I came across a fun problem in an open math forum, and the work being done is similar to other projects I've been working on. No one had attempted it, perhaps because the poser of the question let the thread die, but I figured I would offer a solution for anyone still following the thread.

#### The Problem

Given the positive numbers 1, 2, 3, 4, 5, 6, and 7, how many combinations of these numbers (using every number one time exactly) add up to 100? You are allowed to combine numbers, so long as each number is only used once.

For example: 12 + 3 + 4 + 6 + 75 = 100.

#### An Answer

Consider the meaning of the decimal expansion of a number. For instance, the problem can be made simpler by realizing 12 = 10 + 2.To get a head start on problems like this one before answers are published, follow the Math Club Facebook Page!

You can then try different combinations of your set where different elements have been multiplied by 10.

Then you can look at general cases such as: 10, 2, 3, 4, 5, 6, 70 = 100.

This case is congruent to 12, 3, 4, 5, 6, 70 and 13, 2, 4, 5, 6, 70 and 14, 2, 3, 5, 6, 70 and 15, 2, 3, 4, 6, 70 and 16, 2, 3, 4, 5, 70.

Repeat for the one's-place in 70.

Since you can't use zeros, you have 5 possibilities for the one's-place corresponding to the 10, and 4 more possibilities for the one's place corresponding to the 70. The resulting sum will still be 100.

For the case where 1 and 7 become tens, you have 20 possibilities.

For the case where 2 and 6 become tens, you have 20 possibilities.

For the case where 3 and 5 become tens, you have 20 possibilities.

For the case where 1, 2, and 5 become tens, you have 24 possibilities.

For the case where 1, 3, and 4 become tens, you have 24 possibilities.

No other combination of different elements being multiplied by 10 will result in a sum of 100, primarily because the ones digits don't sum to a multiple of ten.

The total combinations comes to 108 possible combinations.

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