SuprDooprPoopr
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#1
Report Thread starter 6 months ago
#1
777x ≡ 1312 (mod 2021).
In each case, express your answer in the form x ≡ a (mod n) with
0 ≤ a < n.

I have no idea how to tackle this question, maybe x ≡ 1312/777 (mod 2021)?
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ish101
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#2
Report 6 months ago
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basically find the multiplicative inverse $k$, that is the number such that 771kx\equiv 1\times x \pmod{2021}, so 771k\equiv 1\pmod {2021}\implies 771k+2021m=1 which has a solution by Bezouts identity as \gcd(777,2021)=1. once you've found k, just multiply your congruence by k on both sides and you'll end up with x\equiv 1312k\pmod {2021}, calculate 1312k and reduce that modulo 2021
Last edited by ish101; 6 months ago
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DFranklin
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#3
Report 6 months ago
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(Original post by SuprDooprPoopr)
777x ≡ 1312 (mod 2021).
In each case, express your answer in the form x ≡ a (mod n) with
0 ≤ a < n.

I have no idea how to tackle this question, maybe x ≡ 1312/777 (mod 2021)?
Can you explain why it is that you're attempting a question you have no idea how to tackle? What is it you're studying?
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SuprDooprPoopr
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#4
Report Thread starter 5 months ago
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(Original post by DFranklin)
Can you explain why it is that you're attempting a question you have no idea how to tackle? What is it you're studying?
nah I know how to do it, I was being silly.
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