7$s at most half the factors end in $3$ so there can't be any factors ending in $7$. The same argument applies to factors ending in $9$. The ones ending in $1$ don't matter. If we have at least one $3$ again not more than half the factors end in $3$ and we are done.