AMC 10 · 2012 · #8
Grade 6 countingPick an answer.
The question asks for a largest and a smallest value, so this is a max/min problem — Tool #14 (Extreme Principle) is the spine. But you cannot compare orderings until you know how much each slot actually counts. So first use Tool #4 (Introduce a Variable): call the chosen order a,b,c,d,e and write the chain of averages. Then Tool #5 (Look for a Pattern): unfold the nested averages to reveal a fixed weight on each slot. Once the weights are exposed, Tool #14 makes the extremes obvious — to make the total as large as possible you put the biggest numbers on the heaviest slots, and to make it smallest you do the reverse. Tool #7 (Identify Subproblems) keeps the finish clean: compute the max, compute the min, then subtract.
Name the order and chain the averages
The chain of averages unfolds into weights.
Naming the order with letters turns a vague "some arrangement" into an exact expression you can push on.
6.SP.A.3Introduce A VariableUnfold the nesting into slot weights
Each later position is twice as heavy.
The number folded in last never gets re-halved, so later slots weigh far more — the weights double down the line: 1,1,2,4,8.
The number folded in last never gets halved again, so the slot weights double down the line.
▸ Why?
Each averaging step halves everything already inside, so a weight is multiplied by the same factor each time.
▸ Why?
An average is a total shared over a count, so each step is exactly one halving of the running total.
Heaviest slots get the biggest numbers
Big numbers on big weights give the largest value.
A dollar counts most where the multiplier is biggest — so spend your largest values on your heaviest weights.
6.NS.C.7Extreme PrincipleReverse it for the smallest value
Reversing gives the smallest.
Flipping the pairing sends the least value to the heaviest slot, dragging the weighted average as low as it can go.
6.NS.C.7Extreme PrincipleSubtract to get the gap
Subtracting gives 17/8, choice (C).
Same denominators means the gap is just the numerator difference over 16, then reduce.
5.NF.A.1Identify SubproblemsThis chained averaging is really a weighted average where the last number counts most: the slot weights are 1,1,2,4,8 over 16. Put the biggest numbers on the heaviest slots for the max (65/16), flip it for the min (31/16), and the gap is 17/8.
- Name the order and chain the averages
- Unfold the nesting into slot weights
- Heaviest slots get the biggest numbers
- Reverse it for the smallest value
- Subtract to get the gap