The Ring Problem

Hello,

Try solving this problem:
You have 10 stacks of rings in front of you. Each stack has 10 rings. In one stack, the rings weigh 0.9 grams each. In the other nine stacks the rings weigh 1 gram each. You have a weighing scale that measures in grams. You have one chance to place however many rings on the weighing scale. You have to place all the rings on simultaneously, though. From this single measurement, you can tell which stack has the odd rings. How do you accomplish this?

(Please use the spoiler tags for your answer.)

Jam it back in, in the dark.

Spoiler:

Yes.

Here is how it works.

What you must do is put 1 Ring from stack one, 2 Rings from stack two, etc, up to 10 rings from stack ten.

Now, weigh it. The final digit will tell you which stack has the rings that weigh less than the others. The it goes is 10 - X where X is the final digit. So in other words, weigh the rings which will give a number yyy.x, and use x to subtract from 10.

For example. Suppose the rings that weigh less are in stack one. That means it would be .9 (since you get one ring from stack one) + 10*2 (two rings from stack two) + ... + 10 * 10 (ten rings from stack ten). When you add it all up, you will find the result to be something yyy.x where the .x will be a .9 in this case. By the formula above, 10 - 9 = 1, which is the correct stack it came from.

Here is another example. Suppose the lighter rings are from stack two. Then the formula would be 10 (1 ring from stack 1) + .9 * 2 (two rings from stack 2) which equals 1.8 + 10 * 3 (three rings from stack three) + ... + 10 * 10 (ten rings from stack ten). How, if you add it up, the result will be something like yyy.8. Now, do the formula again, namely 10 - x (where x = 8 in this case), which is 10 - 8 = 2. That, again, gets you the correct result.

There's nowhere I can't reach.

Good job.

This thing is sticky, and I don't like it. I don't appreciate it.