[Mailbag] What Do You Do with the Ideas You Used to Call “Mistakes”

Guillaume ParĂ©, in the very interesting comments of my last post where I
urged us to reconsider mistakes: I do agree with what is written, but I am
still w...
1 week ago
The answer is 02.
ReplyDeleteNote that (sqrt(3)+sqrt(2))^2 = 5+sqrt(24).
If n is a positive integer then (5+sqrt(24))^n + (5sqrt(24))^n is an integer, because the terms involving radicals cancel in the binomial expansion. Since (5sqrt(24))^n is less than 1/2, it follows that (5+sqrt(24))^n + (5sqrt(24))^n is the integer nearest to (5+sqrt(24))^n.
Let n = (10^100)/2. In the expansion of (5+sqrt(24))^n + (5sqrt(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.