Difference between revisions of "1985 AHSME Problems/Problem 30"
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==Problem== | ==Problem== | ||
− | Let <math> \lfloor x \rfloor </math> be the greatest integer less than or equal to <math> x </math>. Then the number of real solutions to <math> 4x^2-40\lfloor x \rfloor +51=0 </math> is | + | Let <math>\left\lfloor x\right\rfloor</math> be the greatest integer less than or equal to <math>x</math>. Then the number of real solutions to <math>4x^2-40\left\lfloor x\right\rfloor+51 = 0</math> is |
<math> \mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 3 \qquad \mathrm{(E) \ }4 </math> | <math> \mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 3 \qquad \mathrm{(E) \ }4 </math> | ||
==Solution== | ==Solution== | ||
− | We | + | We rearrange the equation as <math>4x^2 = 40\left\lfloor x\right\rfloor-51</math>, where the right-hand side is now clearly an integer, meaning that <math>4x^2 = n</math> for some non-negative integer <math>n</math>. Therefore, in the case where <math>x \geq 0</math>, substituting <math>x = \frac{\sqrt{n}}{2}</math> gives |
+ | <cmath>40\left\lfloor\frac{\sqrt{n}}{2}\right\rfloor-51 = n.</cmath> To proceed, let <math>a</math> be the unique non-negative integer such that <math>a \leq \frac{\sqrt{n}}{2} < a+1</math>, so that <cmath> | ||
− | <cmath> | + | The above inequalities therefore become <cmath>4a^2 \leq 40a-51 < 4a^2+8a+4 |
+ | \iff 4a^2-40a+51 < 0 \text{ and } 4a^2-32a+55 > 0,</cmath> where the first inequality can now be rewritten as <math>(2a-10)^2 \leq 49</math>, i.e. <math>\left\lvert 2a-10\right\rvert \leq 7</math>. Since <math>(2a-10)</math> is even for all integers <math>a</math>, we must in fact have <cmath>\begin{align*}\left\lvert 2a-10\right\rvert \leq 6 &\iff \left\lvert a-5\right\rvert \leq 3 \ &\iff 2 \leq a \leq 8.\end{align*}</cmath> The second inequality similarly simplifies to <math>(2a-8)^2 > 9</math>, i.e. <math>\left\lvert 2a-8\right\rvert > 3</math>. As <math>(2a-8)</math> is even, this is equivalent to <cmath>\begin{align*}\left\lvert 2a-8 \right\rvert \geq 4 &\iff \left\lvert a-4\right\rvert \geq 2 \ &\iff a \geq 6 \text{ or } a \leq 2,\end{align*}</cmath> so the values of <math>a</math> satisfying both inequalities are <math>2</math>, <math>6</math>, <math>7</math>, and <math>8</math>. Since <math>n = 40a-51</math>, each of these distinct values of <math>a</math> gives a distinct solution for <math>n</math>, and thus for <math>x = \frac{\sqrt{n}}{2}</math>, giving a total of <math>4</math> solutions in the <math>x \geq 0</math> case. | ||
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− | + | As <math>4</math> is already the largest of the answer choices, this suffices to show that the answer is <math>\text{(E)}</math>, but for completeness, we will show that the <math>x < 0</math> case indeed gives no other solutions. If <math>x = -\frac{\sqrt{n}}{2}</math> (and so <math>n > 0</math>), we require <cmath>40\left\lfloor -\frac{\sqrt{n}}{2}\right\rfloor-51 = n,</cmath> and recalling that <math>\left\lfloor -x\right\rfloor = -\left\lceil x\right\rceil</math> for all <math>x</math>, this equation can be rewritten as <cmath>-40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51 = n.</cmath> 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>, but this means <cmath>\begin{align*}n &= -40\left\lceil\frac{\sqrt{n}}{2}\right\rceil-51 \ &\leq -40 \cdot 1 - 51 \ &= -91,\end{align*}</cmath> which is a contradiction. Therefore the <math>x < 0</math> case indeed gives no further solutions, confirming that the total number of solutions is precisely <math>\boxed{\text{(E)} \ 4}</math>. | |
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− | <cmath> 40\lfloor -\frac{\sqrt{n}}{2}\rfloor-51=n </cmath> | ||
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− | <cmath> -40\lceil \frac{\sqrt{n}}{2}\rceil-51=n </cmath> | ||
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− | Since <math> n </math> is positive, the | ||
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==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}} |
Latest revision as of 02:57, 20 March 2024
Problem
Let be the greatest integer less than or equal to . Then the number of real solutions to is
Solution
We rearrange the equation as , where the right-hand side is now clearly an integer, meaning that for some non-negative integer . Therefore, in the case where , substituting gives To proceed, let be the unique non-negative integer such that , so that and our equation reduces to
The above inequalities therefore become where the first inequality can now be rewritten as , i.e. . Since is even for all integers , we must in fact have The second inequality similarly simplifies to , i.e. . As is even, this is equivalent to so the values of satisfying both inequalities are , , , and . Since , each of these distinct values of gives a distinct solution for , and thus for , giving a total of solutions in the case.
As is already the largest of the answer choices, this suffices to show that the answer is , but for completeness, we will show that the case indeed gives no other solutions. If (and so ), we require and recalling that for all , this equation can be rewritten as Since is positive, the least possible value of is , but this means which is a contradiction. Therefore the case indeed gives no further solutions, confirming that the total number of solutions is precisely .
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.