Much of few is many, at least in Spanish:
15*POCO=MUCHO
Try to solve analytically.
Source: MIT Technology Review.
O must be 0 or 5.
Suppose O=0
We can ignore the terminal zeros so that
15*P0C=MUCH
Now H must be 5 and M is odd
15*P0C=MUC5
But now we require 15*0C ends in C5 which means C=1
15*P01=MU15
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.
Suppose O=5
15*P5C5=MUCH5
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.
15*P595=MU975
P must be 6 or less. Checking each gives a single solution:
15*4595=68925

Posted by Jer
on 20160428 11:59:32 