Let n = 20²⁶ . What is the remainder when 49^n + 41^n + 10n is divided by 100?

Step-by-step solution

Given n = 20²⁶ . We need to find the remainder when 49^n + 41^n + 10n is divided by 100 . First, analyze the value of n modulo 100 : n = 20²⁶ = (20^2)¹³ = 400¹³ . Since 400 is a multiple of 100 , 400¹³ 0 100 . Thus, n 0 100 . This implies 10n 0 100 . Next, evaluate 49^n 100 : 49^n = (50 - 1)^n = ^ n C₀ 50^n - ^ n C₁ 50^ n-1 + - ^ n C_ n-1 50 + ^ n C_ n (-1)^n . Since n = 20²⁶ is an even integer, (-1)^n = 1 . Also, n is a multiple of 100 , so n = 100k for some integer k . The term ^ n C_ n-1 50 = 50n = 50(100k) = 50