Difference between revisions of "1995 AIME Problems/Problem 13"
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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>. | 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 n 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 == |
Revision as of 13:36, 24 July 2019
Contents
[hide]Problem
Let be the integer closest to Find
Solution
When , then . 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 n 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 |
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