Maths Olympiad Prep

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Number theory Difficulty 5.1 AIME, harder Prove it United States

Problem:

The positive integer nn is such that the numbers 2n2^{n} and 5n5^{n} start with the same digit when written in decimal notation; determine this common leading digit.

Solution

Solution:

Answer: 33. Note 1=12<22<32<10<42<<92<102=1001 = 1^{2} < 2^{2} < 3^{2} < 10 < 4^{2} < \cdots < 9^{2} < 10^{2} = 100. Divide 2n2^{n} and 5n5^{n} by 1010 repeatedly until each is reduced to a decimal number less than 1010 but at least 11; call the resulting numbers xx and yy. Since (5n)(2n)=10n(5^{n})(2^{n}) = 10^{n}, either xy=1x y = 1 or xy=10x y = 10. Because 2n2^{n} and 5n5^{n} begin with the same digit, xx and yy are bounded by the same pair of adjacent integers. It follows that either x=y=1x = y = 1 or 3x,y<43 \leq x, y < 4. Because nn is positive, neither 2n2^{n} nor 5n5^{n} is a perfect power of 1010, so the former is impossible.

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