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

  • A lot of “little” things add up to “a lot”!! I love this alphametic! – El-Guest Oct 16 at 17:33
  • Ah! I missed a couple of little things. Don't waste any time yet! – JGibbers Oct 16 at 17:33
  • Alright, should be good to go now @El-Guest – JGibbers Oct 16 at 17:34
up vote 10 down vote accepted

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.

  • I had it backwards. His timestamp came later. My comment has been deleted. – Brendon Shaw Oct 16 at 18:33

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.

  • Yours was more thorough, but his answer came about 30 seconds before, so i'm not quite sure what the etiquette is on that here? I gave him the checkmark but maybe I should give you a bounty? – JGibbers Oct 17 at 14:05
  • That’s entirely up to you as the puzzle-setter, @JGibbers! I don’t think there’s a standard etiquette, but it’s left generally to the discretion of the populace. – El-Guest Oct 17 at 14:07
  • 1
    @JGibbers: I'd personally give it to the one with the best explanation. This site isn't just about solving puzzles, its about helping others to do so better too and this answer does that better. – Chris Oct 22 at 22:13
  • Thanks, @Chris! I appreciate the kind words. – El-Guest Oct 23 at 2:50
  • Couldn't agree more! In 16 hours it's his! – JGibbers Oct 23 at 2:53

Assuming that

all the letters represent unique digits,

the one answer I find is:

POCO = 4595 and MUCHO = 68925

But if you allow for

more than one letter representing the same digit,

then there are even more solutions:

POCO = 2000 and MUCHO = 30000, POCO = 4000 and MUCHO = 60000, POCO = 6000 and MUCHO = 90000, POCO = 1000 and MUCHO = 15000, POCO = 3000 and MUCHO = 45000, POCO = 5000 and MUCHO = 75000, POCO = 2010 and MUCHO = 30150, POCO = 4010 and MUCHO = 60150, POCO = 6010 and MUCHO = 90150, POCO = 1010 and MUCHO = 15150, POCO = 3010 and MUCHO = 45150, POCO = 5010 and MUCHO = 75150, POCO = 1595 and MUCHO = 23925, POCO = 3595 and MUCHO = 53925, POCO = 5595 and MUCHO = 83925, POCO = 2595 and MUCHO = 38925, POCO = 4595 and MUCHO = 68925, POCO = 6595 and MUCHO = 98925

And if you allow for

P and M both both able to represent 0

then these solutions also open up:

POCO = 0000 and MUCHO = 00000, POCO = 0010 and MUCHO = 00150, POCO = 0595 and MUCHO = 08925

  • 1
    Good answer! If you check out the alphametic tag info, you’ll notice that standard assumptions are that different letters mean different numbers, and leading letters are non-zero. – El-Guest Oct 16 at 20:32
  • Thanks, El-Guest! That's good to know. – J-L Oct 16 at 21:00
  • Based on your MUCHO I assume your POCO has a typo. Right now MUCHO/15 ≠ your provided POCO. – Maksim Vi. Oct 17 at 0:04
  • Whoops! You are correct, Maksim Vi. I corrected my answer. The answer I mistakenly posted was a four-digit sequence that appears frequently in my life, so it was likely finger memory that was the culprit. So... thanks for the correction! – J-L Oct 17 at 16:44

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