6
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Every letter a decimal digit, different letters for different digits:

    BASE
  + BALL
 --------
   GAMES

Which digit does each letter represent?

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  • 1
    $\begingroup$ I saw a similar puzzle like this: send + more = money. The g has to be more than or equal to one. $\endgroup$ – Daedric Jan 25 '16 at 18:42
  • $\begingroup$ this whole thing seems to originate from: quora.com/… and the answer is probably copied from there aswell, quite similar.... suspicious $\endgroup$ – Daedric Jan 25 '16 at 18:55
  • $\begingroup$ @Daedric more than? It has to be 1 $\endgroup$ – Beastly Gerbil Jan 25 '16 at 18:56
  • $\begingroup$ I changed it for some reason, it originally said equal to one, I had a mental break down and changed my mind lol. $\endgroup$ – Daedric Jan 25 '16 at 18:57
  • $\begingroup$ @Daedric my answer starts similarly to that one, but I considered possible values for A while they tried values for SE and B. $\endgroup$ – f'' Jan 25 '16 at 19:01
12
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Assuming numbers can't start with 0, G is 1 because two four-digit numbers can't sum to 20000 or more.

SE+LL=ES or 1ES.

If it is ES, then LL must be a multiple of 9 because SE and ES are always congruent mod 9. But LL is a multiple of 11, so it would have to be 99, which is impossible.

So SE+LL=1ES. LL must be congruent to 100 mod 9. The only multiple of 11 that works is 55, so L is 5.

SE+55=1ES. This is possible when E+5=S. The possibilities for ES are 27, 38, or 49.

BA+BA+1=1AM. B must be at least 5 because B+B (possibly +1 from a carry) is at least 10.

If A is less than 5, then A+A+1 does not carry, and A must be even. Inversely, if A is greater than 5, it must be odd. The possibilities for A are 0, 2, 4, 7, or 9.

  • 0 does not work because M would have to be 1.
  • 2 and 7 don't work because M would have to be 5.
  • 9 doesn't work because M would also have to be 9.

So A is 4, M is 9, and B is 7. This leaves 38 as the only possibility for ES. The full equation is:

    7483
  + 7455
 --------
   14938
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  • $\begingroup$ Yeah, that's the answer I got by just fiddling with numbers. $\endgroup$ – Ian MacDonald Jan 25 '16 at 18:53
5
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There are in total three solutions (with unique numbers):

G A M E S B L
-------------
0 4 9 1 6 2 5
0 4 9 3 8 2 5
1 4 9 3 8 7 5

So apart from the solution in @f''s answer, we have

  2483
+ 2455
--------
 04938

and

  2461
+ 2455
--------
 04916
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2
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I found an answer by looking at the lowest numbers:

E + L must equal S or 10 + S

S + L must equal E or 10 + E

I set L to zero which means that there won't be a one to carry. This allows me to set E = S ("es", not "five"). Let's pick a number:

E = 5
S = E = 5
L = 0

This gives:

  BA55
+ BA00
------
 GAM55

Let's pick a number for A. This also gives us M. Note that if 2*A is < 10, A has to be even because it equals 2B - 10. If instead 2*A >= 10, the one carries over and A has to be odd because 2B + 1 - 10 has to equal A.

A = 2
M = 4

This gives:

  B255
+ B200
------
 G2455

Now I see that B + B has to equal 12. B = 6:

  6255
+ 6200
------
 12455

Using this method, I also found these results:

B = 7, A = 5, S = 5, L = 0, G = 1, E = 5, and M = 0
B = 8, A = 7, S = 1, L = 0, G = 1, E = 1, and M = 4

This method won't give every result because I'm assuming that L is always zero.

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  • $\begingroup$ It does say "different letters for different digits", which I'm taking to mean that E is not equal to S as you have here. $\endgroup$ – Duncan Jan 25 '16 at 22:58

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