Choose two different integers from 1 to 98, requiring their sum to be less than 100. How many different choices can you make?

different choices

Choose the smaller integer first and then list all the possible larger integers, starting at 1 for the smaller integer.

If the smaller integer is 1, the larger integer can be 2, 3, 4, …, or 98.

If the smaller integer is 2, the larger integer can be 3, 4, 5, …, or 97.

If the smaller integer is 3, the larger integer can be 4, 5, 6, …, or 96.

….

If the smaller integer is 49, the largest integer must be 50.

We notice that there are 97 possibilities for the larger integer when the smaller integer is 1, and as we increase the smaller integer by 1 all the way up to 49, the number of possibilities for the larger integer goes down by 2 each time. It is reasonable to assume this pattern is true, saving a lot of work for analyzing the rest of the cases.

So, you can make 97 + 95 + 93 + … + 3 + 1 different choices. But how do we calculate this huge sum? There are 2 main formulas that are useful for the problems of this section, making calculations much easier.

1) Sum of an arithmetic series. An arithmetic series is the sum of a sequence of numbers that follow a rule: after the first number, the following numbers each go down or up by the same amount (called the common difference) . A sequence like this is called an arithmetic sequence. For example, an arithmetic sequence could be 1, 2, 3, and 4 (the common difference is 1) or perhaps 11, 8, 5, 2, and - 1 (the common difference is 3) . The formula for the sum of an arithmetic series is

.

Applying this to the arithmetic sequence 1, 2, 3, and 4 gives

, which is true.

2) Number of terms of an arithmetic sequence. The formula is

.

Applying this to the arithmetic sequence 11, 8, 5, 2, and - 1 gives

terms,

which is indeed the number of terms of the sequence.

So, using the formulas for the sum of an arithmetic series and the number of terms of an arithmetic sequence to aid in computation, you can make

different choices.

Recognizing the pattern and applying the formulas were the two main parts of this problem’s solution.