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| Twenty metal blocks (Posted on 2005-06-02) |
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Twenty metal blocks are of the same size and external appearance; some are aluminum, and the rest are duraluminum, which is heavier. Using a pan balance to determine how many blocks are aluminum, what is the minimum number of weighings to be done?
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Submitted by pcbouhid
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Rating: 3.5000 (2 votes)
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Solution:
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ELEVEN.
The first weight trial is to place one block on each pan. There are two possible outcomes :
1) On the first trial, one of the pans is heavier.
In this event, one of the blocks must be aluminum and the other duraluminum.
We now place both blocks on one pan, and weigh them against pairs of remaining blocks (those being divided into nine pairs arbitrarily).
Any pair of blocks wich outweighs the first pair must consist of two duraluminum blocks. If the first pair is the heavier, then both blocks of the second pair are aluminum. If both pairs balance, the second pair contains one aluminum and one duraluminum block. Thus, for this event, the number of aluminum (or duraluminum) blocks can be determined by at most TEN weigh trials.
2) On the first weigh trial the pans balance.
In this event, both blocks are aluminum or both are duraluminum.
As before, we now place both blocks on one pan and weigh them against pairs of remaining blocks.
Assume that the first k pairs of blocks from the nine pairs have the same weight as the test blocks, and that the (k + 1)st pair tested have a different weight (if k = 9, then all the blocks weigh the same, and so there are no duraluminum blocks. The event in wich k = 0 falls into the general case).
Suppose, for definiteness, that the (k + 1)st pair is heavier than the test blocks (the reasoning wich follows will be quite analogous if the (k + 1)st pair is lighter).
Then the original two blocks, as well as all those of the first k pairs tested, must be aluminum. Therefore, in the 1 + (k + 1) = (k + 2) weight trials already made, we have found (k + 1) pairs of aluminum blocks.
Now we compare the two blocks of the (k + 1)st (heavier) pair. [This is the (k + 3)rd weight trial].
If both blocks are of the same weight, they must both be duraluminum; if they are not of the same weight, one is aluminum and the other is duraluminum.
In either event, we are able, after (k + 3) weight trials, to display a pair of blocks, of wich one is aluminum and the other duraluminum.
By using this pair, we can determine in, at most (8 - k) weight trials, how many duraluminum blocks remain among the 20 - 2(k + 2) = (16 - 2k) unweighed ones, using the tecnique employed in the first event.
The number of weight trials used in this second event is then equal to (k + 3) + (8 - k) = 11.
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