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

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The remainder when \(121^{103}\) is divided by 101 is calculated using Fermat's Little Theorem, which applies since 101 is a prime number. By simplifying \(121\) modulo \(101\), it is found that \(121 \equiv 20 \pmod{101}\). Further calculations show that \(121^{101} \equiv 20 \pmod{101}\) and \(121^2 \equiv 97 \pmod{101}\). Ultimately, combining these results leads to \(121^{103} \equiv 21 \pmod{101}\). Therefore, the final answer is that the remainder is 21.
maxkor
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What is the remainder of the division number $121^{103}$ by 101
 
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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.
 
maxkor said:
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 $$ 21 $$
 
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