+355 (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 \(4x \equiv 71 \pmod{75}\).
+1892 (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! :)
+1892 (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! :)
