How many solutions are there?

Question

The problem from some math competition (multiple-choice test):

Consider an equation $\lfloor\sqrt{12}x\rfloor=\lfloor\frac{7}{2}x\rfloor$, where $\lfloor x\rfloor$ denotes integer part of $x$ (entier). This equation:

A. has exactly 1 real solution

B. for $x>150$ has at least 1 integer solution

C. has exactly 28 integer solutions

D. has at least 7 natural solutions

My try:

I found out that there are 15 integer solutions:

-14, -12, -10, -8, -6, -4, -2, 0, 1, 3, 5, 7, 9, 11, 13

So I answered "D". However it turned out that it is not correct. I have no idea why. Can anyone help? Thanks in advance. :)

Well, I tried several integers values - I noticed that for $x>13 \vee x<-14$ the difference between $\sqrt{12}x$ and $\frac{7}{2}x$ is too large. — Paweł Orliński, Dec 18 '13 at 20:04
@BrianM.Scott But it is possible that C is correct also (so it could be preferred, because it is more precise -- yes I agree that would be dumb...) — Igor Rivin, Dec 18 '13 at 20:06
for(var x=-100;x<=100;x++) if(Math.floor(Math.sqrt(12)*x) == Math.floor(7/2*x)) console.log(x); There are 28 solutions between -26 to 13. — JiminP, Dec 18 '13 at 20:07
Bad question, unless the instructions allow marking more than one choice. Neither (C) nor (D) implies the other, so hypothetical instructions to choose the strongest true answer wouldn’t help. — Brian M. Scott, Dec 18 '13 at 20:10

score 2 · Accepted Answer · answered Dec 18 '13 at 20:08

2

There are 28 integer solutions: -26, -24, -22, -20, -18, -16, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 3, 5, 7, 9, 11, 13

answered Dec 18 '13 at 20:08

JiminP

1,670

Tim Ratigan · Answer 2 · 2013-12-18T20:43:44.097

To supplement JiminP, one way you to start this is to place bounds on $x$ is to check when it's absolutely false:

$$\begin{align} \left|\frac72x-\sqrt{12}x\right|&\ge 1\\ x^2\left(\frac72-\sqrt{12}\right)^2&\ge1\\ x^2\left(\frac{97}4-7\sqrt{12}\right)&\ge 1\\ x^2&\ge\frac{1}{\frac{97}{4}-7\sqrt{12}}\\ x^2&\ge\frac{1\left(\frac{97}4+7\sqrt{12}\right)}{\frac1{16}}\\ |x|&\ge4\sqrt{\frac{97}4+7\sqrt{12}} \end{align}$$ With one or two upward approximations (in particular, $\sqrt{12}\to \frac72$ and $\sqrt{195}\to 14$), we get $|x|\ge 28$. Obviously that's a pretty weak bound but it gives you a range to check (i.e. $-28<x<28$), instead of just dealing with infinity.

score 1 · Answer 3 · answered Dec 18 '13 at 20:06

1

The answer is pretty clear. There are at least $7$ integer solutions, hence $(D)$. If it is wrong, then it was not $\sqrt{12}$ but something else. Indeed, the first question had $\sqrt{2}$ instead.

answered Dec 18 '13 at 20:06

Dietrich Burde

130,978

1

$\sqrt{12}$ makes heuristic sense because $\sqrt{12}=\frac{\sqrt{48}}{2}\approx \frac72$. OP was the one that corrected it from $\sqrt 2$ anyway. – Tim Ratigan Dec 18 '13 at 20:08
3

It seems that both C and D are correct. How can D not be correct ? – Dietrich Burde Dec 18 '13 at 20:14

How many solutions are there?

3 Answers3