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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
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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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Don't know much about this.....but....maybe   if   x  = 4

 

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

 

 

cool cool cool

 Jun 17, 2021
 #3
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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

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