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avatar+363 

(a) Find a number 0x<347 that solves the congruence 346x129(mod347).

 

(b) Find a number 0x<75 that solves the congruence 4x71(mod75).

 Jul 24, 2024
 #1
avatar+1952 
+1

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

 

x=1;346346(mod347)x=2;3462345(mod347)x=3;3463344(mod347)x=4;3464343(mod347)etc.

 

Thus, we have 

347x=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;346218129(mod347)

 

So 218 is the answer. 

 

Thanks! :)

 Jul 24, 2024
edited by NotThatSmart  Jul 24, 2024
 #2
avatar+1952 
+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=4x71, where y is another integer. 

 

Isolating x from this equation, we get

x=75y+714

 

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=75(3)+714=2964=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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