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 
)
+6250 \(8x + 1 \equiv 5 \pmod{12}\\ 8x \equiv 4 \pmod{12}\\ 8x = 12k + 4,~k \in \mathbb{Z}\\ 2x = 3k + 1,~k \in \mathbb{Z}\\ 2x \equiv 1 \pmod{3} \\ \text{now the tricky part}\\ 2^{-1} \pmod{3} = 2\\ x \equiv 2 \pmod{3}\\ \text{the problem states }a<2, \text{ this is an error, it should be }a\leq 2\)
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