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'Weekly Problem 24 - 2009' printed from http://nrich.maths.org/

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$UK$ means $10U + K$ and $SMC$ means $100S + 10M + C$, so we have $$10U+K+4=100S+10M+C$$ The left hand side is at most $$ 10 \times 9 + 8 + 4 = 102$$ so $$100S+10M+C \leq 102$$ Therefore $S \leq 1$, so $S=1$ (since it can't be zero). So $$10M+C \leq2$$ So $M=0$

$M$ has the lowest value.

This problem is taken from the UKMT Mathematical Challenges.

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