I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!

Monday, November 30, 2009

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There are 12 balls.

11 of the balls weigh the same.

1 of the balls is either heavier or lighter than the rest.

You have an unmarked balance scale.

Using the scale only 3 times, determine which ball is different and whether it is heavier or lighter than the rest.

I think that first we have to mark the balls 1 through 12.

Weigh balls 1, 2, 3, and 4 against 5, 6, 7, and 8. If they balance: Weigh 9 and 10 against 11 and 8 If they balance, 12 is the odd one. Weigh 12 against any other to find out if it is heavy or light.

If 9 and 10 and 11 and 8 do not balance: suppose 11 and 8 are heavier than 9 and 10; then either 11 is heavy, or 9 is light, or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they do not balance, weigh 9 against 10. The lighter is the odd ball.

(This can also be done in reverse, with 9 and 10 being the heavier, using the same argument and reversing the numbers.)

If the first weigh (1,2,3,4 against 5,6,7,8) do not balance: Suppose 5,6,7,8 is heavier than 1,2,3,4. This means that one of 1,2,3,4 is light, or one of 5,6,7,8 is heavy. Weigh 1, 2, and 5 against 3, 6, and 9. If they balance, then either 7 is heavy or 8 is heavy, or 4 is light. Weigh 7 against 8. If they balance, 4 is the odd ball, otherwise the heavier of 7 and 8 is the odd ball.

If 1,2,5 and 3,6,9 do NOT balance. Suppose 1, 2, and 5 are lighter, then either 6 is heavy, or 1 is light, or 2 is light. Weigh 1 against 2 to find out which one of the three choices is true. Suppose 1, 2, and 5 are heavier, then either 3 is light, or 5 is heavy. Weigh 3 against 2 to find out which of the two choices is true.

This one was great by the way. I feel like I have done something similar in class before, but it was fun. I used a lot of paper to figure this one out and ended up drawing a process chart. I would scan it in if I knew how... That would probably make my answer easier to understand.

1) Weigh 6 balls against 6 balls 2) Which ever set of six is the heaviest, take those 6 and weigh 3 balls against three balls 3) Which ever set of three is heaviest, take those three and weigh 1 ball against one ball. If they don't balance, the heavier ball is the one. If they do balance, the one in your hand is the one.

In step 2, what if you weigh the three vs three and they weigh the same? (ie, the odd ball isn't heavy, it's light and contained in the 6 balls you didn't weigh?

That's what makes the puzzle so challenging, you don't know if, when you weigh balls and one side is heavier, if the odd ball is on the heavy side or the light side.

However....With one piece of additional equipment, I can locate the odd ball from a set of thirteen (13) - and can even tell when I have a set of 13 identical ones. Anyone interested in finding out?

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I think that first we have to mark the balls 1 through 12.

ReplyDeleteWeigh balls 1, 2, 3, and 4 against 5, 6, 7, and 8.

If they balance:

Weigh 9 and 10 against 11 and 8

If they balance, 12 is the odd one.

Weigh 12 against any other to find out if it is heavy or light.

If 9 and 10 and 11 and 8 do not balance: suppose 11 and 8 are heavier than 9 and 10; then either 11 is heavy, or 9 is light, or 10 is light.

Weigh 9 against 10.

If they balance, 11 is heavy.

If they do not balance, weigh 9 against 10. The lighter is the odd ball.

(This can also be done in reverse, with 9 and 10 being the heavier, using the same argument and reversing the numbers.)

If the first weigh (1,2,3,4 against 5,6,7,8) do not balance:

Suppose 5,6,7,8 is heavier than 1,2,3,4. This means that one of 1,2,3,4 is light, or one of 5,6,7,8 is heavy.

Weigh 1, 2, and 5 against 3, 6, and 9.

If they balance, then either 7 is heavy or 8 is heavy, or 4 is light.

Weigh 7 against 8.

If they balance, 4 is the odd ball, otherwise the heavier of 7 and 8 is the odd ball.

If 1,2,5 and 3,6,9 do NOT balance.

Suppose 1, 2, and 5 are lighter, then either 6 is heavy, or 1 is light, or 2 is light.

Weigh 1 against 2 to find out which one of the three choices is true.

Suppose 1, 2, and 5 are heavier, then either 3

is light, or 5 is heavy.

Weigh 3 against 2 to find out which of the two choices is true.

This one was great by the way. I feel like I have done something similar in class before, but it was fun. I used a lot of paper to figure this one out and ended up drawing a process chart. I would scan it in if I knew how... That would probably make my answer easier to understand.

Impressive, but too much work...

ReplyDelete1) Weigh 6 balls against 6 balls

2) Which ever set of six is the heaviest, take those 6 and weigh 3 balls against three balls

3) Which ever set of three is heaviest, take those three and weigh 1 ball against one ball. If they don't balance, the heavier ball is the one. If they do balance, the one in your hand is the one.

Hi StanW. I'm not sure, but I don't think your strategy will work because we are unsure if the odd ball out is lighter or heavier.

ReplyDeleteAnonymous, well written and thought out. I hadn't realized how much work it was going to be to figure it all out, but I'm glad you did it!

StanW,

ReplyDeleteIn step 2, what if you weigh the three vs three and they weigh the same? (ie, the odd ball isn't heavy, it's light and contained in the 6 balls you didn't weigh?

That's what makes the puzzle so challenging, you don't know if, when you weigh balls and one side is heavier, if the odd ball is on the heavy side or the light side.

the simplest way. while placing the balls on the scale to measure, you happen to notice that one is heavier than another...

ReplyDelete1234/5678

ReplyDelete125/346

1/2 -> 1H

1\2 -> 2H

1-2 -> 6L

125\346

3/4 -> 3H

3\4 -> 4H

3-4 -> 5L

125-346

7/8 - 8L

7\8 - 7L

AND SO ON.......

However....With one piece of additional equipment, I can locate the odd ball from a set of thirteen (13) - and can even tell when I have a set of 13 identical ones. Anyone interested in finding out?