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Algebra Difficulty 4.5 AIME Prove it Ireland

Prove that if nn is a positive integer either 3n3n or 7n7n contains an odd digit.

Solution

Suppose a number nn exists with the property that 3n3n and 7n7n have only even digits and let NN be the smallest such. If NN is divisible by 1010, N10\frac{N}{10} is a smaller such number. Hence NN ends in one of 2,4,6,82, 4, 6, 8. But 3N+7N=10N3N + 7N = 10N and 10N10N has the tens digit odd as a result of the carry from the unit digits. But this implies NN ends in an odd digit which is impossible.

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