Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Find the answer

In the base 10 arithmetic problem HMMT+GUTS=ROUNDH M M T+G U T S=R O U N D, each distinct letter represents a different digit, and leading zeroes are not allowed. What is the maximum possible value of ROUNDR O U N D?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Clearly R=1R=1, and from the hundreds column, M=0M=0 or 9. Since H+G=9+OH+G=9+O or 10+O10+O, it is easy to see that OO can be at most 7, in which case HH and GG must be 8 and 9, so M=0M=0. But because of the tens column, we must have S+T10S+T \geq 10, and in fact since DD cannot be 0 or 1,S+T121, S+T \geq 12, which is impossible given the remaining choices. Therefore, OO is at most 6. Suppose O=6O=6 and M=9M=9. Then we must have HH and GG be 7 and 8. With the remaining digits 0,2,3,40,2,3,4, and 5, we must have in the ones column that TT and SS are 2 and 3, which leaves no possibility for NN. If instead M=0M=0, then HH and GG are 7 and 9. Since again S+T12S+T \geq 12 and N=T+1N=T+1, the only possibility is S=8,T=4S=8, T=4, and N=5N=5, giving ROUND=16352=7004+9348=9004+7348R O U N D=16352=7004+9348=9004+7348.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.