AMC 10 · 2007 · #18

Grade 9 number-theoryalgebra
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
place-valueperfect-squaresdigit-decomposition convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: place-valueperfect-squares
📏 Medium solution 💡 3 insights
Problem
A three-digit number sits one third of the way between consecutive squares, and the same digits with the last two swapped sit two thirds of the way. Find the sum of the digits.

Pick an answer.

(A)
10
(B)
13
(C)
16
(D)
18
(E)
21
How to solve
Strategy Introduce a Variable

The problem never says which two squares are involved, so Tool #4 (Introduce a Variable) supplies the missing name: let k be the smaller root. Once the stretch is k² to (k+1)², its length is a single quantity, 2k+1. The real leverage is that the distance between the two marked numbers can be measured in two completely different ways — as one third of the gap between the squares, and directly from place value as 9(c-b) — so Tool #13 (Convert to Algebra) turns the word problem into the single equation 2k+1=27(c-b). That equation has two unknowns, so Tool #3 (Eliminate Possibilities) closes it with two independent squeezes: a size bound from "three digits" and a parity bound from "2k+1 is odd". Those steps only prove what k must be; the last step builds the actual numbers and checks the digit-swap condition really holds, which is what makes the answer exist rather than merely be forced. Tool #2 (Make a Systematic List) provides the independent cross-check in review, since only eight values of k are even eligible.

1STEP 1

Name the smaller root, measure the gap

The gap between consecutive squares is an odd number.

(k+1)²-k²=(k²+2k+1)-k²=2k+1
2STEP 2

Cut the gap into three whole thirds

Cutting it into whole thirds is already a restriction.

t=(2k+1)/3, abc=k²+t, acb=k²+2t, acb-abc=t
3STEP 3

Read the same difference off the digits

The digits give the same difference a second way.

9(c-b)=acb-abc=(2k+1)/3 ⟹ 2k+1=27(c-b)
4STEP 4

Squeeze the digit difference down to 1

Size and parity squeeze the digit gap down to one.

k ≤ 31 → 27d=2k+1 ≤ 63 → d ≤ 2; 27d odd → d odd; d=1 → 2k+1=27 → k=13
5STEP 5

Build the numbers and check the swap

Building the numbers confirms the swap, so the sum is 16, choice (C).

13²=169, 14²=196, (196-169)/3=9; abc=169+9=178, acb=169+18=187 ✓; a+b+c=1+7+8=16 → (C)
Answer
16
Check the two placements directly against the definition. From 169 to 196: 178-169=9 and 196-178=18, a 1:2 split, so 178 really is one third of the way. And 187-169=18 with 196-187=9, a 2:1 split, so 187 really is two thirds of the way. The digits behave too: 178 has a=1 ≠ 0, and reversing its last two digits gives exactly 187. The digit sum 1+7+8=16 is choice (C). The trap worth naming is choice (A): the smallest eligible k is 10, which gives 107 and 114; a solver who stops at "the gap is divisible by 3" without testing the digit swap reads off 1+0+7=10 and picks (A). The swap test is what rules it out, since reversing the last two digits of 107 gives 170, not 114.
💡Key takeaway

Measure the same difference two ways — once from the squares and once from the digits — to pin down the only candidate, then build the number and check the condition you had been assuming.

  • Name the smaller root, measure the gap
  • Cut the gap into three whole thirds
  • Read the same difference off the digits
  • Squeeze the digit difference down to 1
  • Build the numbers and check the swap

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