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2003 AMC 10B Problem 7

2003 AMC 10B Problem 7AMC 10 Step by Step

Name: Date: September 13, 2026

Number Properties · difficulty 2 of 5 · about 2 min · position P1-10

The symbolism $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$ . For example, $\lfloor 3 \rfloor = 3,$ and $\lfloor 9/2 \rfloor = 4$ . Compute $\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor.$

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

Answer

Key insight

Floor of sqrt(n) equals k for the 2k+1 integers from k^2 to (k+1)^2 - 1, so group the sum as 3 ones, 5 twos, 7 threes, and one 4.

Solution

$\lfloor \sqrt{n} \rfloor = k$ exactly when $k^2 \le n < (k+1)^2$ . Split $1$ through $16$ by the perfect squares $1, 4, 9, 16$ :

$n = 1, 2, 3$ : three terms, each equal to $1$ ;
$n = 4, \dots, 8$ : five terms, each equal to $2$ ;
$n = 9, \dots, 15$ : seven terms, each equal to $3$ ;
$n = 16$ : one term equal to $4$ .

Adding,
$3 \cdot 1 + 5 \cdot 2 + 7 \cdot 3 + 1 \cdot 4 = 3 + 10 + 21 + 4 = 38.$

The answer is $\boxed{\textbf{(B)}\ 38}$ .

Why this works

A floor-of-root sum is constant on the blocks between consecutive perfect squares, and the block for value $k$ has $(k+1)^2 - k^2 = 2k + 1$ members. Grouping equal terms turns sixteen evaluations into four multiplications, and the same block structure handles sums up to any bound (with a short final block when the bound is not one less than a square).

The trap

Miscounting a block's length as 2k instead of 2k+1, or treating 16 as a 3 rather than a 4.

Common mistakes

Miscounting a block's length as $2k$ instead of $2k+1$ , or treating $16$ as a $3$ rather than a $4$ .
Rounding square roots to the nearest integer instead of rounding down; for example, $\lfloor \sqrt{8} \rfloor = 2$ , not $3$ .

Techniques

Organized listing / direct enumeration · Compute small cases, spot the pattern, generalize