Difference between revisions of "1995 AIME Problems/Problem 13"
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== Problem == | == Problem == | ||
+ | Let <math>f(n)</math> be the integer closest to <math>\sqrt[4]{n}.</math> Find <math>\sum_{k=1}^{1995}\frac 1{f(k)}.</math> | ||
== Solution == | == Solution == | ||
+ | When <math>\left(k - \frac {1}{2}\right)^4 \leq n < \left(k + \frac {1}{2}\right)^4</math>, <math>f(n) = k</math>. Thus there are <math>\left \lfloor \left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4 \right\rfloor</math> values of <math>n</math> for which <math>f(n) = k</math>. Expanding using the [[binomial theorem]], | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4 &= \left(k^4 + 2k^3 + \frac 32k^2 + \frac 12k + \frac 1{16}\right) - \left(k^4 - 2k^3 + \frac 32k^2 - \frac 12k + \frac 1{16}\right)\ &= 4k^3 + k. \end{align*}</cmath> | ||
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+ | Thus, <math>\frac{1}{k}</math> appears in the summation <math>4k^3 + k</math> times, and the sum for each <math>k</math> is then <math>(4k^3 + k) \cdot \frac{1}{k} = 4k^2 + 1</math>. From <math>k = 1</math> to <math>k = 6</math>, we get <math>\sum_{k=1}^{6} 4k^2 + 1 = 364 + 6 = 370</math> (either adding or using the [[perfect square|sum of consecutive squares formula]]). | ||
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+ | But this only accounts for <math>\sum_{k = 1}^{6} (4k^3 + k) = 4\left(\frac{6(6+1)}{2}\right)^2 + \frac{6(6+1)}{2} = 1764 + 21 = 1785</math> terms, so we still have <math>1995 - 1785 = 210</math> terms with <math>f(n) = 7</math>. This adds <math>210 \cdot \frac {1}{7} = 30</math> to our summation, giving <math>{400}</math>. | ||
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+ | ==Solution 2== | ||
+ | This is a pretty easy problem just to bash. Since the max number we can get is <math>7</math>, we just need to test <math>n</math> values for <math>1.5,2.5,3.5,4.5,5.5</math> and <math>6.5</math>. Then just do how many numbers there are times <math>\frac{1}{\lfloor n \rfloor}</math>, which should be <math>5+17+37+65+101+145+30 = \boxed{400}</math> | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1995|num-b=12|num-a=14}} | |
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:25, 12 December 2019
Contents
[hide]Problem
Let be the integer closest to Find
Solution
When , . Thus there are values of for which . Expanding using the binomial theorem,
Thus, appears in the summation times, and the sum for each is then . From to , we get (either adding or using the sum of consecutive squares formula).
But this only accounts for terms, so we still have terms with . This adds to our summation, giving .
Solution 2
This is a pretty easy problem just to bash. Since the max number we can get is , we just need to test values for and . Then just do how many numbers there are times , which should be
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.