500 th Problem! 500 \text{th} ~ \text{Problem!}

Logic Level 5

M A S S + W E I G P H Y S \begin{array}{ccccc} & & & M& A &S&S\\ + & & & W& E & I &G \\ \hline & & & P & H &Y & S\\ \hline \end{array}

In the cryptogram above, each letter represents a distinct single digit non-negative integer. What is the largest possible value of P H Y S \overline{PHYS} ?


The answer is 9817.

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1 solution

Ivan Koswara
Nov 8, 2017

Looking at the units column, clearly G = 0 G = 0 and there is no carry to the tens column. Now ignore the units column, and remember that none of the digits may be 0 (since it's used for G G ).

There are 9 letters and 9 digits, so all digits get used. Adding up all the digits gives M + A + S + W + E + I + P + H + Y M+A+S+W+E+I+P+H+Y , which must be equal to 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 1+2+3+4+5+6+7+8+9 = 45 . Now consider the addition modulo 9. M A S + W E I + P H Y 45 ( m o d 9 ) \overline{MAS} + \overline{WEI} + \overline{PHY} \equiv 45 \pmod 9 , since we can obtain the remainder modulo 9 by adding up the digits. However, M A S + W E I = P H Y \overline{MAS} + \overline{WEI} = \overline{PHY} , so we have 2 P H Y 45 0 ( m o d 9 ) 2 \cdot \overline{PHY} \equiv 45 \equiv 0 \pmod 9 and so P H Y 0 ( m o d 9 ) \overline{PHY} \equiv 0 \pmod 9 . In other words, P H Y \overline{PHY} is divisible by 9. The largest possible number for P H Y \overline{PHY} is hence 981, and the largest possible S S after this is 7, giving max P H Y S = 9817 \max \overline{PHYS} = \boxed{9817} . This is achievable, by for example 3277 + 6540 = 9817 3277 + 6540 = 9817 .

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