2003 AMC 10B Problems/Problem 7

Revision as of 17:55, 26 November 2011 by Gina (talk | contribs) (Solution)

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

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

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AMC 10 Problems and Solutions