Crypt.2018.1

Algebra Level 3

S E N D + M O R E M O N E Y \large{\begin{array}{ccccccc} && & & & S & E & N & D\\ && + & & & M & O & R & E\\ \hline && & & M & O & N & E & Y\\ \end{array}}

For the above alphametic problem in base 10, there is only one probable solution. Each letter represents different digit. Whats the value of S E N D M O R E SEND - MORE ?


The answer is 8482.

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1 solution

Marta Reece
Jan 25, 2018

M = 1 M=1

S + 1 = 10 + O S+1=10+O

Two options S = 9 , O = 0 S=9, O=0 which works

S = 8 , 0 = 2 S=8, 0=2 which does not

E + 0 = n E+0=n no way, therefore E + 1 = N E+1=N

N + R = 10 + E = 10 + N 1 , R = 9 N+R=10+E=10+N-1, R=9 no way, therefore N + R + 1 = 10 + E = 10 + N 1 , R = 8 N+R+1=10+E=10+N-1, R=8

D + E = 10 + Y D+E=10+Y but E has to have N above it and only digits up to 7 are left, so D = 7 , Y = 2 , E = 5 , N = 6 D=7, Y=2, E=5, N=6

This is an old problem. It's a good one and I am glad to see it, but the author might want to specify that it is not original with him.

Marta Reece - 3 years, 4 months ago

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Yup, this one is dated back to 1924, I guess. In 1924, July Henry Dudeney published a classic example in the Strand Magazine associating with the story of a kidnapper’s ransom demand. This doesn't belong to me, for sure.

Sayan Chaudhuri - 3 years, 4 months ago

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