IQ Question!!

[kingbitch]

You are given 10 standard coins of exactly the same appearance. All of them have the same weight except for 1, which is of a different weight though it is unknown if heavier or lighter. With only a standard balance weighing scale, and using only the weighing method, how would you identify the culprit coin in just three steps?

Comments

[chrishansenhome]

Number the coins from 1 to 10.

Weigh 1, 2, 3, 4 against 5, 6, 7, 8

If they balance, then weigh 1 against 9. If they do not balance, 9 is the culprit. Otherwise, 10 is.

If 1, 2, 3, 4 and 5, 6, 7, 8 do not balance, then weigh 1, 5, 6 against 2, 7, 8. If they balance, then the culprit is 3 or 4. So we then weigh 1 against 3. If they balance, then 4 is the culprit. Otherwise, 3 is.

If 1, 5, 6 and 2, 7, 8 do not balance and 1, 2, 3, 4 was the heavy side, before, and 1, 5, 6 is the heavy side now, then 1 is the culprit. If 2, 7, 8 is the heavy side now, then 2 is the culprit. If 5, 6, 7, 8 was the heavy side before, and 1, 5, 6 is the heavy side now, then 5 or 6 is the culprit. Weigh 1 against 5. If they balance, then 6 is the culprit. Otherwise, 5 will be the culprit. If 2, 7, 8 is the heavy side now, then 7 or 8 is the culprit. Weigh 2 against 7. If they balance, then 8 is the culprit, otherwise, 7 is.

Whew! I think that's right.

[kingbitch]

1) Wow!! No one has ever described to me this method before. I thought your strategy of weighing 1,5,6 with 2,7,8, and then proceeding on to the last paragraph, was ingenious.

2) But this method only works if the culprit coin is know to be heavier. Your last paragraph uses that assumption. Let's break it down:

"If 1, 5, 6 and 2, 7, 8 do not balance and 1, 2, 3, 4 was the heavy side, before, and 1, 5, 6 is the heavy side now, then 1 is the heavier coin." This also means that 7 or 8 might be a lighter coin. I don't see a solution to solve from coins 1,7,8 in just one step.

"If 2, 7, 8 is the heavy side now, then 2 is the culprit." This also means that 5 might be a lighter coin. Easy to solve from coins 2,5 in one step.

The same train of thought can be used for the rest of the last paragraph.

[chrishansenhome]

Oh, yes. It was late, and I guess my brain wasn't totally engaged. It waited until I went to sleep and then engaged with a vengeance.

[kingbitch]

Based on xiaogou's answer, I actually found a solution to my own doubt.

"I don't see a solution to solve from coins 1,7,8 in just one step."

Yes it's possible, since 1 can only be heavier and 7,8 can only be lighter, so weight 7 with 8 to see results, which should be apparent.

[thoburn]

how abt..ermmm...get the density of one coin by measureing the weight. Mass over volume.. then get a liquid of density slighly higher than the 1st coin and poured it into a clearless tank. put all the coins in...spreading them.

volume of all coins should be the same.

the denser the object, the higher it floats.

[kingbitch]

only with the balance!!

[thoburn]

a balance weighing scale can be used to measure mass wot

[kingbitch]

ok how do u do it in 3 steps then :P