# Tutoring math, you realize that arithmetic is the foundation. The math tutor brings forward a quick method for determining divisibility by 11, with a couple of applications.

Consider the following problem:

Reduce 55/528 to lowest terms.

Of course, the key is knowing the number(s) that divide into both the top and the bottom. In this case, 11 does:

55/528=(5×11)/(48×11) = 5/48

Since knowing that 11 divides into a number is obviously useful, how do we do so?

Multiply the ones digit (8, in the case of 528) by 1, then the tens digit (2, in 528) by -1, then the hundreds digit (5) by 1. Add up the three products: 8(1) +2(-1) + 5(1) = 11. If the sum is divisible by 11, the number is. Since, for 528, the sum is 11, 528 is divisible by 11.

231 is also divisible by 11: 1(1) +3(-1) + 2(1) = 0. Since 0 is divisible by 11 (0÷11=0), 231 also is. Indeed, 231=11(21).

The pattern continues for numbers in the thousands. Consider 1936. 6(1) + 3(-1) + 9(1) + 1(-1) = 11. Therefore, 1936 is divisible by 11. In particular, 1936=11(176).

Now consider a polynomial factoring example:

**Factor x ^{2} – 4x – 165**

Solution: From my post on easy trinomials, you know that we must find two numbers that multiply to make -165, but add to make -4. Knowing by sight that 11 divides into 165, we easily recognize our numbers as 11 and -15.

x^{2} – 4x – 165 = (x+11)(x-15).

I’ll have more to say about divisibility in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.