The scales puzzle
My first though is to figure out 4 numbers from 1 to 40 that can be added together to obtain any whole numbers from 1 to 40. So, one of the 4 weights must be 1g, which is the smallest weight. Should the next be 2g? There is only one whole number can be obtained by combining 1g and 2g, which is 3g. What about 3g? The combinations of 1g and 3g give us 2 whole numbers, which are 2g (3g-1g) and 4g (1g+3g). What about 4g? But any combination of 1g and 4g won't give me 2g. Now, 1g and 3g are the two of the 4 weights in my mind. And using a similar pattern, I figure out 9g is one of the 4 weights:
Combination of 1, 3, 5:
3-1=2
3+1=4
5-1=4
5-3=2
Obviously, 5g is not a choice since numbers obtained by combining 1,3 and 5 can be obtained by combinations of 1 and 3 as well.
New numbers produced by combining 1, 3 and 6
6-1=5
6+1=7
6+3+1=10
Now we have 1,2,3,4,5,6,7 and 10
New numbers produced by combining 1, 3 and 7
7+3=10
7-1=6
7+1=8
7+3+1=11
7+3-1
7+1-3
Now we have 1,2,3,4,5,6,7,8,9,10,11
New numbers produced by combining 1, 3 and 8
8+1=9
8+3=11
8-1=7
8-3=5
8+3+1=12
8+3-1=10
8+1-3=6
Now we have 1,2,3,4,5,6,7,8,9,10,11,12
New numbers produced by combining 1, 3 and 9
9+1=10
9+3=12
9-1=8
9-3=6
9-3-1=5
9-3+1=7
9-1+3=11
9+1+3=12
New numbers produced by combining 1, 3 and 10
10+1=11
10+3=13
10-1=9
10-3=7
10-3-1=6
10-3+1=8
10-1+3=12
In this case, 5 can not be produced by the combinations of 1, 3 and 10. So 9g should be one of the 4 weights.
Now, I have 1g, 3g and 9g in my mind. The missing one should be 40-(1+3+9) = 27.
So, my answer is 1g, 3g, 9g and 27g.
According to the process above, I don't think we can come up with an alternative solution but I could be wrong because I don't have a very confident way to prove it.
This puzzle may be challenging to some of the high school students. For the students who are not very confident doing math puzzle, I will alter this problem by using smaller number to start with. For example, change the problem to have exactly 2 weights of different amounts that allows them to weigh out 1 to 4 grams. And I might increase the difficulty by increasing the number according to challenge my students according to their ability.
Combination of 1, 3, 5:
3-1=2
3+1=4
5-1=4
5-3=2
Obviously, 5g is not a choice since numbers obtained by combining 1,3 and 5 can be obtained by combinations of 1 and 3 as well.
New numbers produced by combining 1, 3 and 6
6-1=5
6+1=7
6+3+1=10
Now we have 1,2,3,4,5,6,7 and 10
New numbers produced by combining 1, 3 and 7
7+3=10
7-1=6
7+1=8
7+3+1=11
7+3-1
7+1-3
Now we have 1,2,3,4,5,6,7,8,9,10,11
New numbers produced by combining 1, 3 and 8
8+1=9
8+3=11
8-1=7
8-3=5
8+3+1=12
8+3-1=10
8+1-3=6
Now we have 1,2,3,4,5,6,7,8,9,10,11,12
New numbers produced by combining 1, 3 and 9
9+1=10
9+3=12
9-1=8
9-3=6
9-3-1=5
9-3+1=7
9-1+3=11
9+1+3=12
New numbers produced by combining 1, 3 and 10
10+1=11
10+3=13
10-1=9
10-3=7
10-3-1=6
10-3+1=8
10-1+3=12
In this case, 5 can not be produced by the combinations of 1, 3 and 10. So 9g should be one of the 4 weights.
Now, I have 1g, 3g and 9g in my mind. The missing one should be 40-(1+3+9) = 27.
So, my answer is 1g, 3g, 9g and 27g.
According to the process above, I don't think we can come up with an alternative solution but I could be wrong because I don't have a very confident way to prove it.
This puzzle may be challenging to some of the high school students. For the students who are not very confident doing math puzzle, I will alter this problem by using smaller number to start with. For example, change the problem to have exactly 2 weights of different amounts that allows them to weigh out 1 to 4 grams. And I might increase the difficulty by increasing the number according to challenge my students according to their ability.
Good work! Thoughtful process, and it's good to hold back from declaring a proof until you really have a firm, logical basis for it.
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