Differences of Large Sums - An Advanced Math Strategy

Differences of Large Sums –

An Advanced Math Strategy

Today I would like to show you how to solve math problems where you are asked to find two large sums and then subtract them.

These problems always remind me of a little story about a young boy named Karl Gauss. The story actually directly relates to the strategy I will be discussing, so it’s worth taking this little detour.

The Boy Karl Gauss

In a math class, a teacher set her students to a task to keep them busy for a few minutes.

In an attempt to keep the class quiet, the teacher asked the students to take out a piece of paper and add up the integers from 1 to 100 by hand.

The teacher was hoping for the students to painstakingly write down 1, 2, 3, 4, 5 etc… to 100, and then add the numbers together to get an answer. The teacher knew that this problem would take some time to complete as she, herself, took several minutes to get the answer.

However, one of her students, Karl, handed the teacher a piece of paper with the correct answer almost immediately. In the top left corner of the sheet, he had neatly written 5,050.

Nothing else.

Surprised that the student had the correct answer so quickly, the teacher asked how he did it.

“Oh, this was really easy” said Karl, “1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, 4 + 97 = 101… and so on. I could see that there are 100 pairs of numbers that add up to 101. So I multiplied 101 by 100 to get 10,100. I then divided by 2 because I added each number twice. So I wrote down the answer 5,050.”

Let’s take one last look at Karl’s computation in an easy to read format as we will be using this format to solve math problems in just a moment.

1 + 2 + 3 + …+ 100
100+99+98+ … + 1
101+101+101+…+ 101

101+101+101+…+101

= 100(101)

= 10,100

10,100/2 = 5050

We are going to simulate Karl Gauss’s method above to get the answer to some seemingly difficult math questions quickly and without making careless computational errors. Don’t worry – I will explain everything step by step.

This method is best understood with examples So let’s jump right in with some problems.

Example 1

If x denotes the sum of the integers from 10 to 70 inclusive, and y denotes the sum of the integers from 80 to 140 inclusive, what is the value of y – x?

First, we write out each sum formally and line them up with y above x.

80 + 81 + 82 + … + 140
10 + 11 + 12 + … + 70

Next subtract term by term.

80 + 81 + 82 + … + 140
10 + 11 + 12 + … + 70
70 + 70 + 70 + … + 70

Now notice that we’re adding 70 to itself

70 – 10 + 1 = 61

times (by the fence-post formula – see yesterday’s blog post).

This is the same as multiplying 70 by 61. So we get

(70)(61) = 4270.