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Greedy for Cash

Fit numbers on a grid with required sums for squaresSPEND LESS MONEY alphameticNot the “SEND MORE MONEY” alphameticDecipher this puzzleHow can my friend take the item for only $3,000?Find the numbersAdding up numbers in Portuguese is strangeFind the values of U, V, C based on the given relationship…useful for upcoming puzzles

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

1

$begingroup$

Solve this Cryptarithm

 GIVE
 + MORE
 ---------
 MONEY

_{Referenced from the Britannica Encyclopaedia No. 25 - "Number Prague"}

mathematics alphametic

share|improve this question

edited 4 hours ago

Bass

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1

$begingroup$

Solve this Cryptarithm

 GIVE
 + MORE
 ---------
 MONEY

_{Referenced from the Britannica Encyclopaedia No. 25 - "Number Prague"}

mathematics alphametic

share|improve this question

edited 4 hours ago

Bass

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asked 8 hours ago

Quark-epochQuark-epoch

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$endgroup$

add a comment | 

$begingroup$

Solve this Cryptarithm

 GIVE
 + MORE
 ---------
 MONEY

_{Referenced from the Britannica Encyclopaedia No. 25 - "Number Prague"}

mathematics alphametic

share|improve this question

edited 4 hours ago

Bass

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asked 8 hours ago

Quark-epochQuark-epoch

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$endgroup$

 GIVE
 + MORE
 ---------
 MONEY

share|improve this question

edited 4 hours ago

Bass

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asked 8 hours ago

Quark-epochQuark-epoch

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share|improve this question

edited 4 hours ago

Bass

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asked 8 hours ago

Quark-epochQuark-epoch

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edited 4 hours ago

Bass

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edited 4 hours ago

Bass

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asked 8 hours ago

Quark-epochQuark-epoch

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2 Answers
2

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3

$begingroup$

Looks like the solution isn't unique.

 9486
 + 1076
 ---------
 10562
 

I could also be 3 if N is 4, and obviously V and R can be switched.

If, on the other hand, the puzzle had been SEND + MORE = MONEY, the puzzle would have been well known, the solution would have been unique, and the phrase more common.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

BassBass

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$endgroup$

add a comment | 

2

$begingroup$

It's clear that the V and the R are interchangeable so solutions are in pairs.

Here's a solution:

9376+1086 = 10462 or 9386+1076 = 10462

Here's another:

9476+1086 = 10562 or 9486+1076 = 10562

I believe these are the only solutions:

1. M is clearly 1.

2. O is therefore 0. It can't be 1 because it would then be a repeat and it can't be more than 1 because you can't get to 12 when you're adding M=1 to G.
3. G is therefore 8 or 9 with 1 or 0 carried.

4. I + 0 = N, so there must be a 1 carried in the I,O column and N=I+1. Which means we'll be looking for a solution that leaves two adjacent numbers unaccounted for.

5. Because of the carry, the V,R column must be bigger than 10. So the V,R column must come to 12,13,14,15,16,17.

We can then deal with these cases:

12: E=2 (so no carry). VR=(3,9),(4,8),(5,7)

We can eliminate (4,8) because Y will be 4. (3,9) and (5,7) don't work because G would have to be 8 in the former and you won't get the carry and you won't get two adjacent numbers with the latter.

Similar arguments eliminate other options until only:

16: E=6, VR=(7,8). There are then 3,4,5 as the only adjacent numbers left over leading to the two pairs of solutions shown.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

Dr XorileDr Xorile

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add a comment | 

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3

$begingroup$

Looks like the solution isn't unique.

 9486
 + 1076
 ---------
 10562
 

I could also be 3 if N is 4, and obviously V and R can be switched.

If, on the other hand, the puzzle had been SEND + MORE = MONEY, the puzzle would have been well known, the solution would have been unique, and the phrase more common.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

BassBass

36.1k44 gold badges8989 silver badges209209 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Looks like the solution isn't unique.

 9486
 + 1076
 ---------
 10562
 

I could also be 3 if N is 4, and obviously V and R can be switched.

If, on the other hand, the puzzle had been SEND + MORE = MONEY, the puzzle would have been well known, the solution would have been unique, and the phrase more common.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

BassBass

36.1k44 gold badges8989 silver badges209209 bronze badges

$endgroup$

add a comment | 

$begingroup$

Looks like the solution isn't unique.

 9486
 + 1076
 ---------
 10562
 

I could also be 3 if N is 4, and obviously V and R can be switched.

