+0  
 
+1
37
3
avatar+241 

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?

 
RektTheNoob  Jan 12, 2018
Sort: 

3+0 Answers

 #1
avatar
+1

x = 506,629

506,629 mod 105 =2^2

 
Guest Jan 12, 2018
 #2
avatar
+1

(3 + x) mod 3^3 = 2^2

(5 + x) mod 5^3 = 3^2

(7 + x) mod 7^3 = 5^2

By simple iteration:

A*27 + 4 - 3=B*125 + 9 - 5 =C*343 + 25 - 7

A=18,764, B =4,053, C=1,477, Therefore:

x=18,764 *27 + 1 =506,629

506,629 mod 105 =4 =2^2

 
Guest Jan 12, 2018
 #3
avatar+241 
0

Thanks!

 
RektTheNoob  Jan 13, 2018

9 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details