What is the Remainder When \(121^{103}\) is Divided by 101?

[maxkor]

What is the remainder of the division number $121^{103}$ by 101

 

[GJA]

Hi maxkor,

Here are two quick hints to point you in the right direction:

1) Note that 101 is prime

2) Fermat's Little Theorem might be useful here

Fermat's little theorem - Wikipedia, the free encyclopedia

See what you can come up with using these notes. Let me know if anything is still unclear/not quite right.

 

[Mathick]

What is the remainder of the division number $121^{103}$ by 101

Hi maxkor,

Fermat's little theorem states that if $$p$$ is a prime number, then for any integer $$a$$, the number $$a^p − a$$ is an integer multiple of $$p$$. In the notation of modular arithmetic, this is expressed as

$$ a^p \equiv a \pmod p. $$

Therefore, we have that

$$ 121^{101}\; \equiv \;121\; \equiv \;20\; (mod \;101) $$
$$ 121\; \equiv \;20\; (mod \;101) $$
$$ 121^{2}\; \equiv \;20^{2} \;\equiv\; 400 \;\equiv \;97 \;(mod \; 101) $$
$$ 121^{103}\; \equiv \;121^{2}\; \cdot\; 121^{101}\; \equiv \;97 \;\cdot \;20\; \equiv \;1940\; \equiv \;21\; (mod \;101) $$

The remainder of the division number $$ 121^{103} $$ by $$ 101 $$ is $$ 21 $$