AMC 10 · 2003 · #7

Grade 8 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation
floor-functionperfect-squarespattern-recognition identify-subproblems ↑ Prerequisites: perfect-squares
📏 Medium solution 💡 2 insights
Problem
The symbol ⌊ x ⌋ means the largest integer that is not bigger than x (round down). Add up ⌊ √(1) ⌋ + ⌊ √(2) ⌋ + … + ⌊ √(16) ⌋, one term for each whole number from 1 to 16, and choose which value the total equals.

Pick an answer.

(A)
35
(B)
38
(C)
40
(D)
42
(E)
136

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Writing out all sixteen square roots and rounding each one is slow and error-prone, so Tool #5 (Look for a Pattern) is the spine: ⌊ √(n) ⌋ does not change until n reaches the next perfect square, so the terms come in flat blocks of equal value. Tool #7 (Identify Subproblems) then splits the sum into those blocks — a block of 1s, a block of 2s, a block of 3s, and the single 4 — each an easy 'value times how many' piece. Tool #3 (Eliminate Possibilities) guards the finish: the trap choice (E) 136 is what you get by adding 1+2+…+16 (forgetting the square root entirely), and other near-misses come from miscounting a block, so grouping keeps the count honest.

1STEP 1

See where the value stays flat

⌊ √(n) ⌋ is the biggest whole number whose square is at most n, so it holds steady until n hits the next perfect square.

⌊ √(n) ⌋ = k ⇔ k² ≤ n ＜ (k+1)²
2STEP 2

Mark the block boundaries with perfect squares

The squares up to 16 are 1, 4, 9, 16, so the value is 1 for n=1–3, 2 for n=4–8, 3 for n=9–15, and 4 at n=16.

1,4,9,16 = 1²,2²,3²,4²
3STEP 3

Count the terms in each block

Each block runs from one square to just below the next, so the counts are 3 ones, 5 twos, 7 threes, and a single 4.

3 ones, 5 twos, 7 threes, 1 four
4STEP 4

Multiply each block and add

Value times count in each block: 1×3=3, 2×5=10, 3×7=21, 4×1=4, and 3+10+21+4=38, choice (B).

1·3 + 2·5 + 3·7 + 4·1 = 3+10+21+4 = 38 → (B)
Answer
38
The sixteen terms are all between 1 and 4, so the total must sit between 16×1=16 and 16×4=64; the answer 38 lands squarely in that range, while (E) 136 is far too big and is exactly 1+2+…+16 — the trap you fall into by forgetting the square root. A quick recount of the blocks 3+5+7+1=16 terms confirms every number from 1 to 16 is used exactly once, so no term was dropped or double-counted, and 38 is choice (B).
💡Key takeaway

A rounded-down square root only jumps at perfect squares, so group the sum into flat blocks between the squares and add value times count.

  • See where the value stays flat
  • Mark the block boundaries with perfect squares
  • Count the terms in each block
  • Multiply each block and add

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