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

Revision as of 17:31, 17 February 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

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 $\boxed{\text{(C)}~199}$ . (Mudkipswims42)

Alternative Solution (Please excuse me for bad latex)

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

We can write x as floor(x)+{x}. Expanding everything, we get a quadratic in {x} in terms of floor(x): {x}^2+ (2*floor(x)-10,000){x} + [floor(x)]^2 = 0

We use the quadratic formula to solve for {x}: {x} = [ -2*floor(x)+10,000 +/- sqrt( [-2*floor(x)+10,000]^2- 4*[floor(x)]^2 ) ]/2

Since 0=<{x}<1, we get an inequality which we can then solve. After simplifying a lot, we get that [floor(x)]^2 + 2*floor(x) - 9999 < 0

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

~Alex_z_Awesome

This solution needs some editing. Please edit if you have time.

2018 AMC 10B (Problems • Answer Key • Resources)
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2018 AMC 12B (Problems • Answer Key • Resources)
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Difference between revisions of "2018 AMC 10B Problems/Problem 25"

Revision as of 17:31, 17 February 2018

Contents

Problem

Solution

Alternative Solution (Please excuse me for bad latex)

See Also

@@ Line 46: / Line 46: @@
 We can write x as floor(x)+{x}. Expanding everything, we get a quadratic in {x} in terms of floor(x):
 {x}^2+ (2*floor(x)-10,000){x} + [floor(x)]^2 = 0
-We use the quadratic formula to solve for {x}.
+We use the quadratic formula to solve for {x}:
 {x} = [ -2*floor(x)+10,000 +/- sqrt( [-2*floor(x)+10,000]^2- 4*[floor(x)]^2 ) ]/2