2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 25 Hard

Let \left\lfloor {x} \right\rfloor denote the greatest integer less than or equal to x. How many real numbers x satisfy the equation x^{2}+10,000\left\lfloor {x} \right\rfloor=10,000x? (2018 AMC 10B Problem, Question#25)

  • A.

    197

  • B.

    198

  • C.

    199

  • D.

    200

  • E.

    201

Answer:C

This rewrites itself to x^2=10000\left\{x\right\}, Graphing y=10000\left\{x\right\} and y=x^2 we see that the former is a set of line segments with slope 10000 from 0 to 1 wth 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\left\{x\right\} for visualization.

Now notice that when x=\pm100 then graph has a hole at (\pm100,10000) 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 199.

Note: from the graph we can clearly see there are 4 solution on the negative side and only 2 on the positive side. So the solution really should be from -100 to 98, which still counts to 199. A couple of the alternative solutions also seem to have the same flaw.

Same as the first soution，x2=10000\left\{x\right\}.

We can wrte x as \left\lfloor {x} \right\rfloor+\left\{x\right\} , Expanding everything, we get a quadratic in x in terms of \left\lfloor {x} \right\rfloor:\left\{x\right\}^2+(2 \left\lfloor {x} \right\rfloor-10,000)\left\{x\right\}+ \left\lfloor {x} \right\rfloor^2=0, We use the quadratic formula to solve for  \left\{x\right\}:\left\{ x \right\} = \frac{-2 \left\lfloor {x} \right\rfloor +10000 \pm \sqrt{(-2 \left\lfloor {x} \right\rfloor +10000^{2}-4 \left\lfloor {x} \right\rfloor ^{2})}}{2},

Since 0\leqslant\left\{x\right\}<1, we get an inequalty which we can then solve. After simplfying a lot, we get that \left\lfloor {x} \right\rfloor^2+2 \left\lfloor {x} \right\rfloor-9999<0. Solving over the integers, -101< \left\lfloor {x} \right\rfloor<99, and since \left\lfloor {x} \right\rfloor is an integer, there are 199 solutions. Each vaue of \left\lfloor {x} \right\rfloor shoud correspond to one value of x,  so we are done.

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= \pm 100 \sqrt{k},

Because 0\leqslant k<1,  we know that a+k cannot be less than or equal to -100 nor greater than or equal to100. Therefore： -99\leqslant x\leqslant 99.

There are 199 elements in this range,  so the answer is 199.

Notice the given equation is equvilent to \left(\left\lfloor {x} \right\rfloor+\left\{x\right\}\right)^2=10000\left\{x\right\}, Now we now that \left\{x\right\}<1 so plugging in 1 for \left\{x\right\} we can find the upper and lower bounds for the values.

(\left\lfloor {x} \right\rfloor+1)^2=10000(1),

(\left\lfloor {x} \right\rfloor+1)=\pm 100,

\left\lfloor {x} \right\rfloor=99, -101.

And just ike Solution 2, we see that -101<\left\lfloor {x} \right\rfloor<99, and since \left\lfloor {x} \right\rfloor is an integer, there are 199 solutions. Each value of \left\lfloor {x} \right\rfloor should correspond to one value of x, so we are done.

First, we can let \left\{x\right\}=b, \left\lfloor {x} \right\rfloor=a. we know that a+b=x by definition, We can rearrange the equation to obtain x^2=10^4(x-a). By taking square root on both side, we obtain x=\pm 100\sqrt{b} (because x-a=b),  We know since b is the fractional part of x, it must be tat 0\leqslant b<1, Thus, x may take any value in the inteval -100<x<100, Hehnce, we know that there are 199 potential values for \left\lfloor {x} \right\rfloor in that  range and we are done.

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