Problem
NT-B1-M10-P024 A Multiple of \(2026\) with Digits \(0\) and \(1\)
#24
★★★★★ Level 5 of 5
Prove that there exists a positive integer consisting only of digits \(0\) and \(1\) that is divisible by \(2026\).
First construct a number of ones divisible by \(1013\), then append a final digit \(0\).
We have \(2026=2\cdot1013\), and \(\gcd(1013,10)=1\). By the repunit construction, there exists \(R_n\) divisible by \(1013\). Then \(10R_n\) is obtained by appending a zero to a number of ones, so it consists only of digits \(0\) and \(1\). It is divisible by \(10\), hence by \(2\), and also by \(1013\). Therefore it is divisible by \(2026\).
Strong task on adapting the construction when the modulus shares a factor with \(10\).