AMC 10 · 2011 · #21

Grade 7 logicalgebra
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
systems-of-equationspair-countinglogical-deduction systematic-enumeration ↑ Prerequisites: systems-of-equations
📏 Medium solution 💡 3 insights
Problem
Brian has four integers in strict decreasing order, w ＞ x ＞ y ＞ z, that add up to 44. If you take the positive difference of every pair of them, the six results are exactly 1, 3, 4, 5, 6, and 9. Add up every value of w that is possible.

Pick an answer.

(A)
16
(B)
31
(C)
48
(D)
62
(E)
93

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The four numbers only matter through their gaps, so Tool #4 (Introduce a Variable) names the three neighbor gaps a=w-x, b=x-y, c=y-z. Every one of the six pairwise differences is then a run of consecutive gaps, which turns the puzzle into a couple of small equations. Adding all six differences pins down the middle gap b instantly. Tool #2 (Make a Systematic List) then sorts the leftover differences to find the outer gaps, and Tool #3 (Eliminate Possibilities) rules out the arrangements that don't give integers, leaving just the two number sets whose w values we add.

1STEP 1

Name the three neighbor gaps

Track the gaps, not the numbers: a = w-x, b = x-y, c = y-z. The widest span is w-z, so a+b+c = 9.

a=w-x, b=x-y, c=y-z; w-z=a+b+c=9
2STEP 2

Add up all six differences

In gap form the six differences add to 3(a+b+c) + b = 27 + b, and the given six add to 28, so the middle gap is b = 1.

3(a+b+c)+b = 27+b = 28 → b = 1
3STEP 3

Find the two outer gaps

So a + c = 8, and a, c, a+1, c+1 must fill the leftovers 3, 4, 5, 6 — only a and c equal 3 and 5 fits.

a+c=8, {a, c, a+1, c+1}={3,4,5,6} → {a,c}={3,5}
4STEP 4

Turn each gap pattern into numbers

Put x=w-a, y=w-a-b, z=w-9 into the total 4w-(3a+2b+c)=44: the order 3,1,5 gives w=15 and 5,1,3 gives w=16.

4w-(3a+2b+c)=44: w=15 or w=16
5STEP 5

Add the possible values of w

Both (15,12,11,6) and (16,11,10,7) really check out, so the possible w add to 15 + 16 = 31, choice (B).

15 + 16 = 31
Answer
31
Both candidate sets pass every test: strictly decreasing integers, sum 44, and pairwise differences exactly {1,3,4,5,6,9}. The two tops 15 and 16 differ by 1, which matches b=1 being the only gap that changed roles between the two orderings. Their sum 31 is choice (B); the trap answer 16 (choice A) is just one of the two w values, and 48, 62, 93 are far too big for a top number that can be at most about 44/4+9.
💡Key takeaway

Track the gaps between neighbors instead of the numbers themselves, and every pairwise difference becomes a short sum you can solve.

  • Name the three neighbor gaps
  • Add up all six differences
  • Find the two outer gaps
  • Turn each gap pattern into numbers
  • Add the possible values of w

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