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2003 AMC 10B Problem 7

Problem 7 of 25EasierAlgebraNumber Theory

The symbolism ⌊x⌋\lfloor x \rfloor denotes the largest integer not exceeding x.x. For example, ⌊3⌋=3,\lfloor 3 \rfloor = 3, and ⌊92⌋=4.\lfloor \frac{9}{2} \rfloor = 4. Compute ⌊1⌋+⌊2⌋+⌊3⌋+⋯+⌊16⌋. \begin{gathered} \lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor \\ {}+ \cdots + \lfloor \sqrt{16} \rfloor. \end{gathered}

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Solution

The value is 11 for n=1,2,3;n=1,2,3; it is 22 for n=4,…,8;n=4,\ldots,8; it is 33 for n=9,…,15;n=9,\ldots,15; and it is 44 for n=16.n=16. The sum is 3⋅1+5⋅2+7⋅3+1⋅4=38.3\cdot 1 + 5\cdot 2 + 7\cdot 3 + 1\cdot 4 = 38. Thus, the correct answer is B.
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Tagged: floor and ceiling functions · perfect square · summation

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