If, on the other hand, the puzzle had been SEND + MORE = MONEY, the puzzle would have been well known, the solution would have been unique, and the phrase more common.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

BassBass

36.1k44 gold badges8989 silver badges209209 bronze badges

$endgroup$

share|improve this answer

edited 6 hours ago

answered 7 hours ago

BassBass

36.1k44 gold badges8989 silver badges209209 bronze badges

answered 7 hours ago

BassBass

36.1k44 gold badges8989 silver badges209209 bronze badges

answered 7 hours ago

BassBass

36.1k44 gold badges8989 silver badges209209 bronze badges

2

$begingroup$

It's clear that the V and the R are interchangeable so solutions are in pairs.

Here's a solution:

9376+1086 = 10462 or 9386+1076 = 10462

Here's another:

9476+1086 = 10562 or 9486+1076 = 10562

I believe these are the only solutions:

1. M is clearly 1.

2. O is therefore 0. It can't be 1 because it would then be a repeat and it can't be more than 1 because you can't get to 12 when you're adding M=1 to G.
3. G is therefore 8 or 9 with 1 or 0 carried.

4. I + 0 = N, so there must be a 1 carried in the I,O column and N=I+1. Which means we'll be looking for a solution that leaves two adjacent numbers unaccounted for.

5. Because of the carry, the V,R column must be bigger than 10. So the V,R column must come to 12,13,14,15,16,17.

We can then deal with these cases:

12: E=2 (so no carry). VR=(3,9),(4,8),(5,7)

We can eliminate (4,8) because Y will be 4. (3,9) and (5,7) don't work because G would have to be 8 in the former and you won't get the carry and you won't get two adjacent numbers with the latter.

Similar arguments eliminate other options until only:

16: E=6, VR=(7,8). There are then 3,4,5 as the only adjacent numbers left over leading to the two pairs of solutions shown.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

Dr XorileDr Xorile

14.7k33 gold badges3333 silver badges9090 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

It's clear that the V and the R are interchangeable so solutions are in pairs.

Here's a solution:

9376+1086 = 10462 or 9386+1076 = 10462

Here's another:

9476+1086 = 10562 or 9486+1076 = 10562

I believe these are the only solutions:

1. M is clearly 1.

2. O is therefore 0. It can't be 1 because it would then be a repeat and it can't be more than 1 because you can't get to 12 when you're adding M=1 to G.
3. G is therefore 8 or 9 with 1 or 0 carried.

4. I + 0 = N, so there must be a 1 carried in the I,O column and N=I+1. Which means we'll be looking for a solution that leaves two adjacent numbers unaccounted for.

5. Because of the carry, the V,R column must be bigger than 10. So the V,R column must come to 12,13,14,15,16,17.

We can then deal with these cases:

12: E=2 (so no carry). VR=(3,9),(4,8),(5,7)

We can eliminate (4,8) because Y will be 4. (3,9) and (5,7) don't work because G would have to be 8 in the former and you won't get the carry and you won't get two adjacent numbers with the latter.

Similar arguments eliminate other options until only:

16: E=6, VR=(7,8). There are then 3,4,5 as the only adjacent numbers left over leading to the two pairs of solutions shown.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

Dr XorileDr Xorile

14.7k33 gold badges3333 silver badges9090 bronze badges

$endgroup$

add a comment | 

$begingroup$

It's clear that the V and the R are interchangeable so solutions are in pairs.

Here's a solution:

9376+1086 = 10462 or 9386+1076 = 10462

Here's another:

9476+1086 = 10562 or 9486+1076 = 10562

I believe these are the only solutions:

1. M is clearly 1.

2. O is therefore 0. It can't be 1 because it would then be a repeat and it can't be more than 1 because you can't get to 12 when you're adding M=1 to G.
3. G is therefore 8 or 9 with 1 or 0 carried.

4. I + 0 = N, so there must be a 1 carried in the I,O column and N=I+1. Which means we'll be looking for a solution that leaves two adjacent numbers unaccounted for.

5. Because of the carry, the V,R column must be bigger than 10. So the V,R column must come to 12,13,14,15,16,17.

We can then deal with these cases:

12: E=2 (so no carry). VR=(3,9),(4,8),(5,7)

We can eliminate (4,8) because Y will be 4. (3,9) and (5,7) don't work because G would have to be 8 in the former and you won't get the carry and you won't get two adjacent numbers with the latter.

Similar arguments eliminate other options until only:

16: E=6, VR=(7,8). There are then 3,4,5 as the only adjacent numbers left over leading to the two pairs of solutions shown.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

Dr XorileDr Xorile

14.7k33 gold badges3333 silver badges9090 bronze badges

$endgroup$

share|improve this answer

edited 6 hours ago

answered 7 hours ago

Dr XorileDr Xorile

14.7k33 gold badges3333 silver badges9090 bronze badges

answered 7 hours ago

Dr XorileDr Xorile

14.7k33 gold badges3333 silver badges9090 bronze badges

answered 7 hours ago

Dr XorileDr Xorile

14.7k33 gold badges3333 silver badges9090 bronze badges