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The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
4
down vote

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1












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










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  • A lot of “little” things add up to “a lot”!! I love this alphametic!
    – El-Guest
    1 hour ago










  • Ah! I missed a couple of little things. Don't waste any time yet!
    – JGibbers
    1 hour ago










  • Alright, should be good to go now @El-Guest
    – JGibbers
    1 hour ago















up vote
4
down vote

favorite
1












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










share|improve this question









New contributor




JGibbers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • A lot of “little” things add up to “a lot”!! I love this alphametic!
    – El-Guest
    1 hour ago










  • Ah! I missed a couple of little things. Don't waste any time yet!
    – JGibbers
    1 hour ago










  • Alright, should be good to go now @El-Guest
    – JGibbers
    1 hour ago













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





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










share|improve this question









New contributor




JGibbers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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







mathematics no-computers alphametic






share|improve this question









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JGibbers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question









New contributor




JGibbers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question




share|improve this question








edited 1 hour ago





















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asked 2 hours ago









JGibbers

75537




75537




New contributor




JGibbers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





JGibbers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • A lot of “little” things add up to “a lot”!! I love this alphametic!
    – El-Guest
    1 hour ago










  • Ah! I missed a couple of little things. Don't waste any time yet!
    – JGibbers
    1 hour ago










  • Alright, should be good to go now @El-Guest
    – JGibbers
    1 hour ago

















  • A lot of “little” things add up to “a lot”!! I love this alphametic!
    – El-Guest
    1 hour ago










  • Ah! I missed a couple of little things. Don't waste any time yet!
    – JGibbers
    1 hour ago










  • Alright, should be good to go now @El-Guest
    – JGibbers
    1 hour ago
















A lot of “little” things add up to “a lot”!! I love this alphametic!
– El-Guest
1 hour ago




A lot of “little” things add up to “a lot”!! I love this alphametic!
– El-Guest
1 hour ago












Ah! I missed a couple of little things. Don't waste any time yet!
– JGibbers
1 hour ago




Ah! I missed a couple of little things. Don't waste any time yet!
– JGibbers
1 hour ago












Alright, should be good to go now @El-Guest
– JGibbers
1 hour ago





Alright, should be good to go now @El-Guest
– JGibbers
1 hour ago











2 Answers
2






active

oldest

votes

















up vote
4
down vote














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.







share|improve this answer






















  • I had it backwards. His timestamp came later. My comment has been deleted.
    – Brendon Shaw
    52 mins ago

















up vote
2
down vote













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.







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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote














    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.







    share|improve this answer






















    • I had it backwards. His timestamp came later. My comment has been deleted.
      – Brendon Shaw
      52 mins ago














    up vote
    4
    down vote














    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.







    share|improve this answer






















    • I had it backwards. His timestamp came later. My comment has been deleted.
      – Brendon Shaw
      52 mins ago












    up vote
    4
    down vote










    up vote
    4
    down vote










    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.







    share|improve this answer















    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.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 58 mins ago

























    answered 1 hour ago









    Excited Raichu

    1,484120




    1,484120











    • I had it backwards. His timestamp came later. My comment has been deleted.
      – Brendon Shaw
      52 mins ago
















    • I had it backwards. His timestamp came later. My comment has been deleted.
      – Brendon Shaw
      52 mins ago















    I had it backwards. His timestamp came later. My comment has been deleted.
    – Brendon Shaw
    52 mins ago




    I had it backwards. His timestamp came later. My comment has been deleted.
    – Brendon Shaw
    52 mins ago










    up vote
    2
    down vote













    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.







    share|improve this answer
























      up vote
      2
      down vote













      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.







      share|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        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.







        share|improve this answer












        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.








        share|improve this answer












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        share|improve this answer










        answered 1 hour ago









        El-Guest

        15.7k13475




        15.7k13475




















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