## Puzzle #182: Weights and Balance Once More

Somehow puzzles on weights and balances just never end! I found this great one in Martin Gardner’s The Colossal Book of Short Puzzles and Problems.

Five objects, no two the same weight, are to be ranked in order of increasing weight. You have available a balance scale but no weights. How can you rank the objects correctly in no more than seven separate weighings?

For two objects, of course, only one weighing is required. Three objects require three weighings.

For younger children, please do try the problem with four objects, where you are allowed no more than 5 weighings.

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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### 3 Responses to Puzzle #182: Weights and Balance Once More

1. Prakhar says:

Alok Case of four weights:

My son Anahat and I worked on it together – with 75% of work done by him for the case of four weights. We started with the case of 3 weights named A, B, and C. It takes three trials on the balance to get it. So let us say we now know their ranking which is A>B>C.

We introduce the fourth weight D and knowing that two trials are left. First, we compare D with B. If D is less than B, we compare again with C to be certain that D is either > than or less than C. This will provide a solution such as A,B,C,D or A,B,D,C. Likewise, D could turn out to be > than B, in which case the argument can be similarly made involving A, leading to solutions such as D>A>B>C or A>D>B>C

Anahat suggested another approach for this case with A/B and C/D as two trials. Say A>B and C>D.

Next, B will be compared with D.

If B is less than D, then we know that the order is A or C >D>B . In the fourth trial C could be compared with A. If A > C, then the answer is A>C>D>B. If C is > than A, we need to compare A with D to see if it is C>A>D>B or C>D>A>B

If B is more than D, the possibilities are A>B>C>D or A>C>B>D or C>A>B>D. Therefore B is compared with C in trial number 4. If it is greater than C, we have found a solution in four trials, else in 5 trials by comparing A with C

Case of five weights

This has been tried by Anahat .

Step 1: C> D

Step 2: A>B

Step 3: A> C

We now know
A>C>D

Step 4 :B> E

Step 5

Compare B with C

If B> C, then A>B>C>D

Step 6,7

Compare E with C and D to get the right answer within 7 steps

If E is greater than C and the previous solution in step 5 was A,B,C,D no further comparison is done and the answer is A,B,E,C,D. But if B is greater than A in Step 5, we need to compare E with A for a final answer in 7 steps

If E is less than C, then from the previous solution in step 5, we need to compare E with D and get the final answer in 7 steps.

If B< C, then

Step 6,7

Compare B with D in Step 6 to get A,C,B,D or A,C,D,B

2. Pratik Poddar says:

For 2 weights A and B, you need 1 weighing.

For 3 weights A, B and C, you need 1 weighing to arrange A and B. To place C, you would need 2 more weighings.

For 4 weights, A, B, C and D
Lets arrange A, B and C – It would take 3 weighings. Without loss of generality, lets say it was A>B>C.

To place D, compare D and B. and then depending on > or B>C>D.

To place E, compare E and B. Depending on result, compare E and A, or E and C. Depending on result of E and C, you might need to do E and D as well.

Number of weighings needed = 8

I do not see how it can be done in 7 in all cases!

3. Pratik Poddar says:

Some wordpress formatting error

For 2 weights A and B, you need 1 weighing.

For 3 weights A, B and C, you need 1 weighing to arrange A and B. To place C, you would need 2 more weighings.

For 4 weights, A, B, C and D
Lets arrange A, B and C – It would take 3 weighings. Without loss of generality, lets say it was A>B>C.

To place D, compare D and B. and then depending on gt or lt, compare A and D or B and D.

For 5 weights, A, B, C, D and E.
Lets arrange A, B, C and D – It would take 5 weighings. Without loss of generality, lets say A>B>C>D.

To place E, compare E and B. Depending on result, compare E and A, or E and C. Depending on result of E and C, you might need to do E and D as well.

Number of weighings needed = 8

I do not see how it can be done in 7 in all cases!