Difference between revisions of "1985 AHSME Problems/Problem 30"
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| − | We can rearrange the equation into <math> 4x^2=40\lfloor x \rfloor | + | We can rearrange the equation into <math> 4x^2=40\lfloor x \rfloor-51 </math>. Obviously, the RHS is an integer, so <math> 4x^2=n </math> for some integer <math> n </math>. We can therefore make the substitution <math> x=\frac{\sqrt{n}}{2} </math> to get |
| − | <cmath> 40\lfloor \frac{\sqrt{n}}{2}\rfloor | + | <cmath> 40\lfloor \frac{\sqrt{n}}{2}\rfloor-51=n </cmath> |
(We'll try the case where <math> x=-\frac{\sqrt{n}}{2} </math> later.) Now let <math> a\le\frac{\sqrt{n}}{2}<a+1 </math> for an integer <math> a </math>, so that <math> \lfloor \frac{\sqrt{n}}{2}\rfloor=a </math>. | (We'll try the case where <math> x=-\frac{\sqrt{n}}{2} </math> later.) Now let <math> a\le\frac{\sqrt{n}}{2}<a+1 </math> for an integer <math> a </math>, so that <math> \lfloor \frac{\sqrt{n}}{2}\rfloor=a </math>. | ||
| − | <cmath> 40a | + | <cmath> 40a-51=n </cmath> |
| − | Going back to <math> a\le\frac{\sqrt{n}}{2}<a+1 </math>, this implies <math> 4a^2\le n<4a^2+8a+4 </math>. Making the substitution <math> n=40a | + | Going back to <math> a\le\frac{\sqrt{n}}{2}<a+1 </math>, this implies <math> 4a^2\le n<4a^2+8a+4 </math>. Making the substitution <math> n=40a-51 </math> gives the system of inequalities |
| − | <cmath> 4a^2-40a | + | <cmath> 4a^2-40a+51\le0 </cmath> |
| − | <cmath> 4a^2-32a | + | <cmath> 4a^2-32a+55>0 </cmath> |
| − | Approximating the roots of these two quadratics gives two integral solutions for <math> a | + | Approximating the roots of these two quadratics gives two integral solutions for <math> a </math>. Each of these gives a distinct solution for <math> n </math>, and thus <math> x </math>, for a total of <math> 2 </math> positive solutions. |
Now let <math> x=-\frac{\sqrt{n}}{2} </math>. We have | Now let <math> x=-\frac{\sqrt{n}}{2} </math>. We have | ||
| − | <cmath> 40\lfloor -\frac{\sqrt{n}}{2}\rfloor | + | <cmath> 40\lfloor -\frac{\sqrt{n}}{2}\rfloor-51=n </cmath> |
Since <math> \lfloor -x\rfloor=-\lceil x\rceil </math>, this can be rewritten as | Since <math> \lfloor -x\rfloor=-\lceil x\rceil </math>, this can be rewritten as | ||
| − | <cmath> -40\lceil \frac{\sqrt{n}}{2}\rceil | + | <cmath> -40\lceil \frac{\sqrt{n}}{2}\rceil-51=n </cmath> |
Since <math> n </math> is positive, the only possible value of <math> \lceil \frac{\sqrt{n}}{2}\rceil </math> is <math> 1 </math>, meaning <math> n=11 </math>. However, this would make <math> \lceil \frac{\sqrt{n}}{2}\rceil=2 </math>, a contradiction. Therefore, there are no negative roots. | Since <math> n </math> is positive, the only possible value of <math> \lceil \frac{\sqrt{n}}{2}\rceil </math> is <math> 1 </math>, meaning <math> n=11 </math>. However, this would make <math> \lceil \frac{\sqrt{n}}{2}\rceil=2 </math>, a contradiction. Therefore, there are no negative roots. | ||
Revision as of 22:04, 14 February 2014
Problem
Let 
be the greatest integer less than or equal to 
. Then the number of real solutions to 
is
Solution
We can rearrange the equation into 
. Obviously, the RHS is an integer, so 
for some integer 
. We can therefore make the substitution 
to get
![\[40\lfloor \frac{\sqrt{n}}{2}\rfloor-51=n\]](http://latex.artofproblemsolving.com/9/8/1/9819b3678e1719b2ae691bda6e66612023660813.png)
(We'll try the case where 
later.) Now let 
for an integer 
, so that 
.
![\[40a-51=n\]](http://latex.artofproblemsolving.com/f/3/e/f3ee783f5d23ccce683e34125a87b2cf51ba39c9.png)
Going back to , this implies 
. Making the substitution 
gives the system of inequalities
![\[4a^2-40a+51\le0\]](http://latex.artofproblemsolving.com/8/2/c/82c891088dfaf3c7aa7066c89ae69f5766e3aaea.png)
![\[4a^2-32a+55>0\]](http://latex.artofproblemsolving.com/2/8/3/283da2eac8e3a3fed42cdbbbf9e4c27d39648aca.png)
Approximating the roots of these two quadratics gives two integral solutions for . Each of these gives a distinct solution for , and thus , for a total of 
positive solutions.
Now let . We have
![\[40\lfloor -\frac{\sqrt{n}}{2}\rfloor-51=n\]](http://latex.artofproblemsolving.com/c/0/a/c0aa1c2f2248d2c6d688909b446db377363d4a1f.png)
Since 
, this can be rewritten as
![\[-40\lceil \frac{\sqrt{n}}{2}\rceil-51=n\]](http://latex.artofproblemsolving.com/b/8/7/b87d33331217df650af01f5ee06bad90124243cf.png)
Since is positive, the only possible value of 
is 
, meaning 
. However, this would make 
, a contradiction. Therefore, there are no negative roots.
The total number of roots to this equation is thus 
.
See Also
| 1985 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Last Problem | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
