AMC 10 · 2002 · #13

Grade 8 number-theory
sequences-arithmeticperfect-squaresfactors convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: perfect-squaressequences-arithmetic
📏 Long solution 💡 3 insights
Problem
Pick a run of 18 whole numbers in a row, all positive, and add them up. Among all runs whose total is a perfect square, what is the smallest total?

Pick an answer.

(A)
169
(B)
225
(C)
289
(D)
361
(E)
441
How to solve
Strategy Introduce a Variable

Every run of 18 consecutive integers is decided by one number: where it starts. Tool #4 (Introduce a Variable) names that starting number a and turns the whole family of runs into a single formula for the total. Tool #15 (Organize Information in More Ways) then rewrites that formula in factored form, which is where the perfect-square condition becomes readable. Tool #9 (Solve an Easier Related Problem) shrinks the question "is this total a square?" to a much smaller question about one factor. Tool #14 (Extreme Principle) picks the smallest legal value of that factor, and Tool #6 (Guess and Check) confirms the winner by adding the actual run.

1STEP 1

Name the starting number

With a as the smallest, the eighteen terms total 18a + 153.

S = a + (a+1) + … + (a+17) = 18a + 153
2STEP 2

Factor out the nine

Both parts are multiples of nine, so the total is 9(2a+17).

S = 18a + 153 = 9(2a+17)
3STEP 3

Shrink the square condition

Since nine is already a square, the total is a square exactly when 2a+17 is a square.

9m = k² ⇔ m = j² (k = 3j)
4STEP 4

Take the smallest legal square

That value is odd and at least 19, so the smallest legal square is 25.

2a+17 odd, 2a+17 ≥ 19 → 2a+17 = 25
5STEP 5

Check the winning run

Then a = 4 and the run 4 through 21 totals 225, choice (B).

a = 4: 4+5+…+21 = 9 · 25 = 225 = 15² → (B)
Answer
225
Test the answer choices directly against S = 9(2a+17). A valid sum must be 9 times an odd number. Choice (A) 169 = 13² is not even a multiple of 9, and it is also smaller than the minimum possible sum 171, so it fails twice. Choice (B) 225 = 9 · 25 and 25 is odd, giving a = 4 — valid. Of the rest, 289 = 17² and 361 = 19² are not multiples of 9 at all, so no run produces them, while 441 = 9 · 49 is reachable with a = 16. All five choices are perfect squares, but only 225 and 441 are actually sums of such a run, and 225 is the smaller. The size is sensible too: 18 numbers near 12 each should total roughly 216, and 225 sits right there.
💡Key takeaway

Eighteen numbers in a row always add to 9 times an odd number, so the sum is a square exactly when that odd number is one.

  • Name the starting number
  • Factor out the nine
  • Shrink the square condition
  • Take the smallest legal square
  • Check the winning run