Remainder Problems - Part 3

Remainder Problems – Part 3

Welcome to the third and final part of this thread on remainder problems. In the first part we discussed some calculator algorithms for finding remainders. You can review that article here: SAT Remainder Problems – Part 1

Yesterday we discussed the cyclical nature of remainders. You can review that article here: SAT Remainder Problems – Part 2

I decided to spend three posts on this subject because many of my students in a mid-level scoring range seem to have a difficult time with ACT and GRE remainder questions.

Solution To Last Week’s Remainder Problem

So where were we? I left you with an ACT math problem to solve using the information I provided regarding the cyclical nature of remainders. Let’s find out how you did.

The problem again:

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

Solution: We first find the least positive integer greater than 4 that is divisible by both 6 and 8. This is the least common multiple of 6 and 8 which is 24.

We now simply add the remainder.

24 + 4 = 28.

Thus, the answer is 28.

If you don’t know how to find the least common multiple of two numbers, it’s okay. If you’re taking the ACT, you can just let your TI-84 calculator do it for you. In your calculator press MATH, scroll right to NUM, and press 8 for lcm. Then type 6,8) and you will get an output of 24. So lcm(6,8)=24.

If you are taking the GRE, or you just want to learn other ways to compute the lcm, see the following article: Greatest Common Divisor (GCD) And Least Common Multiple (LCM)

Remainders In Disguise

Test makers’ love to put their remainder problems in disguise. In fact, many remainder problems on standardized tests do not even have the word “remainder” in the question! So how can you be expected to know when to find a remainder? The answer is quite simple. If the problem mentions some kind of sequence that keeps repeating over and over, then the problem may be asking you to find a remainder.

Remember this and you will recognize a remainder problem every time. In fact, this is such a key point I am going to repeat this in bold for you!

If the problem mentions some kind of sequence that keeps repeating over and over, then the problem may be asking you to find a remainder.

Let’s take a look at a standard example of such a question:

Example

Cards numbered from 1 through 2013 are distributed, one at a time, into nine stacks. The card numbered 1 is placed on stack 1, card number 2 on stack 2, card number 3 on stack 3, and so on until each stack has one card. If this pattern is repeated, each time beginning with stack one, on which stack will the card numbered 2013 be placed?

A. 3^rd stack
B. 4^th stack
C. 5^th stack
D. 6^th stack
E. 7^th stack

Solution: We are actually being asked to find the remainder when 2013 is divided by 9. Let’s do this using the first calculator algorithm that I showed you.

Step 1: Perform the division in your calculator

2013/9 ~ 223.667

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

9*223 = 2007

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

2013 – 2007 = 6.

So the card numbered 2013 will be placed on the 6^th stack, choice D.

Notice that the word remainder is never mentioned in this problem. The giveaway that this is a remainder problem in disguise is at the beginning of the last sentence where it says “If this pattern is repeated…”

So this is the end of this three part thread on remainders.