AMC 10 · 2012 · #13
Grade 6 algebraAn iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The five numbers used are exactly $1, 2, 3, 4, 5$, each once; "Average" of two numbers means their mean: $\frac{x+y}{2}$; The averaging is done in a chain: first two, then bring in the 3rd, then the 4th, then the 5th; The order in which the five numbers are placed is ours to choose; Answer choices: (A) $\frac{31}{16}$, (B) $2$, (C) $\frac{17}{8}$, (D) $3$, (E) $\frac{65}{16}$
Unknowns: The difference (largest final value) $-$ (smallest final value)
Understand
Restated: 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.
Givens: The five numbers used are exactly $1, 2, 3, 4, 5$, each once; "Average" of two numbers means their mean: $\frac{x+y}{2}$; The averaging is done in a chain: first two, then bring in the 3rd, then the 4th, then the 5th; The order in which the five numbers are placed is ours to choose; Answer choices: (A) $\frac{31}{16}$, (B) $2$, (C) $\frac{17}{8}$, (D) $3$, (E) $\frac{65}{16}$
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #5 Look for a Pattern, #7 Identify Subproblems
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.
Execute — Answer: C
6.SP.A.3 Step 1 Name the order and chain the averages
- Let the chosen order be $a, b, c, d, e$.
- The procedure builds four running means.
- Averaging two numbers just means adding them and halving.
- Writing each step in order: the mean of the first two is $m_1=\frac{a+b}{2}$; then $m_2=\frac{m_1+c}{2}$; then $m_3=\frac{m_2+d}{2}$; and finally $m_4=\frac{m_3+e}{2}$, which is the value the procedure returns.
💡 Naming the order with letters turns a vague "some arrangement" into an exact expression you can push on.
5.NF.B.4 Step 2 Unfold the nesting into slot weights
- Substitute each mean into the next.
- Every time you fold in a new number, everything already accumulated gets halved again, while the new number enters at half.
- So $m_2=\frac{a+b}{4}+\frac{c}{2}$, then $m_3=\frac{a+b}{8}+\frac{c}{4}+\frac{d}{2}$, and finally $m_4=\frac{a+b}{16}+\frac{c}{8}+\frac{d}{4}+\frac{e}{2}$.
- Reading off the coefficients, the five slots carry weights $\frac{1}{16},\frac{1}{16},\frac{2}{16},\frac{4}{16},\frac{8}{16}$ — and these add to $1$, as a weighted average should.
💡 The number folded in last never gets re-halved, so later slots weigh far more — the weights double down the line: $1,1,2,4,8$.
6.NS.C.7 Step 3 Heaviest slots get the biggest numbers
- Tool #14 (Extreme Principle): the total is $\frac{1a+1b+2c+4d+8e}{16}$ with fixed weights $1,1,2,4,8$.
- To make it as large as possible, hand the biggest numbers to the heaviest slots.
- Put $5$ on the weight-$8$ slot ($e$), $4$ on weight-$4$ ($d$), $3$ on weight-$2$ ($c$), and $1,2$ on the two weight-$1$ slots (their order does not matter).
💡 A dollar counts most where the multiplier is biggest — so spend your largest values on your heaviest weights.
6.NS.C.7 Step 4 Reverse it for the smallest value
- Same weights, opposite assignment.
- To make the total as small as possible, hand the smallest numbers to the heaviest slots: $1$ on weight-$8$ ($e$), $2$ on weight-$4$ ($d$), $3$ on weight-$2$ ($c$), and $4,5$ on the two weight-$1$ slots.
💡 Flipping the pairing sends the least value to the heaviest slot, dragging the weighted average as low as it can go.
5.NF.A.1 Step 5 Subtract to get the gap
- The largest and smallest possible values are $\frac{65}{16}$ and $\frac{31}{16}$.
- They already share the denominator $16$, so subtract the numerators and simplify.
- The difference is $\frac{34}{16}=\frac{17}{8}$, which is choice (C).
💡 Same denominators means the gap is just the numerator difference over $16$, then reduce.
6.SP.A.3 Let the chosen order be $a, b, c, d, e$. The procedure builds four running means 5.NF.B.4 Substitute each mean into the next. Every time you fold in a new number, everyth 6.NS.C.7 Tool #14 (Extreme Principle): the total is $\frac{1a+1b+2c+4d+8e}{16}$ with fixe 6.NS.C.7 Same weights, opposite assignment. To make the total as small as possible, hand 5.NF.A.1 The largest and smallest possible values are $\frac{65}{16}$ and $\frac{31}{16}$ Review
Reasonableness: Sanity of the weights: $\frac{1}{16}+\frac{1}{16}+\frac{2}{16}+\frac{4}{16}+\frac{8}{16}=\frac{16}{16}=1$, so $m_4$ is a genuine weighted average and must land between $1$ and $5$ — both $\frac{65}{16}\approx4.06$ and $\frac{31}{16}\approx1.94$ 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 $\frac{17}{8}=2.125$ is comfortably less than the full spread $5-1=4$, which makes sense since the two light slots dampen the extremes.
Alternative: Tool #2 (Make a Systematic List) as a spot-check: only the last three slots (weights $2,4,8$) matter for the extremes, since the two weight-$1$ slots always soak up whatever two numbers are left. For the max, testing $e=5$ against $e=4$ directly shows $e=5$ wins by $\frac{8}{16}$, and cascading the same logic down fixes $d=4$, $c=3$ — confirming $\frac{65}{16}$ without algebra. The mirror check confirms $\frac{31}{16}$, and the difference is again $\frac{17}{8}$.
CCSS standards used (min grade 6)
6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Reading the procedure as repeated means — averaging two numbers is $\frac{x+y}{2}$ — and chaining those means with named slots $a,b,c,d,e$.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Unfolding the nested averages, where each fold halves everything accumulated so far, to expose the fixed slot weights $\frac{1}{16},\frac{1}{16},\frac{2}{16},\frac{4}{16},\frac{8}{16}$.)6.NS.C.7Understand ordering and absolute value of rational numbers (Applying the Extreme Principle: ordering the slot weights and pairing the largest numbers with the heaviest slots to maximize, and the reverse pairing to minimize.)5.NF.A.1Add and subtract fractions with unlike denominators (Subtracting the two extreme values $\frac{65}{16}-\frac{31}{16}$ and reducing $\frac{34}{16}$ to $\frac{17}{8}$.)
⭐ 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 ($\frac{65}{16}$), flip it for the min ($\frac{31}{16}$), and the gap is $\frac{17}{8}$.
⭐ 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 ($\frac{65}{16}$), flip it for the min ($\frac{31}{16}$), and the gap is $\frac{17}{8}$.
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