Suppose you want to create a set of weights so that any object with an integer weight from 1 to 40 pounds can be balanced on a two-sided scale by placing a certain combination of these weights onto that scale. what is the fewest number of weights you need, and what are their weights?

You need 4 weights, and the weights are 1; 3; 9 and 27.

Step-by-step explanation:

Since the scale has two plates, we can place weights on either side and also the object so it can be balanced.

This is a key part of the problem, it's not only on the other side of the scale, but on both sides.

Let's do the math now.

If i get two weights, 1 and 3. I can form this combinations.

Object of 1lb = 1

Object of 2lb + 1 weight = 3 weight.

Object of 3lb = 3 weight

Object of 4lb = 1 weight + 3 weight.

So what if i want to add the next weight and that weight to add me the maximum amount of objects. The weight would have to have a difference with the last object plus one. So if i grab 9. 9 minus 4 is 5. And that is a difference with the last object plus 1.

With a weight of 9, now i can add all the integers up to 13lb.

And the next step? Lets add one more. Keeping the last rule, the weight would have to have a difference with the last object plus one. So if i grab 27, 27 minus 13 is 14. And that is a difference witht the last object plus 1.

The sum of all the weights adds up to 40 pounds. And i can balance any integer in the middle.

The formula we are using is p – n = n + 1

Where p is the new weight. and n is the last object we weighted. And the sum of the weights goes up to the last object we can place on the scale, and in this case is 40.

saving = more money

step-by-step explanation:

if you have 100 bucks, and you want to buy a candy bar, even if you have one at home, if you choose not to buy it, you save.

~courtney