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Given that \(x\) is a positive integer less than 100, how many solutions does the congruence \(x + 13 \equiv 55 \pmod{34}\) have?

 Dec 18, 2018
 #1
avatar+21848 
+8

Given that  \(\large{x}\)  is a positive integer less than 100,
how many solutions does the congruence  have?
\(\large{x + 13 \equiv 55 \pmod{34}}\)

 

\(\begin{array}{|rcll|} \hline x + 13 &\equiv& 55 \pmod{34} \\ x + 13 &\equiv& 55-34 \pmod{34} \\ x + 13 &\equiv& 21 \pmod{34} \quad & | \quad - 13 \\ x &\equiv& 21-13 \pmod{34} \\ x &\equiv& 8 \pmod{34} \\ \mathbf{x} & \mathbf{=} & \mathbf{8 + n \cdot 34}, ~ n \in \mathbb{N} \\ \hline \end{array} \)

 

\(\begin{array}{|c|l|c|} \hline n, ~ n \in \mathbb{N} & \mathbf{x = 8 + n \cdot 34} & ~ x>0,~ x<100 \\ \hline 0 & x = 8+0\cdot 34 \\ & x= 8 & \checkmark \\ \hline 1 & x = 8+1\cdot 34 \\ & x= 42 & \checkmark \\ \hline 2 & x = 8+2\cdot34 \\ & x = 76 & \checkmark \\ \hline 3 & x = 8+3\cdot 34 \\ & x = 110 & x> 100 \\ \hline \ldots & & x> 100 \\ \hline \end{array}\)

 

The congruence  has three solutions: \(\mathbf{x = 8}\) and \(\mathbf{x = 42}\) and \(\mathbf{x = 76}\)

 

laugh

 Dec 18, 2018
edited by heureka  Dec 19, 2018
 #2
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+1

But, isn't 42 + 13  mod 34 = 21 and 76 +13 mod 34 = 21 ????

 Dec 18, 2018
 #4
avatar+99307 
0

Yes that is right, what is the problem ?

 

55( mod 34) also equals 21

Melody  Dec 18, 2018
 #3
avatar+99307 
+1

Did you miss 8 Heureka ?

 Dec 18, 2018
edited by Melody  Dec 18, 2018
edited by Melody  Dec 18, 2018
 #5
avatar+21848 
+7

Hello Melody,

 

of course there is x = 8.

 

laugh

heureka  Dec 19, 2018

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