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by Strong Eagle » Tue, 23 Aug 2005 10:39 am
The trick to this problem is to realize that the largest number that can be carried to the next column is 1. For example, since there are no two numbers that are the same being added together, the largest number that can be added up is 9 + 8 = 17, and with a carry from the previous column, the largest number is 18.
So, this means that A = 1 because S + something had to carry and 1 is the only thing that can carry.
Now, the only number that S can be to force a carry is 9, again because only 1 can be carried from the last column. And if S = 9, then M = 0 because 9 + 1 = 10.
So to recap to this point:
A = 1
S = 9
M = 0
So now we look at the right hand column, Y + S = A. From the numbers we have already, Y + 9 = 1, but since it can't be 1, it must be 11, so Y = 2 and we carry 1 to the next column.
So, now we have:
Y = 2
A = 1
S = 9
M = 0
4 numbers used up and 6 left to go.
Let's go back over to the left column where F + T = E. This has to be greater than 10 in order to carry to add 1 to S to get the A. Actually, since we have already determined 0, 1, and 2, it must be equal to or greater than 13, and E must be equal to or greater than 3.
Leaving out the four numbers already figured out, the only combinations for F + T that is greater than or equal to 13 are 5+7 (with a carry), 5+8, 6+7, 6+8, and 7+8.
Now it becomes a case of plugging in the combinations to see what works. We take a look at the second column from the right. With the carry, we know that T + E + 1 = C. So, we start using the pairs of numbers to get answers for E and C.
We start with F = 5 and T = 7 and assume a carry. Then E = F + T + 1 = 3. We plug that into T + E + 1 = C and get 7 + 3 + 1 = 11, or C = 1, but C can’t be 1 because we already know that A = 1. If we make F = 7 and T = 5 and try again, we get C = 9, and that won’t work either since we already know S = 9.
So, we try with F = 5 and T = 8. This time E = 3 or 4 with a carry. Then T + E + 1 = C becomes 8 + 3 (or 4) + 1 = 12 or 13 (C = 2 or 3). We know C cannot be 2 because Y = 2. So, maybe E = 4 and C = 3.
We move to the next column which is F + T = I. We know we have a carry so it is really F + T + 1 = I. Let’s plug in the numbers. 5 + 8 + 1 = 14, or I = 4. Whoops. I can’t be 4 because we already have E = 4.
We go back and check out the next set of numbers, 6 and 7. Turns out they don’t work either. The only set of numbers for F and T that work are 6 and 8. Once you use 6 and 8 for F and T it is easy to calculate the remainder of the puzzle and you get:
FIFTY
65682
STATES
981849
AMERICA
1047531
F = 6
I = 5
T = 8
Y = 2
S = 9
A = 1
E = 4
M = 0
R = 7
C = 3