If $3x+7 \equiv$ (mod $16$), then $2x+11$ is congruent (mod $16$) to what integer between $0$ and $15$, inclusive?

P.S The answer is NOT $11$, I already tried that.

Guest Jun 17, 2021

#1**0 **

Also, I've been wondering if the answer is $5$ because when we add $2$ to both sides to get rid of the remainder, we get

1. $3x + 9 \equiv 0$ (mod $16$)

2. $3 \cdot 13 + 9 = 48$ which is divisible by $16$ and when $x = 13$ this is the smallest possible value for $x$ that satisfies this. If we then apply this value to $2x + 11$ we get

3. $2 \cdot 13 + 11 = 37$

4. If we divide $37$ by $16$ we get $2$ r $5$. So, I'm wondering if the answer to this is $5.$

Am I correct?

P.S BTW I'm the same guest.

Guest Jun 17, 2021