Difference between revisions of "2018 AMC 10B Problems/Problem 25"

Revision as of 16:40, 18 June 2018

Problem

Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$ ?

$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$

Solution 1

This rewrites itself to $x^2=10,000\{x\}$ .

Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$ , then $1$ to $2$ with a hole at $x=2$ etc.

Here is a graph of $y=x^2$ and $y=16\{x\}$ for visualization.

$[asy] import graph; size(400); xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5})); yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18})); real y(real x) {return x^2;} draw(circle((-4,16), 0.1)); draw(circle((-3,16), 0.1)); draw(circle((-2,16), 0.1)); draw(circle((-1,16), 0.1)); draw(circle((0,16), 0.1)); draw(circle((1,16), 0.1)); draw(circle((2,16), 0.1)); draw(circle((3,16), 0.1)); draw(circle((4,16), 0.1)); draw((-5,0)--(-4,16), black); draw((-4,0)--(-3,16), black); draw((-3,0)--(-2,16), black); draw((-2,0)--(-1,16), black); draw((-1,0)--(-0,16), black); draw((0,0)--(1,16), black); draw((1,0)--(2,16), black); draw((2,0)--(3,16), black); draw((3,0)--(4,16), black); draw(graph(y,-4.2,4.2),green); [/asy]$

Now notice that when $x=\pm 100$ then graph has a hole at $(\pm 100,10,000)$ which the equation $y=x^2$ passes through and then continues upwards. Thus our set of possible solutions is bounded by $(-100,100)$ . We can see that $y=x^2$ intersects each of the lines once and there are $99-(-99)+1=199$ lines for an answer of $\boxed{\text{(C)}~199}$ .

Solution 2

Same as the first solution, $x^2=10,000\{x\}$ .

We can write $x$ as $\lfloor x \rfloor+\{x\}$ . Expanding everything, we get a quadratic in ${x}$ in terms of $\lfloor x \rfloor$ : $\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0$

We use the quadratic formula to solve for {x}: $\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{( -2 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 ) }}{2}$

Since $0 \leq \{x\} < 1$ , we get an inequality which we can then solve. After simplifying a lot, we get that $\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0$ .

Solving over the integers, $-101 < \lfloor x \rfloor < 99$ , and since $\lfloor x \rfloor$ is an integer, there are $\boxed{\text{(C)}~199}$ solutions. Each value of $\lfloor x \rfloor$ should correspond to one value of $x$ , so we are done.

Solution 3

Let $x = a+k$ where $a$ is the integer part of $x$ and $k$ is the fractional part of $x$ . We can then rewrite the problem below:

$(a+k)^2 + 10000a = 10000(a+k)$

From here, we get

$(a+k)^2 + 10000a = 10000a + 10000k$

Solving for $a+k = x$

$(a+k)^2 = 10000k$

$x = a+k = \pm100\sqrt{k}$

Because $0 \leq k < 1$ , we know that $a+k$ cannot be less than or equal to $-100$ nor greater than or equal to $100$ . Therefore:

$-99 \leq x \leq 99$

There are 199 elements in this range, so the answer is $\boxed{\textbf{(C)} \text{ 199}}$ .

Solution 4

The following C++ code confirms that there are $\boxed{\textbf{(C)} \text{ 199}}$ solutions in total.

typedef long double ld;

int main() {
	// x^2=10000{x}
	// (t+b)^2=10000b 
	// b^2+(2t-10000)b+t^2=0 
	
	FOR(t,-200,201) {
	    ld b = 2*t-10000, c = t*t;
	    ld s1 = (-b+sqrt(b*b-4*c))/2;
	    ld s2 = (-b-sqrt(b*b-4*c))/2;
	    if (0 <= s1 && s1 < 1) cout << t+s1 << "\n";
	    if (0 <= s2 && s2 < 1) cout << t+s2 << "\n";
	}
}

@@ Line 108: / Line 108: @@
 {{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}}
 {{MAA Notice}}
+[[Category:Intermediate Algebra Problems]]

2018 AMC 10B (Problems • Answer Key • Resources)
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