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-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 | 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-51=n </cmath> | + | <cmath> 40\left\lfloor \frac{\sqrt{n}}{2}\right\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 | + | (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 a nonnegative integer <math> a </math>, so that <math> \left\lfloor \frac{\sqrt{n}}{2}\right\rfloor=a </math>. |
<cmath> 40a-51=n </cmath> | <cmath> 40a-51=n </cmath> | ||
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<cmath> 4a^2-32a+55>0 </cmath> | <cmath> 4a^2-32a+55>0 </cmath> | ||
− | + | The first inequality simplifies to <math>(2a-10)^2\le 49</math>, or <math>|2a-10|\le 7</math>. Since <math>2a-10</math> is even, we must have <math>|2a-10|\le 6</math>, so <math>|a-5|\le 3</math>. Therefore, <math>2\le a\le 8</math>. The second inequality simplifies to <math>(2a-8)^2 > 9</math>, or <math>|2a-8| > 3</math>. Therefore, as <math>2a-8</math> is even, we have <math>|2a-8|\ge 4</math>, or <math>|a-4|\ge 2</math>. Hence <math>a\ge 6</math> or <math>a\le 2</math> Since both inequalities must be satisfied, we see that only <math>a=2</math>, <math>a=6</math>, <math>a=7</math>, and <math>a=8</math> satisfy both inequalities. Each will lead to a distinct solution for <math>n</math>, and thus for <math>x</math>, for a total of <math>4</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-51=n </cmath> | + | <cmath> 40\left\lfloor -\frac{\sqrt{n}}{2}\right\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-51=n </cmath> | + | <cmath> -40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51=n </cmath> |
− | Since <math> n </math> is positive, the | + | Since <math> n </math> is positive, the least possible value of <math> \left\lceil \frac{\sqrt{n}}{2}\right\rceil </math> is <math> 1 </math>, hence <math> -40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51\le -91</math>, i.e., it must be negative. But it is also equal to <math>n</math>, which is positive, and this is a contradiction. Therefore, there are no negative roots. |
− | The total number of roots to this equation is thus <math> | + | The total number of roots to this equation is thus <math> 4</math>, or <math>\boxed{(\text{E})} </math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=29|after=Last Problem}} | {{AHSME box|year=1985|num-b=29|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:22, 12 December 2017
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
(We'll try the case where later.) Now let for a nonnegative integer , so that .
Going back to , this implies . Making the substitution gives the system of inequalities
The first inequality simplifies to , or . Since is even, we must have , so . Therefore, . The second inequality simplifies to , or . Therefore, as is even, we have , or . Hence or Since both inequalities must be satisfied, we see that only , , , and satisfy both inequalities. Each will lead to a distinct solution for , and thus for , for a total of positive solutions.
Now let . We have
Since , this can be rewritten as
Since is positive, the least possible value of is , hence , i.e., it must be negative. But it is also equal to , which is positive, and this is a contradiction. Therefore, there are no negative roots.
The total number of roots to this equation is thus , or .
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 |
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