Divisibility by Seven

Everyone learns in grade school some simple tests for divisibility by small numbers such as 2, 3, 5, and 9. But far less well-known are some simple divisibility tests for the number 7. Here are a couple: Test #1. Take the digits of the number in reverse order, from right to left, multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary. Add the products. This sum has the same remainder mod 7 as the original number! Example: Is 1603 divisible by seven? Well, 3(1)+0(3)+6(2)+1(6)=21 is divisible by 7, so 1603 is. Test #2. Remove the last digit, double it, subtract it from the truncated original number and continue doing this until only one digit remains. If this is 0 or 7, then the original number is divisible by 7. Example: 1603 -> 160-2(3)=154 -> 15-2(4)=7, so 1603 is divisible by 7. See the reference for more tests and more references. Presentation Suggestions: Do examples as you go! Perhaps remind students of the divisibility test for 9 before presenting these. If you are teaching a course, you may wish to assign their proofs as an exercise! The Math Behind the Fact: Here's a hint on how to prove them. For the first test, note that (mod 7), 1==1, 10==3, 100==2, 1000==6, etc. For the second test, note that (mod7), 10A+B==10*(A-2B). The second trick mentioned here can be modified to check for divisibility by other primes. For example, to check divisibility by 13, take the last digit, multiply by 4 and add to the truncated portion. To check divisibility by 19, double the last digit and add. In fact, for any prime p, there exists some integer k such that divisibility by p can be ascertained by multiplying the unit's digit by k and adding (or subtracting) from the truncated portion of the numeral.