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(a) Find a number \(0 \le x < 347\) that solves the congruence \(346x \equiv 129 \pmod{347}\).

 

(b) Find a number \(0 \le x < 75\) that solves the congruence \(4x \equiv 71 \pmod{75}\).

 Jul 24, 2024
 #1
avatar+1868 
+1

(a) Let's note a pattern in the equation. 

 

\(x=1; 346 \equiv 346\pmod{347}\\ x=2; 346*2 \equiv 345\pmod{347}\\ x=3; 346*3 \equiv 344 \pmod{347}\\ x=4; 346*4 \equiv 343 \pmod{347}\\ etc. \)

 

Thus, we have 

\(347-x=129\) as the pattern for this congruence. 

 

thus, solving for x, we find that \(x=218\)

 

Testing if this is indeed true, we find that 

\(x=218; 346*218 \equiv 129 \pmod{347}\)

 

So 218 is the answer. 

 

Thanks! :)

 Jul 24, 2024
edited by NotThatSmart  Jul 24, 2024
 #2
avatar+1868 
+1

(b) Let's use a different tactic here. 

We want the fact that 4x subtracted by the remainder is divisble by 75. 

We can have the equation \(75y=4x-71\), where y is another integer. 

 

Isolating x from this equation, we get

\(x=\frac{75y+71}{4}\)

 

Now, any value of y we plug in should work. We just need an integer, and we're done. 

When we plug in \(y=3\), we find that

\(x=\frac{75(3)+71}{4}=\frac{296}{4}=74\)

 

Thus, 74 is our answer. Just fit it into the interval! Lol

 

Thanks! :)

 Jul 24, 2024
edited by NotThatSmart  Jul 24, 2024

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