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And 299999999 is divisible by 13 (lemmy.world)

submitted 3 weeks ago by m4m4m4m4 to c/[email protected]

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[–] [email protected] 1 points 3 weeks ago

@m4m4m4m4

Caught me… A pretty quick way to see this and the title is using Fermat’s little theorem which states that k^(p-1) ~ 1 mod p for nonzero k. Using this we can write

3*10^5 ~ 3^6 ~ 1 mod 7
and
3*10^8 ~ 3*(-3)^8 ~ 3^9 ~ 16^9 ~ 2^36 ~ (2^12)^3 ~ 1^3 ~ 1 mod 13