How many solutions are there? 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. :)
 A: 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
A: 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.
A: 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.
