Chapter
Divisibility Rules
Digit rules for divisibility, modular proofs of rules, missing digit problems, and last-digit arguments.
Theory
1. Digit Sums
Because \(10\equiv1\pmod9\), every decimal number has the same remainder modulo \(9\) as the sum of its digits.
2. Last Digits
Modulo \(2,4,5,8,10\), only the last few digits matter.
3. The Rule for 11
Since \(10\equiv-1\pmod{11}\), alternating sums of digits determine divisibility by \(11\).
Examples
Example 1. Why the Rule for Nine Works
This is the core proof behind digit-sum tests.
Problem.
Prove that a decimal integer has the same remainder modulo \(9\) as the sum of its digits.
Solution.
If \(n=a_k10^k+\cdots+a_0\), then \(10^j\equiv1\pmod9\). Hence \(n\equiv a_k+\cdots+a_0\pmod9\).
Example 2. A Missing Digit
This combines two digit rules.
Problem.
Find all digits \(A\) such that \(52A8\) is divisible by both \(3\) and \(4\).
Solution.
For divisibility by \(4\), \(10A+8\equiv2A\pmod4\), so \(A\) is even. For divisibility by \(3\), \(15+A\) is divisible by \(3\), so \(A\in\{0,3,6,9\}\). Thus \(A=0\) or \(6\).
Example 3. Digits for Divisibility by 11 and 5
This is a missing-digit problem using the rule for 11.
Problem.
Find all digits \(A,B\) such that \(3A5B\) is divisible by both \(11\) and \(5\).
Solution.
For divisibility by \(11\), \(8-A-B\) must be divisible by \(11\). If \(B=0\), then \(A=8\). If \(B=5\), then \(A=3\). The numbers are \(3850\) and \(3355\).
Example 4. Divisibility by 72
This is a compact multi-rule digit problem.
Problem.
Find all digit pairs \(A,B\) such that \(72A4B\) is divisible by \(72\).
Solution.
We need divisibility by \(8\) and \(9\). Checking the last three digits \(A4B\) and digit sum \(13+A+B\), the pairs are \((1,4)\) and \((6,8)\). The numbers are \(72144\) and \(72648\).
Problems
Problems
No published problems are available yet.
Ladders
No published ladders were found.