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
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.
Do examples as you go! Perhaps remind students of the
divisibility test for 9 before presenting these. If
you are teaching a number theory 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),
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.
See also Divisibility by Eleven.
How to Cite this Page:
Su, Francis E., et al. "Divisibility by Seven."
Math Fun Facts.