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• 0 ## Question In the following alphametic equation, each of the capital letters denotes a different base ten digit from 0 to 9. None of the numbers can contain any leading zero.

CARRIE + FISHER + 1956 + PP = PPPPPP

Determine the number represented by "SHAPER".

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

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• 0 SHAPER = 901786

P=7, H=0, R=6, E=8, I=5, S=9, A=1, (C,F)=(3,4) or (4,3)

I probably did this the long, hard way, but here goes:

I first realized that E+R+6 must equal 10 or 20, so that the result + P would give us P in the answer. I quickly eliminated the "= 10" solution by looking at the next column. If you carry over a 1, then you will have to solve the same exact equation as the previous column, I+E+6 = 10 or 20. Since I != R, this cannot be possible. So all possible (E,R) pairs are: (5,9), (6,8), (8,6), (9,5). The next column is the same equation. You can eliminate the "= 10" possibility again since E cannot be less than 5. So with this new knowledge, all possible (E,R,I) triplets are: (5,9,8), (6,8,7), (8,6,5), (9,5,4).

Since we know that we are getting 2 carried over from the last column, the next columns equation becomes: R+H+11=X+P, where X is 0, 10, or 20. X cannot be 0 because any positive integer + 11 will give you a number greater than 11. Using substitution with the above triplets, you can see that if R=9, the equation for this column cannot be satisfied. Therefore, we can eliminate the (5,9,8) triplet.

Now here is where a little bit of trial and error came in. I decided to make a list of all possible (P,H) pairs given a value for R. The only possible values for R are 5, 6, and 8. So here are the pairs:

When X = 10 and R = 5, the possible (P,H) pairs are: (6,0) (7,1) (8,2) (9,3).

When X = 10 and R = 6, the possible (P,H) pairs are: (7,0) (8,1) (9,2).

When X = 10 and R = 8, the possible (P,H) pairs are: (9,0).

When X = 20 and R = 5, the possible (P,H) pairs are: (1,5) (2,6) (3,7) (4,8) (5,9).

When X = 20 and R = 6, the possible (P,H) pairs are: (1,4) (2,5) (3,6) (4,7) (5,8) (6,9).

When X = 20 and R = 8, the possible (P,H) pairs are: (1,2) (2,3) (3,4) (4,5) (5,6) (6,7) (7,8) (8,9).

The ones with the strikethrough were eliminated because they suggest that either P or H is the same value as one of the values of the triplet. Now for more trial and error. I noticed that there is only one solution if P=6, so I took all the information I know for when P=6 and plugged it all back in to the original equation:

P=6, H=0, R=5, E=9, I=4

CA5549

F4S095

1956

66

------

666666

I verified that the right three columns checked out (which they do), and got a carry-over of 1 for the next column. So, 1+5+S+1 = 6. The only possible solution for S is 0. But since H already is 0, this is NOT the solution.

Then I noticed that there is also only one solution for if P=8. I similarly eliminated this solution when A had to be 4, and I was already 4.

When P=7, there are two potential solutions. Using the (7,1) pair, I came up with the solution of (7,1,5,9,4,0,3) for (P,H,R,E,I,S,A), respectively. I could have just stopped there and given the answer of SHAPER = 013795, but for two reasons, this is incorrect. First, the original problem stated that none of the numbers had a leading 0. Secondly, even if we assumed that the previous statement doesn't apply to SHAPER, what are the possible solutions for C and F? Well, the remaining numbers that are not assigned to a letter are 2, 6, and 8. There is no combination of two of these three numbers that add to 7.

Using the (7,0) pair gave the following equation and proved to be the right answer:

P=7, H=0, R=6, E=8, I=5

CA6658

F5S086

1956

77

------

777777

S=9, A=1, (C,F)=(3,4) or (4,3)

Sorry about the formatting, I couldn't figure out how to conserve it.

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• 0 In the following alphametic equation, each of the capital letters denotes a different base ten digit from 0 to 9. None of the numbers can contain any leading zero.

CARRIE + FISHER + 1956 + PP = PPPPPP

Determine the number represented by "SHAPER".

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

It took me an hour...

901786

pivoting point is Letter I.

E+R+6+P=P

I+E+5+P+2=P

R+H+9+2=P

R+S+1+1=P

A+I+1=P

C+F=P

C+F < 5

E+R = 14 (from E+R+6+P=P+20)

C,F,P and S can't be 0.

start with the candidates for P from 9 to 0 first, after computing second line, check with 5th line if the result is possible, if not, go to the next smaller candidate and so on.

P=7 solves the equation.

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