Difference between revisions of "2003 AMC 10B Problems/Problem 7"
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The first three values in the sum are equal to <math>1,</math> the next five equal to <math>2,</math> the next seven equal to <math>3,</math> and the last one equal to <math>4.</math> For example, since <math>2^2=4</math> any square root of a number less than <math>4</math> must be less than <math>2.</math> Sum them all together to get | The first three values in the sum are equal to <math>1,</math> the next five equal to <math>2,</math> the next seven equal to <math>3,</math> and the last one equal to <math>4.</math> For example, since <math>2^2=4</math> any square root of a number less than <math>4</math> must be less than <math>2.</math> Sum them all together to get | ||
| − | <cmath>3\cdot1 + 5\cdot2 + 7\cdot3 + 1\cdot4 = 3+10+21+4 = \boxed{\ | + | <cmath>3\cdot1 + 5\cdot2 + 7\cdot3 + 1\cdot4 = 3+10+21+4 = \boxed{\textbf{(B) \ } 38}</cmath> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2003|ab=B|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:09, 4 July 2013
Problem
The symbolism 
denotes the largest integer not exceeding 
. For example, 
and 
. Compute
![\[\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor.\]](http://latex.artofproblemsolving.com/a/5/8/a58d1e1c98abd63840c9b83573e068bb7408aea6.png)
Solution
The first three values in the sum are equal to 
the next five equal to 
the next seven equal to 
and the last one equal to 
For example, since 
any square root of a number less than 
must be less than 
Sum them all together to get
![\[3\cdot1 + 5\cdot2 + 7\cdot3 + 1\cdot4 = 3+10+21+4 = \boxed{\textbf{(B) \ } 38}\]](http://latex.artofproblemsolving.com/c/f/f/cffbd6bff0dc51c1c4487240a68ffad24446e9b3.png)
See Also
| 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. 
