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In the cryptarithm shown, different letters represent different digits. if two letters are the same, they represent the same digit. When A=5 and O=4, what is the least value that WIN couldd be?

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In the cryptarithm shown, different letters represent different digits. if two letters-example-1
User Abuybuy
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1 Answer

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Given A = 5 and O = 4, we have

I + A + O = I + 9

and the only way for I + 9 ≡ I (mod 10) is if a 1 is carried over from the sum in the units columns, since that would make

I + 9 + 1 ≡ I + 10 ≡ I (mod 10)

which means

10 ≤ 2C + E ≤ 19

Now, I ∈ {0, 1, 2, 3, 6, 7, 8, 9}, so we'll have to carry another 1 into the next column's sum.

I + A + O + 1 = I + 10 ===> 3T + 1 = W

This tells us that

3T = W - 1

i.e. W - 1 must be a multiple of 3. Then either W = 4 or W = 7, but the first case is dropped because we already have O = 4. So we know W = 7 and T = 2.

From here, I think it's a matter of checking all the possible cases. Work your way up from the smallest value for I, and determine what possible values N can take.

• Suppose I = 0. Then N ∈ {1, 3, 6, 8, 9}.

•• If N = 1, then 2C + E = 11. Consider the possible integer partitions of 11 that involve an even number (the contribution of 2C):

10 + 1 ===> C = 5, E = 1

8 + 3 ===> C = 4, E = 3

6 + 5 ===> C = 3, E = 5

4 + 7 ===> C = 2, E = 7

2 + 9 ===> C = 1, E = 9

None of these options work, since each of {1, 2, 4, 5, 7} are taken.

•• If N = 3, then 2C + E = 13, which means

12 + 1 ===> C = 6, E = 1 (bingo!)

10 + 3 ===> C = 5, E = 3

8 + 5 ===> C = 4, E = 5

6 + 7 ===> C = 3, E = 7

4 + 9 ===> C = 2, E = 9

The other options don't work, since each of {2, 3, 4, 5, 7} are taken.

So, the least value is WIN = 703 :

TIC = 206

TAC = 256

TOE = 241

TIC + TAC + TOE = 703

User Mosch
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