AMC 10 · 2012 · #13

Grade 6 algebra
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
weighted-averagesequences-geometricoptimization extremal-constructionconvert-to-algebra ↑ Prerequisites: fraction-arithmeticmean-median-mode-range
📏 Medium solution 💡 3 insights
Problem
Take the numbers 1, 2, 3, 4, 5 and line them up in some order. Average the first two. Average that result with the third number. Average that with the fourth. Average that with the fifth. Different orderings give different final numbers. Find the gap between the biggest possible final number and the smallest possible final number.

Pick an answer.

(A)
$\frac{31}{16}$
(B)
2
(C)
$\frac{17}{8}$
(D)
3
(E)
$\frac{65}{16}$

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

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.

1STEP 1

Name the order and chain the averages

Call the order a, b, c, d, e. The first mean is (a+b)/2, and each later mean averages the running value with the next number.

m₁=(a+b)/2, m₂=(m₁+c)/2, m₃=(m₂+d)/2, m₄=(m₃+e)/2
2STEP 2

Unfold the nesting into slot weights

Substituting each mean into the next halves everything already there, so the five slots carry weights 1, 1, 2, 4, 8 over 16.

m₄=a/16+b/16+c/8+d/4+e/2 = (1 · a+1 · b+2 · c+4 · d+8 · e)/16
3STEP 3

Heaviest slots get the biggest numbers

Extreme Principle: give the biggest numbers to the heaviest slots — 5, 4, 3 on weights 8, 4, 2 — so the maximum is 65/16.

Max=(1·1+1·2+2·3+4·4+8·5)/16=(1+2+6+16+40)/16=65/16
4STEP 4

Reverse it for the smallest value

Same weights, opposite pairing: 1, 2, 3 on weights 8, 4, 2 and 4, 5 on the light slots, giving a minimum of 31/16.

Min=(1·4+1·5+2·3+4·2+8·1)/16=(4+5+6+8+8)/16=31/16
5STEP 5

Subtract to get the gap

Both share denominator 16, so the gap is 65/16 - 31/16 = 34/16 = 17/8, choice (C).

65/16-31/16=34/16=17/8 → (C)
Answer
17/8
Sanity of the weights: 1/16+1/16+2/16+4/16+8/16=16/16=1, so m₄ is a genuine weighted average and must land between 1 and 5 — both 65/16≈4.06 and 31/16≈1.94 do. The max is pulled high (near 4) because 5 sits on the heaviest slot, and the min is pulled low (near 2) because 1 does — exactly the expected directions. The gap 17/8=2.125 is comfortably less than the full spread 5-1=4, which makes sense since the two light slots dampen the extremes.
💡Key takeaway

This 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

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