Problem
NT-B2-M11-P014 A Square with the Same Last Digits
#14
★★★★☆ Level 4 of 5
Find all two-digit endings \(x\) modulo \(100\) for which \(x^2\) ends in the same two digits as \(x\).
Solve \(x^2\equiv x\pmod{100}\) separately modulo \(4\) and \(25\).
The condition is \(x(x-1)\equiv0\pmod{100}\). Since \(100=4\cdot25\), with coprime factors, we choose \(x\equiv0\) or \(1\pmod4\) and \(x\equiv0\) or \(1\pmod{25}\). The four systems give:
\(x\equiv0\pmod4, x\equiv0\pmod{25}\): \(x\equiv0\).
\(x\equiv1\pmod4, x\equiv1\pmod{25}\): \(x\equiv1\).
\(x\equiv1\pmod4, x\equiv0\pmod{25}\): \(x\equiv25\).
\(x\equiv0\pmod4, x\equiv1\pmod{25}\): \(x\equiv76\).
The answer is \(00,01,25,76\).
This reinforces CRT on a concrete and memorable object.