Thanks to Suman Saraf for recently pointing me to NSA (National Security Agency’s) monthly puzzle blog. Here is a nice and simple problem that they posted in their April blog. Worth trying with your children!
This puzzle created by Andy F., Applied Research Mathematician, NSA
Mel has four weights. He weighs them two at a time in all possible pairs and finds that his pairs of weights total 6, 8, 10, 12, 14, and 16 pounds. How much do they each weigh individually?
As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.
Happy weighings!
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There will be two solutions
First -1,5,7,9
Second-2,4,6,10
Let us consider a<b<c<d.
So four fixed equations are formed as follows-
a+b=6
a+c=8
b+d=14
c+d=16
But the other two equations are variable
(i)b+c=10 and a+d=12
(ii)a+d=10 and b+c=10
If we solve the set(i) we will get a=2 ,b=4,c=6,d=10
and set(ii) we will get a=1,b=5,c=7,d=9
Two observations that we can make after looking at the weights:
1. All weights are even or odd
2. No weight is repeated
Using brute force further, i observed two possible cases-
1,5,7,9 and 2,4,6,10
Edit: observations that we can make after looking at the sums of weights
Since each of these weights is previous weight + 2, we can choose odd numbers which are multiples of two apart; starting with 1. Once we reach the limit of 9 or a weight of 10, we use the next higher weight to reach the greater weights. Therefore i chose 1, 5, 7, and 9. Cannot use 3 because 1+3 is only 4.
1+5 = 6 (5 is 2*2 more than 1)
1+7 = 8 (7 is 2*3 more than 1)
1+9 = 10 (9 is 2*4 more than 1)
5+7 = 12 (this is a combination of added 2s : 2*2 + 2*3)
5+ 9 = 14 ( similar explanation 2*2+2*3)
7+9 = 16 (similar explanation 2*3+2*4)
We can confirm that four equations are formed as follows
a+b=6
a+c=8
b+d=14
c+d=16
But the other two equations can be as follows
(i)b+c=10 and a+d=12
(ii)a+d=10 and b+c=10
If we solve the set(i) we weill get a=2 ,b=4,c=6,d=10
and set(ii) we will get a=1,b=5,c=7,d=9
We can prove that there are only 2 possible cases i.e (1,5,7,9) and (2,4,6,8).
Let the measures of the weights be a,b,c,d and WLOG let a<b<c<d (where < means less than or equal to).
Now using a<b<c<d, we can only have two possibilities. They are:
a) (a+b)<(a+c)<(b+c)<(a+d)<(b+d)<(c+d)
b) (a+b)<(a+c)<(a+d)<(b+c)<(b+d)<(c+d)
In case a) we get the following system
a+b=6….(i)
a+c=8…..(ii)
b+c=10….(iii)
a+d=12….(iv)
b+d=14….(v)
c+d=16….(vi)
Subtracting (ii) from (iii) and adding the resulting equation to (i) we get b=4. From here we can easily calculate value a,c,d. giving us (2,4,6,8)
In case b) we get a similar system as case a) but this time we have
a+d=10….(iii)
b+c=12…..(iv)
Subtracting (ii) from (iv) and adding this to (i) we get b=5. Then we easily get a,c,d. The final answer is (1,5,7,9).
However, It should also be noted that parity arguments do not work in this case. It has not been specified in the question that the magnitudes of the weights are natural numbers. Hence making a statement like 'the weights are all even or all odd' do not hold any weight, since a solution like
a=1.5, b=4.5, c=6.5 d=8.5 needs to also be taken into consideration. The answer conveniently only has natural number solutions but it may not be the case in similar questions.
Let the four weights be x1, x2, x3, x4 in ascending order.
The sums of pairs arranged in ascending order would be:
x1+x2 < x1+x3 < x1+x4 , x2+x3 < x2+x4 < x3+x4
So, x1+x2 = 6 and x2+x4 = 14 and x3+x4=16 and x1+x3=8
x1+x4 is one of 10 or 12.
x2+x3 is one of 10 or 12 as well.
x1+x2+x2+x4 = 20
2*x2 + (x1+x4) = 20
So, x2 is one of 4 or 5.
When x2=4, numbers are 2,4,6,10
When x2=5, numbers are 1,5,7,9
These are the only two solutions possible.
The four weighta a,b,c,d are obviously either just odd or just even. If we try the first consecutive evens because the difference in weights is 2, The 2,4,6,8 pair will never sum more than 14. However, the set 1,5,7,9 weighs correctly to each pair and is the first set when I stop looking for other answers starting with 9,7 as non repeated weights adding to 16 with 1,5 = 6 and 1,7 = 8 and so on. Thus 1,5,7,9 is an answer. Other sets of smaller or higher numbers will have lower or higher pair weights that do not satisfy the answer.
The four weights a a,b,c,d are obviously either just odd or just even. If we try the first consecutive evens because the difference in weights is 2, The 2,4,6,8 pair will never sum more than 14. However, the set 1,5,7,9 weighs correctly to each pair and is the first set when I stop looking for other answers starting with 9,7 as non repeated weights adding to 16 with 1,5 = 6 and 1,7 = 8 and so on. Thus 1,5,7,9 is an answer. Other sets of smaller or higher numbers will have lower or higher pair weights that do not satisfy the answer.