mjhjmtu

Each letter represents a digit, so what is Mucho and Poco (although who's to say?)

 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
 POCO 
+POCO 
-----
MUCHO

As far as I know there's only one unique answer

Puzzle courtesy of Bob High - MIT Technology Review Sep/Oct 2014

Poco = 4595, Mucho = 68925

Method:

I figured out O had to be 0 or 5 pretty quickly, because those are the only two numbers that keep their last digit when multiplied by 15. Then, I honestly just lucked into it using the tried and true (get it? hehe) method of trial and error.

Okay, so we know

15 x O ends in O. Therefore O must be 0 or 5. Let O be 0. Then 15C ends in H. Further, if 15C equals 10 x A + H for some number A, then A = C because of the 3rd column. Therefore 15C = 10C + H, so H = 5C. There is only one combination that does that: C = 1 and H = 5. Then we must have 15P = 10M + U. But 15P ends in 0 or 5, and we already used a 0 and a 5. Contradiction! So O is 5. Then 15O = 75, so 15C + 7 = 10xA + H. Therefore H must be either 2 or 7. Further, 75 + A = 10xB + C. Let C be 0. Then H is 7, and A is 0. Contradiction. Let C be 1. Then H is 2, and A is 2. Contradiction. Let C be 2. Then H is 7, and A is 3. Contradiction. Let C be 3. Then H is 2, and A is 5. Contradiction. Let C be 4. Then H is 7, and A is 6. Contradiction. Let C be 6. Then H is 7, and A is 9. Contradiction. Let C be 7. Then H is 2, and A is 11. Contradiction. Let C be 8. Then H is 7, and A is 12. Contradiction. Let C be 9. Then H is 2, and A is 14. This works! 15 x 9 = 135 + 7 = 142. Then 15 x 5 = 75 + 14 = 89. We therefore have -595 x 15 = - -925. Finally, 15 x P + 8 = MU. M,U,P can’t be 9,2,5. So 15 x 4 + 8 = 68, which means P = 4, M = 6, and U = 8.

We therefore have

4595 x 15 = 68925, ie. POCO = 4595 and MUCHO = 68925.