Bossnass

I agree with the logic that you posted.

So working with P,R=1,0, there can be no carryover to the fifth colum, to be less then 5 because the sum must be less then ten and already we know that it must be even.

So O is either 2 or 4, making S equal to 1,6, or 7. 1 is already taken, so it must be six or seven and then T must be odd, from 12 or 14.

With T being odd, we know it isn't 1 or 3, and 9 is taken. So T = 5 or 7. T can only be 5 if U = 2.

Now 1,2/4, 5/7, 6/7 and 9 are taken.

So I tried O=2, implying E= 4 or 5, and S= 6. Since we are assuming O=2, U can't be 2, so T = 7. Accordingly, U= 3 Also, since T=7, S is going to be even (no carried one)

So 1,2,3,(4or5),6,7, and 9 are taken. So C must be the an even number, C=4or8. If C=4, then S=8. But S is already 6, and (8+8)= 16. Further, if C was 4, then since O is 2, E would be 4, and C can't = E.

Now you plug it in to double check.
H=9, O=2, C=8, U=3, S=6,
P=1, O=2, C=8, U=3, S=6,
P=1 R=0, E=5, S=6, T=7, O=2.

And it works.

This took my about 15 minutes. I'm an adult university student; a programming course I took didn't directly give me credit for a programming course I need, so I'm currently in a 1 credit, 1 hour a week, 'Logic 273' course. This is simple now, but is almost exactly like the questions we started with in January. There are brillaint people in class that come up with elegant solutions. I'm more of a brute force or methodical worker myself. Must be a pretty hardcore grade 4 teacher.

I guess it took me more then 15 minutes. Mal hadn't posted the solution yet when I started. I had to use paper and I was eating lunch at the time, so I didn't rush back. I should have refreshed.