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

tertre  Mar 23, 2017
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 #1
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If \(3x+7\equiv 2\pmod{16}\)\(3x+5\) is divisible by 16.

The smallest integer solution for x is 9. where 3(9) + 5 = 32 is divisible by 16.

Substitute x = 9, 2x + 11 = 29.

29 mod 16 = 13.

Let's test for the 2nd smallest integer solution for x, which is 25. where 3(25) + 5 = 80 is divisible by 16.

2x + 11 = 61.

61 mod 16 = 13.

Good!!

Therefore, the answer is 13 :D

MaxWong  Mar 23, 2017
 #2
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0

Thanks so much!

tertre  Mar 23, 2017

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