# Divisibility Tests: A History and User's Guide - The Beginnings of Divisibility

Author(s):
Eric L. McDowell (Berry College)

The earliest known divisibility test we have found comes from the Babylonian Talmud.  It was used to calculate the given year within a Sabbatical Cycle; that is, it provides the remainder obtained upon dividing an integer $$N$$ by $$7$$:

... put aside the hundreds as Jubilee Cycles and convert the remainder into Sabbatical Cycles [of seven years each] after adding thereto two years for every complete century; what is left over will give him the number of the given year in the current Sabbatical Cycle .

Translated into the language of mathematics, the passage implies that for integers $$N,$$ $$a,$$ and $$b$$ with $$N=100a+b,$$ we have that

$100a+b\equiv 2a+b\,\,({\rm{mod}}\,\,7).$

In other words, $$100a+b$$ and $$2a+b$$ yield the same remainder upon division by $$7.$$  For example, $$1866=100(18)+66\equiv 2(18)+66 = 102\,\,({\rm{mod}}\,\,7),$$ and $$102\equiv 2(1)+2=4\,\,({\rm{mod}}\,\,7).$$  Therefore, $$1866\equiv4\,\,({\rm{mod}}\,\,7).$$  Consequently, since $$4$$ is not divisible by $$7,$$ neither is $$1866.$$

The Talmud test is a minor variant of a class of divisibility tests that was described by Blaise Pascal (1623-1662) in an essay entitled De Numeris Multiplicibus  that was likely written prior to 1654.  As we shall see, Pascal's method can be used to devise divisibility tests of a variety of styles for any integer $$d$$.  In the case that a number fails to be divisible by $$d,$$ all of these tests have the ability to recognize the remainder obtained upon division by $$d.$$  They are all explained easily by the following theorem:

Theorem 1.1.  Let $$d$$ be a positive integer, and let $$a,$$ $$a^\prime,$$ $$b$$ and $$b^\prime$$ be integers for which $$a\equiv a^\prime\,\,({\rm{mod}}\,\,d)$$ and $$b\equiv b^\prime\,\,({\rm{mod}}\,\,d).$$ Then

(1)  $$a+b\equiv a^\prime+b^\prime\,\,({\rm{mod}}\,\,d)$$

(2)  $$ab\equiv a^\prime b^\prime\,\,({\rm{mod}}\,\,d).$$

Proof.  To prove (1), note first that $$(a+b)-(a^\prime+b^\prime)= (a-a^\prime)+(b-b^\prime).$$  But since $$a-a^\prime=dq_1$$ and $$b-b^\prime=dq_2$$ for some integers $$q_1$$ and $$q_2,$$ it follows that $$(a+b)-(a^\prime+b^\prime)=d(q_1+q_2)$$.  Thus, $$a+b\equiv a^\prime+b^\prime\,\,({\rm{mod}}\,\,d).$$ $$\Box$$

To prove (2), observe that

 $$ab-a^\prime b^\prime$$ $$=ab - a^\prime b + a^\prime b - a^\prime b^\prime$$ $$=(a-a^\prime) b+(b-b^\prime)a^\prime$$ $$=dq_1 b + dq_2a^\prime$$ $$=d(q_1 b+q_2a^\prime).$$

Thus, $$ab\equiv a^\prime b^\prime\,\,({\rm{mod}}\,\,d).$$