Remainder Problems - Part 2

Remainder Problems – Part 2

Last week I introduced the idea of solving remainder problems without using long division. We may as well take advantage of the fact that we can use a calculator on the ACT (a TI-84 or equivalent) and GRE (the basic calculator they give you) to solve seemingly difficult math questions with ease. If you haven’t done so already, please read part 1 before continuing with this article: Remainder Problems – Part 1

Solution To Yesterday’s Problem

Yesterday I left you with a cliffhanger of a problem to solve with your calculator. Below is the solution. Let’s see if you got the problem correct and which method you used to solve the problem.

Here is the question again:

What is the remainder when 15,216 is divided by 73?

If you recall I gave three methods to solve such a problem, the first of which is by using long division. Since this post is about solving problems without using long division, let’s skip that method and jump straight to the much quicker and more efficient calculator algorithms – great for use on test day!

Method 2 – First Calculator Algorithm:

Step 1: Perform the division in your calculator:

15,216/73 ~ 208.438

Step 2: Multiply the integer part of this answer by the divisor:

73*208 = 15,184

Step 3: Subtract this result from the dividend to get the remainder:

15,216 – 15,184 = 32.

And here is an alternative:

Method 3 – Second Calculator Algorithm:

Step 1: Perform the division in your calculator:

15,216/73 ~ 208.438

Step 2: Subtract off the integer part of this result:

ANS – 208 ~ .438

Step 3: Multiply this result by the divisor:

73*ANS = 32.

Did you get this problem correct? Which method did you use? Try using both methods and then choose the one that you will be most comfortable with on the day of your standardized math test. And remember: the best way to get comfortable with a new technique is to practice with it!

The Cyclical Nature Of Remainders

Today let’s talk a bit about the cyclical nature of remainders. If you can understand this, then solving many ACT and GRE remainder problems will become second nature to you.

Let’s see what happens when we divide various positive integers by the number 3.

When we divide 3 by 3 we get a quotient of 1 and a remainder of 0. In other words,

3 = 3(1) + 0.

When we divide 4 by 3 we get a quotient of 1 and a remainder of 1. In other words,

4 = 3(1) + 1.

When we divide 5 by 3 we get a quotient of 1 and a remainder of 2. In other words,

5 = 3(1) + 2.

Now we have to be careful. When we divide 6 by 3, the quotient is now 2 and the remainder is 0. That is, 6 = 3(2) + 0.

So the remainders start over. When dividing by 3, the remainders cycle from 0 to 2 then back to 0.

0, 1, 2, 0, 1, 2, 0, 1, 2…

The easiest way to find a positive integer that has a certain remainder is to start with one that is evenly divisible, and then add the remainder. For example, suppose we want to find a positive integer that has a remainder of 7 when divided by 9. We can start with a number evenly divisible by 9, such as 9 itself, and then add 7. Since 9 + 7 = 16, we see that 16 is such a number. In fact,

16 = 9(1) + 7.

That is, when we divide 16 by 9 the quotient is 1 and the remainder is 7.

Since 18 is also divisible by 9, 18 + 7 = 25 is another positive integer that has a remainder of 7 when divided by 9. There are infinitely many possibilities!

Note that 0 is actually divisible by 9 as well. Indeed, we have

0 = 9(0) + 0.

Thus, 0 + 7 = 7 is a number that gives a remainder of 7 when divided by 9. In other words, we can actually just use the remainder itself!

Exercise

Now let’s see if you can apply this information to an actual ACT math problem. Here is a nice medium difficulty problem I like to give to my own ACT math students:

What is the least positive integer greater than 4 that leaves a remainder of 4 when divided by both 6 and 8?

Do you know how to go about solving this? You can find the answer in part 3 of this article here: Remainder Problems – Part 3