# Lucky Number 7

by January 11, 2013

on**Know Your Numbers**

It doesn’t matter if you are preparing for the SAT, GRE, or GMAT, if you want to excel on the math portion of the exam you have to know your numbers. Memorizing multiplication tables is only half the battle. If you want a top score, you should be able to see the number 41,394 and know that it is divisible by 6 without consulting a calculator. While calculators are allowed on the SAT and GRE, they often reduce momentum. If you become too reliant on your calculator you are liable to second guess yourself during the exam and spend time using it unnecessarily.

Practicing the rules of divisibility will help you factor large numbers and increase your ability to power through the quantitative portion of the exam successfully. By the time exam day rolls around you should know that 41,394 is divisible by 6 because if you add up all the integers you get 21 and if integers add up to a number divisible by 3 then the number itself is divisible by 3. Since the number 41,394 is also even, it is clearly divisible by 2 and if a number is divisible by 2 and 3 then it is divisible by 6.

Divisibility strategies vary from number to number, but are typically simple and (kinda, sorta) fun. But when it comes to simple divisibility solutions, lucky number 7, never feels lucky on exam day. To some mathematical minds the number 7 is so tricky that they believe there is no “trick” to figuring out if something is divisible by 7.

Our GMAT guru, Craig Ryan begs to differ…

**Divisibility by 7**

According to Craig, divisibility by 7 can be tested by a recursive method.

A number of the form 10*x* + *y* is divisible by 7 if and only if *x* − 2*y* is divisible by 7.

In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a small number (below 20 in absolute value) is obtained. The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7.

**For example, the number 371:**

10*x* + *y*

if and only if x − 2y is divisible by 7

x= 37 (the first two digits)

y= 1

37 − (2×1) = 37 − 2 = 35

3 − (2 × 5) = 3 − 10 = −7

Since −7 is divisible by 7, we know 371 is divisible by 7.

How about a larger number?

**Let’s try 2,968:**

x=296

y=8

296 –(2×8) = 280

28 – (2×0) = 28

2 – (2×8) = -14

-1 – (2×4) = -7

If you know your multiplication tables then you should know after the first solution round that 28(0) is divisible by 7 and thus be able eliminate the final 3 steps.

Once you understand the rules of divisibility, factorization of large numbers is much more manageable. With manageability comes speed, with speed efficiency, and with efficiency comes success.