Problem
NT-B2-M11-P001 Digit Sum in Base \(b\)
#1
★★☆☆☆ Level 2 of 5
Let \(b\ge2\), and let \(N=\overline{a_ra_{r-1}\ldots a_0}_b\). Prove that \(N\equiv a_0+a_1+\cdots+a_r\pmod{b-1}\). Find all bases \(b>5\) for which \(\overline{312}_b\) is divisible by \(b-1\).
Use \(b\equiv1\pmod{b-1}\).
Expand the number: \(N=a_rb^r+\cdots+a_1b+a_0\). Since \(b\equiv1\pmod{b-1}\), every power \(b^i\) is congruent to \(1\), so \(N\equiv a_0+\cdots+a_r\pmod{b-1}\).
For \(\overline{312}_b\), the digit sum is \(6\). Thus we need \(b-1\mid6\). Since \(b>5\), \(b-1\ge5\), so the only possibility is \(b-1=6\). The answer is \(b=7\).
A good starting problem: the student should see that digit-sum rules work in any base.