AMC 10 · 2011 · #13

Grade 7 logicalgebra
systems-of-equationspair-countinglogical-deduction systematic-enumeration ↑ Prerequisites: systems-of-equations
📏 Medium solution 💡 3 insights
Problem
Four decreasing whole numbers have a known total and a known set of six pairwise differences. Add the possible largest values.

Pick an answer.

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

Three neighbouring gaps describe everything.

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

Add up all six differences

Adding all six pins the middle gap at 1.

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

Find the two outer gaps

The two outer gaps must be 3 and 5.

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

Each arrangement gives its own largest value.

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

Add the possible values of w

Their sum is 31, choice (A).

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