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Nine weights covering gram weights of 1, 2, 4, 8, 16, 32, 64, 128, 256, and 512 should suffice. In fact, these weights would cover all integral weights in grams up to and including 1023. The weights used to balance a sample correspond to the "1" digits in a binary number system representation of the weight in grams.YoursFalsey wrote:Since this topic has been quiet for a few days, a new puzzle.
Jerrod the Jeweler is going into a new sideline of assaying ore samples. He intends to weigh ore samples with an old fashioned balance scale, placing the samples on one side and his weights on the other side until a balance is achieved. If he wishes to be able to balance all possible samples with an integral weight in grams between 1 and 1000, what is the smallest number of weights he will need?
(Oh, this isn't school, so a stringently rigorous proof isn't necessary, but by the same token, give some explanation so we can follow your answer...)
No, he is still using his pan balance.jonesthecurl wrote:He buys electric scales.
Good outside-the-box thinking, but not what I had in mind. Anyway, how could he ensure the accuracy of the weights of the pieces he cut? And would you deal with someone whose weights weren't marked "legal for trade" by the Bureau of Weights and Measures? No, he will buy a set of less than nine official weights and use them unaltered. He will use the equal-arm pan balance unaltered (no tricks with weights at different distance from the fulcrum). But he will use the balance in a way somewhat different from that described in the previous puzzle.spiesr wrote:He buys one 1023g (or larger) weight and cuts it into the necessary pieces. This could be cheaper...
No, how would you weigh a 3g sample in that scheme?Stroop wrote:I think the 2g and 4g weights can be replaced by one 5g weight. Similar substitutions could be possible with heavier weights, but I'm too lazy to check.
Simple. You take your sample, and put it on one end of the scale together with the 5g weight. Put the 8g weight on the other side of the scale.ender516 wrote:No, how would you weigh a 3g sample in that scheme?Stroop wrote:I think the 2g and 4g weights can be replaced by one 5g weight. Similar substitutions could be possible with heavier weights, but I'm too lazy to check.
Or use the 1 and the 2Stroop wrote:Simple. You take your sample, and put it on one end of the scale together with the 5g weight. Put the 8g weight on the other side of the scale.ender516 wrote:No, how would you weigh a 3g sample in that scheme?Stroop wrote:I think the 2g and 4g weights can be replaced by one 5g weight. Similar substitutions could be possible with heavier weights, but I'm too lazy to check.
Read the bold bit again.jonesthecurl wrote:Or use the 1 and the 2Stroop wrote:Simple. You take your sample, and put it on one end of the scale together with the 5g weight. Put the 8g weight on the other side of the scale.ender516 wrote:No, how would you weigh a 3g sample in that scheme?Stroop wrote:I think the 2g and 4g weights can be replaced by one 5g weight. Similar substitutions could be possible with heavier weights, but I'm too lazy to check.
good point...Stroop wrote:Read the bold bit again.jonesthecurl wrote:Or use the 1 and the 2Stroop wrote:Simple. You take your sample, and put it on one end of the scale together with the 5g weight. Put the 8g weight on the other side of the scale.ender516 wrote:No, how would you weigh a 3g sample in that scheme?Stroop wrote:I think the 2g and 4g weights can be replaced by one 5g weight. Similar substitutions could be possible with heavier weights, but I'm too lazy to check.
Code: Select all
Binary digits: 0, 1
Ternary digits: 0, 1, 2
Signed-ternary digits: 0, +, -
Binary digits Ternary Signed-ternary
16g 8g 4g 2g 1g Decimal 27g 9g 3g 1g 27g 9g 3g 1g
0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 1 0 0 0 1 0 0 0 +
0 0 0 1 0 2 0 0 0 2 0 0 + -
0 0 0 1 1 3 0 0 1 0 0 0 + 0
0 0 1 0 0 4 0 0 1 1 0 0 + +
0 0 1 0 1 5 0 0 1 2 0 + - -
0 0 1 1 0 6 0 0 2 0 0 + - 0
0 0 1 1 1 7 0 0 2 1 0 + - +
0 1 0 0 0 8 0 0 2 2 0 + 0 -
0 1 0 0 1 9 0 1 0 0 0 + 0 0
0 1 0 1 0 10 0 1 0 1 0 + 0 +
0 1 0 1 1 11 0 1 0 2 0 + + -
0 1 1 0 0 12 0 1 1 0 0 + + 0
0 1 1 0 1 13 0 1 1 1 0 + + +
0 1 1 1 0 14 0 1 1 2 + - - -
0 1 1 1 1 15 0 1 2 0 + - - 0
1 0 0 0 0 16 0 1 2 1 + - - +
1 0 0 0 1 17 0 1 2 2 + - 0 -
1 0 0 1 0 18 0 2 0 0 + - 0 0
1 0 0 1 1 19 0 2 0 1 + - 0 +
1 0 1 0 0 20 0 2 0 2 + - + -
1 0 1 0 1 21 0 2 1 0 + - + 0
1 0 1 1 0 22 0 2 1 1 + - + +
1 0 1 1 1 23 0 2 1 2 + 0 - -
1 1 0 0 0 24 0 2 2 0 + 0 - 0
1 1 0 0 1 25 0 2 2 1 + 0 - +
1 1 0 1 0 26 0 2 2 2 + 0 0 -
1 1 0 1 1 27 1 0 0 0 + 0 0 0
Glad to see you taking it with good graces.Stroop wrote:Ahh, should have thought of ternary once I decided to put weights with the ore sample...
At least I get an honourable mention
This looks good provided the wives only object to their husbands being with unescorted women, because after the second step, husband A is on the east side with wives B and C, but husbands B and C are there as well.Stroop wrote:Alright. We have husband A and wife A, husband B and wife B and husband C and wife C.
Husband A and wife A cross.
Husband A crosses back.
Wife B and wife C cross.
Wife A crosses back.
Husband and wife A cross.
Husband A crosses back.
Husband A and husband B cross.
Wife C crosses back.
Husband C and wife C cross.
I think this way no wife has a problem with their jealousy and they all cross.
No, I believe you are wrong, Stroop is correct.YoursFalsey wrote:Stroop is on the right track, but unfortunately, the wives DO object even if their husband and the other wife(s) have other husbands present- they are not just jealous, but insanely jealous.Apparently they are even afraid of spontaneous orgies and the like.
Ender's alternative solution has a lot to recommend it, because I can't see putting up with someone that jealous, but apparently these husbands are more gentlemanly than I am...

Sorry, porken, but as Ender pointed out, after the canoe has gone across and back, wife A is on the far side of the river while her husband is on the original side with the other husbands and (more importantly) the other wives. In each intermediate step, each husband must either be in a group that is male only or their wife must be present. Other wives may be present, but only if the husband's wife is also there.porkenbeans wrote:No, I believe you are wrong, Stroop is correct.YoursFalsey wrote:Stroop is on the right track, but unfortunately, the wives DO object even if their husband and the other wife(s) have other husbands present- they are not just jealous, but insanely jealous.Apparently they are even afraid of spontaneous orgies and the like.
Ender's alternative solution has a lot to recommend it, because I can't see putting up with someone that jealous, but apparently these husbands are more gentlemanly than I am...