Problem
NT-B2-M11-P002 Alternating Digit Sum
Prove the divisibility rule for \(11\) using the alternating digit sum. Then find all four-digit numbers of the form \(\overline{ab37}\) that are divisible by \(11\).
Use \(10\equiv-1\pmod{11}\), then solve the congruence for \(a-b+3-7\).
If \(N=\overline{a_ra_{r-1}\ldots a_0}\), then \(10^i\equiv(-1)^i\pmod{11}\). Hence \(N\equiv a_0-a_1+a_2-\cdots+(-1)^r a_r\pmod{11}\).
For \(\overline{ab37}\), the condition is \(a-b+3-7=a-b-4\equiv0\pmod{11}\). Since \(1\le a\le9\) and \(0\le b\le9\), we must have \(a-b-4=0\) or \(-11\). Thus \(a=b+4\) for \(b=0,1,2,3,4,5\), or \(a=b-7\) for \(b=8,9\). The numbers are \(4037,5137,6237,7337,8437,9537,1837,2937\).
Watch that the student does not mix up the digit order: one may start from either end, but the sign pattern must stay consistent.