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Find the least positive four-digit solution to the following system of congruences.

 

\(\begin{align*} 7x &\equiv 21 \pmod{14} \\ 2x+13 &\equiv 16 \pmod{9} \\ -2x+1 &\equiv x \pmod{25} \\ \end{align*}\)

 Nov 10, 2020
 #1
avatar+102 
0

Is the LaTeX not working?

 

\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

(Test)

 Nov 10, 2020
 #2
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+1

The 3 congruences can be re-written as follows:

 

x mod 2 =1,
x mod 9  =6,
x mod 25   =17, solve for x

 

 

Using CRT plus MMI

 

i=0;j=0;m=0;t=0;a=(2, 9, 25);r= (1, 6, 17);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

 

x =450m + 267, where m=0, 1, 2, 3.........etc.

x =(450 x 2) + 267

x =1167 - This is the least 4-digit solution.

 Nov 10, 2020
 #3
avatar+102 
+1

Thanks anonymous!

 Nov 11, 2020

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