Much of few is many, at least in Spanish:
Try to solve analytically.
Source: MIT Technology Review.
O must be 0 or 5.
We can ignore the terminal zeros so that
Now H must be 5 and M is odd
But now we require 15*0C ends in C5 which means C=1
With the lack of carries, 15*P=MU but both 0 and 5 are taken so there's no choices for U.
So O is not 0.
If C is odd, H=2. If C is even H=7
If H=2, it can easily be shown that no choice of C in POCO puts the same C in MUCHO.
If H=7, the only choice for C that works is C=9.
P must be 6 or less. Checking each gives a single solution:
Posted by Jer
on 2016-04-28 11:59:32