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Suppose that x is an integer that satisfies the following congruences:

\(\begin{align*} 3+x &\equiv 2^2 \pmod{3^3} \\ 5+x &\equiv 3^2 \pmod{5^3} \\ 7+x &\equiv 5^2 \pmod{7^3} \end{align*}\)

What is the remainder when x is divided by 105?

 Jul 21, 2019
 #1
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After a slight re-arrangement of your congruences, you have the following congruences, which are the same.
x mod 3^3 = 1
x mod 5^3 = 4
x mod 7^3 = 18 
Using "Chinese remainder theorem" + "Modular multiplicative inverse", which are inculded in this short computer code, the value of x=
1157625 m +  506629, where m=0, 1, 2, 3......etc.
i=0;j=0;m=0;t=0;a=(27, 125, 343);r= (1, 4, 18);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.
So, the smallest value for x =506,629
506,629 mod 105 =2^2 - the remainder.

 Jul 21, 2019
 #2
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For more detailed explanation of the "Chinese Remainder Theorem + Modular Multiplicative Inverse", see this link here: https://web2.0calc.com/questions/congruences

 Jul 21, 2019
 #3
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Suppose that x is an integer that satisfies the following congruences:
\(\begin{align*} 3+x &\equiv 2^2 \pmod{3^3} \\ 5+x &\equiv 3^2 \pmod{5^3} \\ 7+x &\equiv 5^2 \pmod{7^3} \end{align*}\)
What is the remainder when x is divided by 105?

 

see: https://web2.0calc.com/questions/modular-arithmetic-congruences

 

laugh

 Jul 22, 2019

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