Let’s begin by asking you to answer the following Level 5 math question from the May 2013 SAT:
When 53 is divided by a positive integer n, the remainder is 3. How many values of n are possible?
Your answer: _______________ The correct answer: 4
Don’t feel badly if you missed this question. I tutor outstanding math students, the majority of whom will be taking AP Calculus. However, their advanced mathematical knowledge and high-powered calculators were not enough to help them answer this question. What happened?
First and foremost, the majority of my students did not really understand the concept of a remainder. Several students told me that their math teachers omitted the topic saying that it is irrelevant to real life. Unfortunately, remainder problems are now appearing with increasingly frequency on a real life situation called the SAT.
So let’s begin with a quick review. A remainder is the amount left over after division. For example, six will divide into 19 three times with a remainder of 1.
Students who were not baffled by the concept of remainder correctly realized that the answer had to be a factor of 50 with a remainder of 3. These students listed 1, 2, 5, 25, and 50 as factors of 50. Some
students then wrote down 6 as their answer. Wrong! Pause for just a moment. Would a Level 5 math problem really be that easy? No it would not. Long experience indicates that difficult math problems almost always require at least one additional step.
Some of my students did realize that 1 would divide into 53 and not leave a remainder. So they triumphantly wrote down 5 as their answer. Wrong again! What about 2? Incredibly, several students casually divided 53 by 2 and got 25 with a remainder of 3. Needless to say, 53 divided by 2 gives us 26 with a remainder of 1.
The correct answer to this problem is 4 since 5, 10, 25, and 50 will all divide into 53 and leave a remainder of 3.
It is important to note that this problem was one of two Level 5 math problems on the May 2013 SAT. According to the May 2013 scale -1 question = 770 and -2 = 740. So this “irrelevant” remainder problem was worth 30 points!