$\left.\begin{array}{l}{\left[\begin{array}{l}\text { Riddles }\end{array}\right]} \\ \text { [ CaseAnalysis }\end{array}\right]$

Authors: Tokoreve S.I. Khachatryan A.B.

Find the smallest four-digit number СЕЕМ for which there is a solution to the riddle МЫ + РОЖЬ = СЕЕМ. (Identical letters correspond to identical digits, different letters correspond to different digits.)

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With $C > P$ and in particular, $C > 1$.

## Solution

Since $C > P$, then $C > 1$. As we are looking for the smallest number, let's try $P=1, C=2$, and $E=0$. Then $M=3$. The case $CEEM = 2003$ is possible: $35 + 1968 = 2003$ or $38 + 1965 = 2003$.

In addition to the solutions mentioned, the puzzle has 38 more solutions (this can be verified, for example, using a computer program):

$31 + 4972 = 5003, 32 + 4971 = 5003, 31 + 5972 = 6003, 32 + 5971 = 6003, 81 + 3927 = 4008, 87 + 3921 = 4008, 61 + 2945 = 3006, 65 + 2941 = 3006, 81 + 4927 = 5008, 14 + 2987 = 3001, 17 + 2984 = 3001, 87 + 4921 = 5008, 15 + 2986 = 3001, 16 + 2985 = 3001, 81 + 5927 = 6008, 87 + 5921 = 6008, 41 + 7963 = 8004, 71 + 4936 = 5007, 43 + 7961 = 8004, 76 + 4931 = 5007, 15 + 3986 = 4001, 16 + 3985 = 4001, 61 + 7945 = 8006, 65 + 7941 = 8006, 46 + 1958 = 2004, 48 + 1956 = 2004, 14 + 5987 = 6001, 17 + 5984 = 6001, 57 + 1948 = 2005, 58 + 1947 = 2005, 83 + 6925 = 7008, 85 + 6923 = 7008, 46 + 2958 = 3004, 48 + 2956 = 3004, 24 + 5978 = 6002, 57 + 2948 = 3005, 28 + 5974 = 6002, 58 + 2947 = 3005$.

## Answer

2003. For example, $35 + 1968 = 2003$ or $38 + 1965 = 2003$.