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Eight weight combination

by look416
Tags: combination, weight
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May3-12, 01:48 AM
P: 87
Question :
Please give 8 weights which can weigh from 1 ... to 3280.
you can put these 8 weights on either side of the balance, but you only can use once.

have done few approach, most nearest one is
1,2,3,13,39,117,351,1053 but adds up to 1755 only,which mean i can only weigh up to 1755g only
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May3-12, 03:56 AM
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Hurkyl's Avatar
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I see two different ways to measure a weight of 2. That's probably wasteful. How do you measure 20, by the way?
May3-12, 04:18 AM
P: 688
Quote Quote by Hurkyl View Post
How do you measure 20, by the way?
39 on one side, 1+2+3+13 (plus the 20) on the other. (Thank you, computer.)

Quote Quote by Hurkyl View Post
(Signature:) "Any sufficiently advanced technology..."
You may want to add the Hollywood scriptwriting rule, "Any sufficiently advanced technology is indistinguishable from an arbitrary plot device".

Edit: Dang computer... now I have a solution, but I don't know why it works. :)

May3-12, 05:46 AM
P: 87
Eight weight combination

Quote Quote by Hurkyl View Post
I see two different ways to measure a weight of 2. That's probably wasteful. How do you measure 20, by the way?
mind share your ways to measure them?
for object weights 20,dodo just solve it for me ^^
May3-12, 08:22 AM
P: 688
It's rather obvious, I think; how did you come up with that collection of numbers, otherwise?
One way: 2 on one side, the measured object on the other.
Another: 3 on one side, 1 plus the measured object on the other.

Hurkyl is giving you a way of looking for your solution: strive for numbers such that each measured weight can only be weighted in one way. After a while, a pattern should emerge.
May3-12, 09:13 AM
P: 87
which mean my 2 is not needed?
May3-12, 09:29 AM
P: 688
Quote Quote by look416 View Post
which mean my 2 is not needed?
Probably. (But then you cannot measure 5.) So keep trying until there is only one way of measuring 1,2,3,... maybe up to 20 or 30, unless you find something before that.
May4-12, 07:09 AM
coolul007's Avatar
P: 234
Eureka, 3280 = (3^8 -1)/2. therefor the weights would be powers of 3, 1,3,9,27,..., 2187
May4-12, 02:19 PM
P: 688
I think I know now why this works: because -1,0,1 form a complete set of residues modulo 3.

In other words: when expressing a natural number in base 3, you use the digits 0,1,2. But the digit 2 can be expressed as 10-1 (base 3); that is, 1's on two digit positions. As a result, the number can be expressed by adding and subtracting powers of 3.

For example, take the number 16, which is 121 = 100 + 20 + 1 (base 3). The 20 can be expressed as 100-10, so 121 = 100 + 100 - 10 + 1 = 200 - 10 + 1 (base 3). Now the 200 is 1000-100, so 121 = 1000 - 100 - 10 + 1 (base 3); or, in decimal, 16 = 27 - 9 - 3 + 1, or 16 + 9 + 3 = 27 + 1. So, to measure 16, put it together with 9 and 3 on one side, and put 27 and 1 on the other side.
Jul9-12, 02:13 AM
P: 45
I'm not sure what you mean by "but you only can use once", so this could be wrong.

Would you need to only weigh exactly either only odd or only even weights? For example, if could weight 35 and 37, and the item you were weighing was 36, you could try it on the 35 balance, see it's too heavy, try it on the 37 balance, see it's too light, and conclude it weighs 36?

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