Seeing Should not be Believing

$\begin{aligned} 101^{113} & \equiv 101^{\color{#3D99F6}113 \text{ mod }\phi (113)} \text{ (mod 113)} & \small \color{#3D99F6} \text{Since }\gcd(101,113) = 1 \text{ Euler's theorem applies.} \\ & \equiv 101^{\color{#3D99F6}113 \text{ mod }112} \text{ (mod 113)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(113) = 112 \\ & \equiv 101^{\color{#3D99F6} 1} \equiv 101 \text{ (mod 113)} \end{aligned}$

$\begin{aligned} {\color{#3D99F6}100!} + 100 & \equiv {\color{#3D99F6} - 1} + 100 \text{ (mod 101)} & \small \color{#3D99F6} \text{By Wilson's theorem: }(p-1)! \equiv -1 \text{ (mod }p) \text{, where }p \text{ is a prime.} \\ & \equiv 99 \text{ (mod 101)} \end{aligned}$

$\implies 17(A+B) = 17(101+99) = \boxed{3400}$

References:

• Euler's theorem
• Wilson's theorem

By Fermat's Little Theorem, when a^p is divided by p, the remainder is a. Therefore A=101. Now, by Wilson's Theorem, when (p-1)! is divided by p, the remainder is-1. Hence, when (p-1)! +100 is divided by p, then the remainder would be -1+100=99. Thus B=99. Hence 17(A+B)=17(101+99)=17 200=3400.Note that ! stands for factorial and n!=1 2 3 .........*n.

By Fermat's Little Theorem 101^113 . cong. 101 (mod 113). Since 101 is a prime, by Wilson's Theorem, 100! + 1 .cong. 0 (mod 101), so 100! + 100 .cong. 99 (mod 101). So A + B = 99 +101 = 200, and 17(A + B) = 3400.