No spanish required

up vote
4
down vote

favorite

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

edited 1 hour ago

asked 2 hours ago

JGibbers

75537

New contributor

A lot of Ã¢Â€ÂœlittleÃ¢Â€Â things add up to Ã¢Â€Âœa lotÃ¢Â€Â!! I love this alphametic!
â€“Â El-Guest
1 hour ago

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

Alright, should be good to go now @El-Guest
â€“Â JGibbers
1 hour ago

add a commentÂ |Â

up vote
4
down vote

favorite

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

edited 1 hour ago

asked 2 hours ago

JGibbers

75537

New contributor

A lot of Ã¢Â€ÂœlittleÃ¢Â€Â things add up to Ã¢Â€Âœa lotÃ¢Â€Â!! I love this alphametic!
â€“Â El-Guest
1 hour ago

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

Alright, should be good to go now @El-Guest
â€“Â JGibbers
1 hour ago

add a commentÂ |Â

up vote
4
down vote

favorite

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

edited 1 hour ago

asked 2 hours ago

JGibbers

75537

New contributor

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

edited 1 hour ago

asked 2 hours ago

JGibbers

75537

New contributor

edited 1 hour ago

asked 2 hours ago

JGibbers

75537

New contributor

edited 1 hour ago

asked 2 hours ago

JGibbers

75537

New contributor

asked 2 hours ago

JGibbers

75537

asked 2 hours ago

JGibbers

75537

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 Ã¢Â€ÂœlittleÃ¢Â€Â things add up to Ã¢Â€Âœa lotÃ¢Â€Â!! I love this alphametic!
â€“Â El-Guest
1 hour ago

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

Alright, should be good to go now @El-Guest
â€“Â JGibbers
1 hour ago

add a commentÂ |Â

A lot of Ã¢Â€ÂœlittleÃ¢Â€Â things add up to Ã¢Â€Âœa lotÃ¢Â€Â!! I love this alphametic!
â€“Â El-Guest
1 hour ago

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

Alright, should be good to go now @El-Guest
â€“Â JGibbers
1 hour ago

A lot of Ã¢Â€ÂœlittleÃ¢Â€Â things add up to Ã¢Â€Âœa lotÃ¢Â€Â!! I love this alphametic!
â€“Â El-Guest
1 hour ago

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

Alright, should be good to go now @El-Guest
â€“Â JGibbers
1 hour ago

add a commentÂ |Â

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.

edited 58 mins ago

answered 1 hour ago

Excited Raichu

1,484120

I had it backwards. His timestamp came later. My comment has been deleted.
â€“Â Brendon Shaw
52 mins ago

add a commentÂ |Â

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 canÃ¢Â€Â™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.

answered 1 hour ago

El-Guest

15.7k13475

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

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.

edited 58 mins ago

answered 1 hour ago

Excited Raichu

1,484120

I had it backwards. His timestamp came later. My comment has been deleted.
â€“Â Brendon Shaw
52 mins ago

add a commentÂ |Â

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.

edited 58 mins ago

answered 1 hour ago

Excited Raichu

1,484120

I had it backwards. His timestamp came later. My comment has been deleted.
â€“Â Brendon Shaw
52 mins ago

add a commentÂ |Â

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.

edited 58 mins ago

answered 1 hour ago

Excited Raichu

1,484120

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.

edited 58 mins ago

answered 1 hour ago

Excited Raichu

1,484120

edited 58 mins ago

answered 1 hour ago

Excited Raichu

1,484120

answered 1 hour ago

Excited Raichu

1,484120

answered 1 hour ago

Excited Raichu

1,484120

I had it backwards. His timestamp came later. My comment has been deleted.
â€“Â Brendon Shaw
52 mins ago

add a commentÂ |Â

I had it backwards. His timestamp came later. My comment has been deleted.
â€“Â Brendon Shaw
52 mins ago

I had it backwards. His timestamp came later. My comment has been deleted.
â€“Â Brendon Shaw
52 mins ago

add a commentÂ |Â

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 canÃ¢Â€Â™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.

answered 1 hour ago

El-Guest

15.7k13475

add a commentÂ |Â

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 canÃ¢Â€Â™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.

answered 1 hour ago

El-Guest

15.7k13475

add a commentÂ |Â

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 canÃ¢Â€Â™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.

answered 1 hour ago

El-Guest

15.7k13475

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 canÃ¢Â€Â™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.

answered 1 hour ago

El-Guest

15.7k13475

answered 1 hour ago

El-Guest

15.7k13475

answered 1 hour ago

El-Guest

15.7k13475

answered 1 hour ago

El-Guest

15.7k13475

add a commentÂ |Â

JGibbers is a new contributor. Be nice, and check out our Code of Conduct.

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