(a) Find a number 0≤x<347 that solves the congruence 346x≡129(mod347).
(b) Find a number 0≤x<75 that solves the congruence 4x≡71(mod75).
(a) Let's note a pattern in the equation.
x=1;346≡346(mod347)x=2;346∗2≡345(mod347)x=3;346∗3≡344(mod347)x=4;346∗4≡343(mod347)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≡129(mod347)
So 218 is the answer.
Thanks! :)
(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=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! :)