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?

xXxTenTacion Jul 21, 2019

#1**+1 **

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.**

Guest Jul 21, 2019

#2**+1 **

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

Guest Jul 21, 2019

#3**+2 **

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

heureka Jul 22, 2019