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# modulo help

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+347

(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
+1657
+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

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

Thanks! :)

Jul 24, 2024
edited by NotThatSmart  Jul 24, 2024
#2
+1657
+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