Sumudu’s approach below is a good one and the best for mine.

]]>3 balls on each side of the scale. if they are even..throw them away.

now you have 6 remaining balls.

same move again, the heavier side of the scale has the heavy ball and you are left with 1 more try on the scale.

1 ball on each side, if they even out, the 1 left out is the heavy 1 ðŸ˜‰

]]>2) Mark all as either 1-4(+) 5-8(-) or 9-12(E) *E for Even

3) Weigh 5+1-9E against 6+2-7+

If Even –> Weigh the two remaining 3-4- (if uneven take the lighter one)

If Left is heavier –> weigh 5+E

If Right is heavier –> weight 6+7+ (if uneven take the heavier one) ]]>

– Step 1: Balance 2 piles of 4 chips

– Case 1: Both Piles Level

o Step 2: Take 3 of the 4 unknown chips and weigh against 3 of the known normals

ï‚§ Case 1: Both Piles Level

â€¢ Step 3: Take the remaining chip and weigh against a known normal to get answer

ï‚§ Case 2: Piles not level

â€¢ Take the 3 potentially light or heavy chips (you know which one they are now) and balance two of them to see which is the lighter or heaver one or if itâ€™s the one in your hand.

– Case 2 (Worst Case): Piles not Level

o Identify the 4 potentially heavy and 4 potentially light chips from step 1

o Step 2

ï‚§ Side 1 of scale: 3 potentially light chips and 1 of the potentially heavy chips

ï‚§ Side 2 of scale: 3 of the 4 known normal chips from step 1 and the 1 leftover potentially light chip

ï‚§ Case 1: Both Piles Level

â€¢ Step 3: Take the remaining 3 potentially heavy chips and balance 2 of them, seeing which is the heaver one or if itâ€™s the 3rd one in your hand

ï‚§ Case 2: Side 1 is higher

â€¢ Step 3: As none of the chips on side 2 can be heavy, it must be one of the 3 potentially light chips on side 1, so do like step 3 above for them to determine the light chip

ï‚§ Case 3: Side 2 is higher

â€¢ Step 3: Since there is only one heavy chip on side 1 and one light chip on side 2, we know that one of them is the culprit, so take one of them and weigh it against a known normal chip and get your answer!

I came to this conclusion by thinking about how I wanted to have 3 chips at the end that were either all potentially heavy or potentially light. This made me want to eliminate 5 of the 8 unknowns in step 2. The key was having only one potentially heavy chip on side 1 and only one potentially light chip on side 2 so that if side 2 becomes higher, it must be one of those two chips. I kinda lucked out picking groups of 4 for step one, but I spent considerable time examining other groupings for step1 and 3 groups of 4 seemed to be getting the closest.

-Chaz Morantz ]]>

This is a really good point – I just assumed it would be easier, but come to think of it the balance scales really do give you more information.

Back to the drawing board ðŸ˜‰

]]>Your very articulate response is much appreciated and I look forward to trying the method out!

Thanks.

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