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What is the largest negative integer \(x\) satisfying \(24x \equiv 15 \pmod{1199}~?\)

Guest Sep 10, 2018
 #1
avatar+93866 
+1

 

What is the largest negative x integer  satisfying    \( 24x \equiv 15 \pmod{1199}~? \)

 

 

15 (mod1199)  is equivalent to  15-1199- 1199k = -1184 - 1199k   (mod1199)

 

24x =  -1184 - 1199k

 

1199k+24x = -1184

 

there does not appear to be any common factors

 

Look for a solution to   1199k+24x=-1

 

1199=50*24-1

so

1199(1)+24(-50)=-1

and

1199(1184   )+24(-50*1184   )=-1184

the general solution is

1199(1184 -24a  )+24(-50*1184 + 1199a   )=-1184      where a is an integer

 

so I think I want   

\(-50*1184+1199a<0\\ 1199a<58200\\ a<49.37\\ \text{but a is an integer so the biggest a is 49}\)


\(1199(1184-24*49) +24(-50*1184+1199*49)=-1184\\ 1184(8)+24(-449)=-1184\)

 

\( 24*-449 \equiv 15 \pmod{1199} \)

 

So I think that the negative integer with the smallest absolute value (that is the biggest one) is 

\(x= -449\)

Melody  Sep 11, 2018
 #2
avatar+93866 
0

I have never learnt to do modulus arithemtic formally (I'm self taught) so it would be good if someone checked this answer.

Thanks :)

Melody  Sep 11, 2018
 #3
avatar+20148 
+3

What is the largest negative integer \(x\) satisfying \(24x \equiv 15 \pmod{1199}~?\)

 

\(\begin{array}{|rclll|} \hline 24x &\equiv& 15 \pmod{1199} \\ \text{or} \\ 24x &=& 15 + 1199k \quad & | \quad k \in Z \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{15 + 1199k}{24}} \\\\ x &=& \dfrac{49\cdot 24k + 23k + 15 }{24} \\\\ x &=& 49k + \underbrace{ \dfrac{23k + 15 }{24} }_{=a} \\\\ && & a = \dfrac{23k + 15 }{24} \\\\ && & 24a = 23k + 15 \\\\ && & 23k = 24a - 15 \\\\ && & \mathbf{ k = \dfrac{24a - 15}{23} }\\\\ && & k = \dfrac{23a+a - 15}{23} \\\\ && & k = a + \underbrace{ \dfrac{a - 15}{23} }_{=b} \\\\ && && b = \dfrac{a - 15}{23} \\\\ && && 23b = a - 15 \\\\ && & & \mathbf{ a = 23b + 15} \\\\ && & \mathbf{ k = \dfrac{24(23b + 15) - 15}{23} }\\\\ && & k = \dfrac{24\cdot 23b + 24\cdot 15 - 15}{23} \\\\ && & k = \dfrac{24\cdot 23b + 23\cdot 15}{23} \\\\ && & \mathbf{k = 24b+15} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{15 + 1199(24b+15)}{24}} \\\\ x & = & \dfrac{15 + 1199\cdot 24b+ 1199\cdot 15}{24} \\\\ x & = & \dfrac{1199\cdot 24b+ 1200\cdot 15}{24} \\\\ x & = & \dfrac{1199\cdot 24b+ 50\cdot 24 \cdot 15}{24} \\\\ x & = & 50\cdot 15 + 1199 b \\\\ x & = & 750 + 1199 b \quad & | \quad b = -1 \\\\ x & = & 750 - 1199 \\\\ \mathbf{ x } &\mathbf{ = } & \mathbf{-449} \\ \hline \end{array}\)

 

laugh

heureka  Sep 11, 2018
edited by heureka  Sep 11, 2018
 #4
avatar
+1

heureka made a minor typo. x =- 449

Guest Sep 11, 2018
edited by Guest  Sep 11, 2018
 #5
avatar+20148 
+3

Thank you, Guest !

laugh

heureka  Sep 11, 2018
 #6
avatar+93866 
+1

Thanks Very much Heureka :)

Melody  Sep 11, 2018

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