Earlier today I set you the following puzzle:

In the sum, each letter stands for a unique digit (i.e no two letters stand for the same digit.) In other words, POCO is a four digit number, and MUCHO is a five digit number.

Can you work out which letter stands for which digit so that the sum makes sense?

**Solution** POCO is 4595 and MUCHO is 68925

**Workings **For fifteen letter Os to add up to a number ending in the letter O, then O is either 0 (zero) or 5. Let’s say that O is 0 (zero). In this case, fifteen Cs must be CH. The only digits that fit are C = 1 and H = 5. The number MU, therefore, is 15 x P, which means that U must be an 0 (zero) or a 5. But both of those numbers are already taken by the letter O and H. So we can eliminate the case that O is 0, and deduce that O = 5.

If O = 5, the carry into the tens column is 7, since 5 x 15 = 75.

We can eliminate some of the candidates for C. To start, C cannot be 1, since then 15 x C + 7 = 22, which means there is a carry of 2 to the hundred’s column, which would mean that the C in MUCHO would be 7. C cannot have two values so we eliminate C = 1.

We can also eliminate values from 2 to 8 for C, shown here for completeness:

If C = 2, then 15 x C + 7 = 37, which means there is a carry of 3 to the hundreds’s column, which would mean the C in MUCHO would be 8. C cannot have two values so we eliminate C = 2.

If C = 3, then 15 x C + 7 = 52, which means there is a carry of 5 to the hundreds’s column, which would mean the C in MUCHO would be 0. C cannot have two values so we eliminate C = 3.

If C = 4, then 15 x C + 7 = 67, which means there is a carry of 6 to the hundreds’s column, which would mean the C in MUCHO would be 1. C cannot have two values so we eliminate C = 4.

If C = 5, then 15 x C + 7 = 82, which means there is a carry of 8 to the hundreds’s column, which would mean the C in MUCHO would be 3. C cannot have two values so we eliminate C = 5.

If C = 6, then 15 x C + 7 = 97, which means there is a carry of 9 to the hundreds’s column, which would mean the C in MUCHO would be 4. C cannot have two values so we eliminate C = 6.

If C = 7, then 15 x C + 7 = 112, which means there is a carry of 1 to the hundreds’s column, which would mean the C in MUCHO would be 6. C cannot have two values so we eliminate C = 7.

If C = 8, then 15 x C + 7 = 127, which means there is a carry of 2 to the hundreds’s column, which would mean the C in MUCHO would be 7. C cannot have two values so we eliminate C = 8.

So C must be 9. If C = 9, then 15 x C + 7 = 142, which means there is a carry of 4 to the hundreds’s column, which means the C in MUCHO is also a 9.

If C = 9, H = 2, and the thousands column has a carry of 7 + 1 = 8. In other words 15P + 8 = MU. P must be between 1 and 6, since MU is less than 100. We have as candidates 1, 3, 4, 6. We can eliminate 1, since this makes MU 23, and 2 is taken. We can eliminate 3, since this makes MU 53 and 5 is taken. We can eliminate 6 because this makes MU 98 and 9 is taken. So P = 4, M= 6 and U = 8.

Gracias!

*I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.*

*The source for today’s puzzle is: Puzzle Corner, MIT News, September 2014.*

*My latest puzzle book is So You Think You’ve Got Problems?*