2003 AMC 10B Problems/Problem 7

Problem

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 $\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor.$

$\textbf{(A) } 35 \qquad\textbf{(B) } 38 \qquad\textbf{(C) } 40 \qquad\textbf{(D) } 42 \qquad\textbf{(E) } 136$

Solution

The first three values in the sum are equal to $1,$ the next five equal to $2,$ the next seven equal to $3,$ and the last one equal to $4.$ For example, since $2^2=4$ any square root of a number less than $4$ must be less than $2.$ Sum them all together to get

$3\cdot1 + 5\cdot2 + 7\cdot3 + 1\cdot4 = 3+10+21+4 = \boxed{\textbf{(B) \ } 38}$

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2003 AMC 10B Problems/Problem 7

Problem

Solution

See Also