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

Eric L. McDowell (Berry College), "Divisibility Tests: A History and User's Guide - The Beginnings of Divisibility," Convergence (May 2018)

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