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1. How many positive integers N from 1 to 5000 satisfy the congruence \(N \equiv 11 \pmod{13}\)?

 

2. How many positive integers N from 1 to 5000 satisfy both congruences, \( N\equiv 5\pmod{12}\) and \( N\equiv 11\pmod{13}\)?

 Dec 17, 2020
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1 - 

N mod 13 = 11

N = 13m  +  11, where m=0, 1, 2, 3.........etc.

5000 / 13 + 1 =385 integers that satisfy the congruence.

 

 

2 -

N mod 13 = 11

N mod 12 = 5, solve for N

 

Using "Chinese Remainder Theorem + Modular Multiplicative Inverse", we have:

N = 156m  +  89, where m =0, 1, 2, 3........etc.

5000 / 156 + 1 =33 integers that satisfy both congruences.

 Dec 17, 2020
edited by Guest  Dec 17, 2020

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