AMC 10 · 2010 · #14

Grade 7 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

optimizationextremal-constructionlinear-equations-one-var extreme-principlebound-inequality-then-enumerate ↑ Prerequisites: optimization

📏 Long solution 💡 3 insights

Problem

Five positive whole numbers add to a fixed total, and the largest neighbouring sum is to be made small. Find that smallest largest sum.

Pick an answer.

(A)

670

(B)

671

(C)

802

(D)

803

(E)

804

How to solve

Strategy Extreme Principle

Tool #14 (Extreme Principle): every "smallest possible maximum" question is really two questions stacked, and answering only one of them is the standard way to get this wrong. You must prove no arrangement beats some number, and separately build an arrangement that reaches it. Tool #4 (Introduce a Variable): treating M as a symbol turns the word "largest" into four inequalities you can add together, which is where all the leverage is. Tool #9 (Easier Problem): temporarily allow the five numbers to be 0. That relaxed version has a clean answer, and comparing the two versions shows exactly which single hypothesis the real answer hangs on — it is the word "positive", nothing else. Tool #15 (Organize Differently): the same total 2010 can be split into blocks in more than one way, and a second, differently shaped split confirms the floor without reusing the first argument's key step. Tool #3 (Eliminate): the five choices fall into two clusters, {670, 671} and {802, 803, 804}, and each cluster is the answer to a specific wrong reading — naming those readings is a real check, not a guess.

1STEP 1

Turn the largest into four caps

The largest becomes four separate caps.

a+b ≤ M, b+c ≤ M, c+d ≤ M, d+e ≤ M

2STEP 2

Solve the version without the rule

Ignoring positivity gives a first, lower guess.

(670,0,670,0,670): 670+0=670, 0+670=670, 670+0=670, 0+670=670

3STEP 3

Positivity caps the middle number

Positivity squeezes the middle number.

2010=(a+b)+c+(d+e) ≤ M+(M-1)+M=3M-1

4STEP 4

Solve the inequality for M

The inequality lifts the floor to 671.

3M ≥ 2011 → M ≥ 2011/3=670.33… → M ≥ 671

5STEP 5

Build an arrangement that hits 671

A real arrangement reaches that floor.

669+1=670, 1+670=671, 670+1=671, 1+669=670

6STEP 6

Check the floor a second way

A second grouping confirms the same floor.

2011 ≤ 2010+d=(a+b)+(c+d)+(d+e) ≤ 3M

7STEP 7

Rule out the other choices

So the answer is 671, choice (B).

M_min=671 → (B)

Answer

671

The answer sits exactly one unit above 670, the zeros-allowed optimum, which is the cheapest penalty positivity could possibly charge — it forces a single unit into a valley, and that unit is counted inside a capped pair. A direct feasibility count confirms both directions at once. For a fixed cap t, the largest total the five numbers can reach is obtained by making the valleys as small as allowed, b=d=1, which lets a, c, e each reach t-1, for a maximum total of (t-1)+1+(t-1)+1+(t-1)=3t-1. At t=670 that ceiling is 2009, short of 2010, so a cap of 670 is impossible; at t=671 it is 2012, and totals can be lowered one at a time from there, so 2010 is reachable. Both facts point at 671, choice (B). The answer is also far below the even-split value 804, which fits the fact that only neighbours are penalised.

💡Key takeaway

Only neighbours get added, so the winning shape is big, small, big, small, big. If zeros were allowed you could reach 670 with (670,0,670,0,670); because every number must be at least 1, that forced unit sits inside a pair that is already capped and pushes the best possible largest sum up by exactly one, to (B) 671 — and (669,1,670,1,669) shows 671 really happens.

Turn the largest into four caps
Solve the version without the rule
Positivity caps the middle number
Solve the inequality for M
Build an arrangement that hits 671
Check the floor a second way
Rule out the other choices