+0

0
42
1

The solution of $$8x+1\equiv 5\pmod{12}$$ is $$x\equiv a\pmod{m}$$ for some positive integers $$m\ge 2$$ and $$a<2$$. Find $$a+m$$.

Hi ya'll, I got 17, because 5 can be a solution for x (actual solution is 3n+2 for some random integer n).

If you plug that in, you get

41 ≡ 5 (mod 12), which works because 41 (mod 12) = 5.

That means a=5 and m=12.

$$12\ge 2$$ and $$5<12$$.

5+12=17, so why is this wrong?  (I checked and it said it was wrong )

Dec 19, 2018

#1
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$$8x + 1 \equiv 5 \pmod{12}\\ 8x \equiv 4 \pmod{12}\\ 8x = 12k + 4,~k \in \mathbb{Z}\\ 2x = 3k + 1,~k \in \mathbb{Z}\\ 2x \equiv 1 \pmod{3} \\ \text{now the tricky part}\\ 2^{-1} \pmod{3} = 2\\ x \equiv 2 \pmod{3}\\ \text{the problem states }a<2, \text{ this is an error, it should be }a\leq 2$$

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Dec 19, 2018