tag:blogger.com,1999:blog-1974366466896776665.post3329004883486708620..comments2022-03-27T05:59:34.842-05:00Comments on Math Problem of the Day: Last two digits of (sqrt(2)+sqrt(3))^(10^100)Davidhttp://www.blogger.com/profile/09232747857608296294noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-1974366466896776665.post-43419785629173131442009-10-08T16:17:05.483-05:002009-10-08T16:17:05.483-05:00The answer is 02.
Note that (sqrt(3)+sqrt(2))^2 =...The answer is 02.<br /><br />Note that (sqrt(3)+sqrt(2))^2 = 5+sqrt(24).<br /><br />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.<br /><br />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)<br /><br />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.Davidhttps://www.blogger.com/profile/09232747857608296294noreply@blogger.com