Monday, October 5, 2009

Last two digits of (sqrt(2)+sqrt(3))^(10^100)

Find the last two digits of the integer closest to (sqrt(2)+sqrt(3))^(10^100).
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  1. DavidOctober 8, 2009 at 4:17 PM

    The answer is 02.

    Note that (sqrt(3)+sqrt(2))^2 = 5+sqrt(24).

    If n is a positive integer then (5+sqrt(24))^n + (5-sqrt(24))^n is an integer, because the terms involving radicals cancel in the binomial expansion. Since (5-sqrt(24))^n is less than 1/2, it follows that (5+sqrt(24))^n + (5-sqrt(24))^n is the integer nearest to (5+sqrt(24))^n.

    Let n = (10^100)/2. In the expansion of (5+sqrt(24))^n + (5-sqrt(24))^n, all terms are multiples of 100, except for the first term 2*5^n and the last term 2*24^(n/2)

    It is easy to check that 5^k = 25 (mod 100) for all k>1, and 24^k = 76 (mod 100) for all even k>1. Therefore, our integer is congruent to 2*25+76 modulo 100, so the last two digits are 02.

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