Given the set of numbers [7, 14, 21, 28, 35, 42], find a subset of these numbers that sums to 100.
First, make sure you understand the terminology: "...sums to 100" means that the object is to find some combination of the numbers in the original set that, when added together, add up to 100. You could spend all day on this seemingly easy question before giving up in frustration.
Why? Because it's a trick question! Many word problems hinge not on understanding the characteristics of adding, subtracting, multiplying, and dividing, but on recognizing the characteristics of the numbers you're given.
Before you even try adding some of these numbers together, in hopes of stumbling on the answer, take a look at the numbers themselves. Do you see anything that these numbers all have in common?
They're all multiples of 7, which means that they can each be represented as a number times 7. Or, because multiplication is really just a shortened form of addition, they can each be represented by a bunch of 7s being added together:
- 7 = 7 x 1 = 7
- 14 = 7 x 2 = 7 + 7
- 21 = 7 x 3 = 7 + 7 + 7
- 28 = 7 x 4 = 7 + 7 + 7 + 7
- 35 = 7 x 5 = 7 + 7 + 7 + 7 + 7
- 42 = 7 x 6 = 7 + 7 + 7 + 7 + 7 + 7
Now notice what happens when you try adding these numbers together. Let's say you add 21 and 28:
21 + 28 = (7 x 3) + (7 x 4) or (7 + 7 + 7) + (7 + 7 + 7 + 7)
The associative property of addition states that the grouping of elements doesn't make a difference; you can simply remove the parentheses when only addition is involved, which gives you this:
21 + 28 = 7 + 7 + 7 + 7 + 7 + 7 + 7 or 7 x 7
Since all multiples of 7 can be written as the sum of a certain number of 7s, whenever you add multiples of 7, the sum itself can also be written as the sum of a certain number of 7s, which is to say that if you add two or more multiples of 7, the sum is also a multiple of 7. This is true for all numbers; for example, if you add two or more multiples of 19, the sum is also a multiple of 19.
Looking back at the original problem, it is now clear that it's a trick question. Since you begin with all multiples of 7, there cannot be a subset of those numbers that sums to 100 because 100 is not a multiple of 7. The closest you can get is either 98 (42 + 35 + 21) or 105 (42 + 35 + 28).