Difference between revisions of "2020 AMC 12A Problems/Problem 25"
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<math>\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332</math> | <math>\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332</math> | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>1<k<2</math> be the unique solution in this range. Note that <math>ck</math> is also a solution as long as <math>ck < c+1</math>, hence all our solutions are <math>k, 2k, ..., bk</math> for some <math>b</math>. This sum <math>420</math> must be between <math>\frac{b(b+1)}{2}</math> and <math>\frac{(b+1)(b+2)}{2}</math>, which gives <math>b=28</math> and <math>k=\frac{420}{406}=\frac{30}{29}</math>. Plugging this back in gives <math>a=\frac{29 \cdot 1}{30^2} = \frac{29}{900} \implies \boxed{\textbf{C}}</math>. | Let <math>1<k<2</math> be the unique solution in this range. Note that <math>ck</math> is also a solution as long as <math>ck < c+1</math>, hence all our solutions are <math>k, 2k, ..., bk</math> for some <math>b</math>. This sum <math>420</math> must be between <math>\frac{b(b+1)}{2}</math> and <math>\frac{(b+1)(b+2)}{2}</math>, which gives <math>b=28</math> and <math>k=\frac{420}{406}=\frac{30}{29}</math>. Plugging this back in gives <math>a=\frac{29 \cdot 1}{30^2} = \frac{29}{900} \implies \boxed{\textbf{C}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | First note that <math>\lfloor x\rfloor \cdot \{x\}<0</math> when <math>x<0</math> while <math>ax^2\ge 0\forall x\in \mathbb{R}</math>. Thus we only need to look at positive solutions (<math>x=0</math> doesn't affect the sum of the solutions). | ||
+ | Next, we breakdown <math>\lfloor x\rfloor\cdot \{x\}</math> down for each interval <math>[n,n+1)</math>, where <math>n</math> is a positive integer. Assume <math>\lfloor x\rfloor=n</math>, then <math>\{x\}=x-n</math>. This means that when <math>x\in [n,n+1)</math>, <math>\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2</math>. Setting this equal to <math>ax^2</math> gives | ||
+ | <cmath>nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}</cmath> | ||
+ | We're looking at the solution with smaller <math>x</math>, which is <math>x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)</math>. Note that if <math>\lfloor x\rfloor=n</math> is the greatest <math>n</math> such that <math>\lfloor x\rfloor \cdot \{x\}=ax^2</math> has a solution, the sum of all these solutions is slightly over <math>\sum_{k=1}^{n}k=\frac{n(n+1)}{2}</math>, which is <math>406</math> when <math>n=28</math>, just under <math>420</math>. Checking this gives | ||
+ | <cmath>\sum_{k=1}^{28}\frac{n}{2a}\left(1-\sqrt{1-4a}\right)=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420</cmath> | ||
+ | <cmath>\frac{1-\sqrt{1-4a}}{2a}=\frac{420}{406}=\frac{30}{29}</cmath> | ||
+ | <cmath>29-29\sqrt{1-4a}=60a</cmath> | ||
+ | <cmath>29\sqrt{1-4a}=29-60a</cmath> | ||
+ | <cmath>29^2-4\cdot 29^2a=29^2+3600a^2-120\cdot 29a</cmath> | ||
+ | <cmath>3600a^2=116a</cmath> | ||
+ | <cmath>a=\frac{116}{3600}=\frac{29}{900} \implies \boxed{\textbf{(C) }929}</cmath> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2020|ab=A|num-b=24|after=Last Problem}} | {{AMC12 box|year=2020|ab=A|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:13, 2 February 2020
Contents
Problem 25
The number , where and are relatively prime positive integers, has the property that the sum of all real numbers satisfying is , where denotes the greatest integer less than or equal to and denotes the fractional part of . What is ?
Solution 1
Let be the unique solution in this range. Note that is also a solution as long as , hence all our solutions are for some . This sum must be between and , which gives and . Plugging this back in gives .
Solution 2
First note that when while . Thus we only need to look at positive solutions ( doesn't affect the sum of the solutions). Next, we breakdown down for each interval , where is a positive integer. Assume , then . This means that when , . Setting this equal to gives We're looking at the solution with smaller , which is . Note that if is the greatest such that has a solution, the sum of all these solutions is slightly over , which is when , just under . Checking this gives
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
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 | |
All AMC 12 Problems and Solutions |
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