+0

0
94
3

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.

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.

Jun 17, 2021
#2
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n mod 16   =   19 mod 16   =  3

3x  +  7 =   2x  + 11

x   =   4

3(4)  +  7  = 19  ( mod 16 )    =   3 mod 16

2(4)  + 11  =  19    (mod 16 )   =  3 mod 16   Jun 17, 2021
#3
0

I reviewed it and realized that the answer was actually $13$ because $2x+10 \equiv 12$ (mod $16$), and because $2x+11$ is one more, we get $12+1 =13$

Guest Jun 17, 2021