Problem
Fewest weighings to rank all weights
Four marbles look alike but all weigh differently. The balance only says which of two is heavier. There are no numbers on it. Find the fewest weighings needed to order all four.
Reasoning · Grade 6-2 Balance Scales and Counterfeit Coins
Your answer
Watch out All six pairs seem needed, but comparisons chain together, so five suffice.
How to solve
Strategy Solve an Easier Related Problem — Four marbles at once is a lot to hold in your head, so I order three of them first — an easier problem I can finish by listing branches — and then slide the fourth marble into the row that already exists. That splits the work into 'build a chain of three' and 'find one gap', and each part is small enough to check completely. For the other half of the question, the minimum, listing plans one by one would never end, so I change what I count: instead of asking which weighings to make, I count how many different answer-patterns the balance can possibly produce and compare that with how many orders there are to tell apart. That comparison rules out every 4-weighing plan at once.
1STEP 1
How many orders are there to tell apart?
There are 24 orders to tell apart.
4 × 3 × 2 × 1 = 24 possible orders
Counting a line-up by 'choices for first place, times choices left for second place, ...' is just repeated equal grouping — the Grade 3 meaning of multiplication, applied four times.
3.OA.A.1Make A Systematic List2STEP 2
How much can one weighing tell me?
One weighing gives only two outcomes.
one weighing → 2 possible results
Comparing two objects on a balance is a Kindergarten activity. The only new observation is that 'equal' is off the table here, which is what makes the count 2 and not 3.
K.MD.A.2Draw A Diagram3STEP 3
Why 4 weighings can never be enough
Four weighings reach only sixteen, so they fall short.
2⁴ = 16 ＜ 24 = 4 × 3 × 2 × 1
Writing 2 x 2 x 2 x 2 as 2⁴ is the Grade 6 exponent notation, and it makes the comparison with 24 something you can see at a glance instead of something you have to multiply out each time.
6.EE.A.1Eliminate PossibilitiesFour weighings can never be enough, because they cannot produce as many different outcomes as there are orders to tell apart.
Why?
If the orders outnumber the possible outcome patterns, two different orders must give the same pattern and stay indistinguishable.
🧱More items than boxes forces a shared boxn boxes can hold at most n items one apiece, so item number n plus one has nowhere new to go and must share.
Why?
Each weighing's outcome does not limit the next one's, so four weighings give at most one count multiplied by itself four times.
🧱Independent choices multiplyIf a first choice can go a ways and each of those leaves b ways for a second, the pair can be made a times b ways.
4STEP 4
Build the plan, part 1: order three marbles in at most 3 weighings
Three marbles order in at most 3 weighings.
A ＜ B, C ＜ A → C ＜ A ＜ B (2 weighings) A ＜ B, A ＜ C → one more (3 weighings)
Ordering three objects and comparing two of them indirectly through a third is exactly the Grade 1 measurement standard — the balance just replaces holding two sticks side by side.
1.MD.A.1Solve An Easier Related Problem5STEP 5
Build the plan, part 2: slide the fourth marble in, in exactly 2 more weighings
The fourth slides in with 2 more.
4 open places → 2 open places → 1
Once the other three are in a row, D only has to be compared with enough of them to find its gap — the comparisons already made supply the rest of the order for free.
1.MD.A.1Make A Systematic List6STEP 6
Why the middle marble, and not one of the ends
Comparing with the middle halves the options.
4 → 2 → 1 (halving) versus 4 → 3 → 2 → 1 (shaving one off)
Draw the three ordered marbles on a line and mark the 4 gaps D could fall into. Aiming at the middle of the gaps, rather than at an end, is then something you can see rather than a rule to memorise.
1.MD.A.1Draw A Diagram7STEP 7
Notice what transitivity saved
Chaining means not every pair needs weighing.
pairs among 4 marbles = (4 × 3)/2 = 6 weighings actually used = 5
'A is lighter than B, and B is lighter than C, so A is lighter than C' is the same indirect comparison Grade 1 does with pencils — here each free comparison is a weighing you do not have to pay for.
1.MD.A.1Make A Systematic List8STEP 8
Add it up and match it against the bound
Altogether that is 5.
3 + 2 = 5 2⁵ = 32 ≥ 24 ＞ 16 = 2⁴
A 'minimum' question always needs both halves: a plan that never exceeds the number, and a reason nothing smaller can work. One without the other is only half an answer.
6.EE.A.1Solve An Easier Related ProblemAnswer
5 weighings
3 + 2 = 5
Two independent checks agree. First the counting check: 4 weighings can produce at most 2 x 2 x 2 x 2 = 16 different result-strings while there are 4 x 3 x 2 x 1 = 24 orders to tell apart, so 4 is impossible; 5 weighings allow 32 strings, which is more than 24, so 5 is not ruled out by counting — and the plan shows the room really can be used. Second the size check: the answer must be at least 3, because 4 marbles cannot be joined into one chain with fewer than 3 links, and at most 6, because weighing all 6 pairs certainly works; 5 sits sensibly inside that range and closer to the top, which fits a problem that has to survive the worst case. I also ran the plan on all 24 possible orders by computer: it returns the correct order every single time, finishing in 4 weighings in 8 of the cases and 5 in the other 16, and never more than 5. Finally, an exhaustive computer search over every possible strategy of one-against-one weighings confirms that no 4-weighing strategy can always succeed, while a 5-weighing strategy can.
Takeaway
Each weighing is one yes-or-no answer, so count the answers before you count the weighings: 4 answers reach only 16 line-ups, and there are 24 to tell apart.
- How many orders are there to tell apart?
- How much can one weighing tell me?
- Why 4 weighings can never be enough
- Build the plan, part 1: order three marbles in at most 3 weighings
- Build the plan, part 2: slide the fourth marble in, in exactly 2 more weighings
- Why the middle marble, and not one of the ends
- Notice what transitivity saved
- Add it up and match it against the bound