A friend of mine sent me a clip of a trick by Penn & Teller, and asked me to explain it.
Since it is mathematical, I decided to use it as one of this month's problems.
You are given 36 numbers arranged in a 6 x 6 square:
You are asked to pick six numbers, and after each number is picked, you remove the remaining numbers in each row and column.
Then you add the numbers together to get the sum.
No matter which numbers are picked, you always wind up with the same sum of 111.
Click here to see the trick
The board above is just an addition table where the thirty-six numbers are the sums.
The problem for you is to determine the values of the twelve numbers which created the table.
In other words, solve for the twelve variables: M, A, T, H, I, S, G, O, L, D, E, N.
For example, the number 15 is the sum of L and T, and the number 26 is the sum of E and A.
The trick works because it forces you to pick one of each of the twelve variables since you cannot pick any other numbers in the same row or column.
Click here to see an explanation of the trick with only sixteen numbers.
Solution to the Problem:
M = 1, A = 2, T = 3, H = 4, I = 5, S = 6,
G = 0,
0 = 6,
L = 12,
D = 18,
E = 24,
N = 30.
and 1 + 2 + 3 + 4 + 5 + 6 + 0 + 6 + 12 + 18 + 24 + 30 = 111.
Here is the completed addition table:
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Davit Banana sent in a different set of numbers which also work!!:
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Brijesh Dave sent in a different set of numbers which also work!!:
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Then, Dr. Kishan sent in his excellent work showing that the solution is not unique and giving a general solution:
From above table we have
G+M=1,
G+A=2,
. . .
G+S=6.
From these relations, we have
A=M+1, T=M+2, H=M+3, I=M+4, S=M+5.
If M=m then A=m+1, T=m+2, H=m+3, I=m+4, S=m+5 and
G=1-m, O=7-m, L=13-m, D=19-m, E=25-m, N=31-m.
This has not a unique solution. For m=0, we have
M, A, T, H, I, S=0, 1, 2, 3, 4, 5 and G, O, L, D, E, N=1, 7, 13, 19, 25, 31.
Similarly, we have
G+M=1,
O+M=7,
. . .
N+M=31.
From these relations, we have
O=G+6, L=G+12, D=G+18, E=G+24, N=G+30.
If G=n then O=n+6, L=n+12, D=n+18, E=n+24, N=n+30 and
M=1-n, A=2-n, T=3-n, H=4-n, I=5-n, S=6-n .
This does not have a unique solution. For n=0, we have
M, A, T, H, I, S=1, 2, 3, 4, 5, 6 and G, O, L, D, E, N=0, 6, 12, 18, 24, 30.
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