In the Math Corner of my March Diary I offered a theorem on divisibility. First I gave a rule for extracting a particular integer k, which might be either positive or negative, from any prime number p:
RULE: Any prime p greater than 2 ends with either 1, 3, 7, or 9.
If p ends with 1 or 9, it's right next to a multiple of 10. It's either 10k+1 or 10k−1. That k will be your multiplier, and you'll sign it with the sign in the previous sentence. So if p ends with 1, k is positive; if p ends with 9, k is negative.
If p ends with 3 or 7, just multiply it by 3 before applying all that. For p = 7, tripling gets you 21, so k = +2. For k = 13, however, you'd be looking at 39, so k = −4.
With that rule established, I stated a theorem, and challenged readers to prove the theorem.
The Stump Divisibility Theorem.
Suppose p is a prime greater than 2, and I've extracted from it a signed number k, according to RULE. Suppose N is any number, with rightmost digit d. Call the rest of N the "stump." Then p divides exactly into N iff p divides into (the stump minus dk).
(I had already told the reader that "iff" is math shorthand for "if and only if.")
Proof: Not much of a stretch. First, by looking at RULE, persuade yourself that p must divide exactly into 10k + 1. If p = 43, for example, then k works out to −13, so 10k + 1 is −129, which is 43 × (−3). From the way I've defined k, in fact, p will always divide into 10k + 1 either ±1 times or ±3 times.
Now decompose N into its decimal digits. We know the rightmost digit is d; so for some positive whole number s the digits are as, as-1, as-2, …, a2, a1, d, and:
N = as10s + as-110s-1 + … + a2102 + 10a1 + d.
= as10s + as-110s-1 + … + a2102 + 10a1 − 10dk + 10dk + d.
= 10[(as10s-1 + as-110s-2 + … + 10a2 + a1) − dk] + 10dk + d.
But what is that thing in the inner parenthesis there? Why, it's the stump! So:
N = 10(stump minus dk) + (10k + 1)d
Since we know that p divides exactly into 10k + 1, it divides exactly into one of the other two things, N or 10(stump minus dk), iff it divides exactly into the other. QED.