Neat. You need 1, 3, 9, 27
Put weights on each side:
1: 1 on left, nothing on right
2: 3 on left, 1 on right
3: 3 on left
4: (3,1)()
5: (9)(3,1)
6: (9)(3)
7: (9,1)(3)
8: (9)(1)
9: (9)()
...
13: (9,3,1)
14: (27)(9,3,1)
...
40: (27,9,3,1)
The general rule is you need powers of 3, and the reason is that each weight can be counted on the left 0 times, 1 time, or -1 times.
Neat.
2006-11-16 22:52:34
·
answer #1
·
answered by sofarsogood 5
·
0⤊
0⤋
1, 2, 4, 8, 16, 32 will weigh up to 63 kg in 1 kg increments
20, 10, 5, 2, 2, 1 will weigh up to 40 kg in 1 kg increments
In general, {1, 2, 4, 8, ....., 2^(n-1)} units will provide (2^n) - 1 incremental measurements, which is the most efficient. A "standard" weight set comprised of 1, 2, 2, 5, 10, 10, 25, 50, however, is easier to use for people unfamiliar with powers of 2. A 50 lb weight or a 25 kg weight are the largest that can handled one-handed, so in practice, the progression has to be abandoned in favor of multiple 25 kg or 50 lb weights (32 kg or 64 lb in the binary progression)
2006-11-17 05:22:06
·
answer #2
·
answered by Helmut 7
·
0⤊
0⤋