Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $[x]$ represents the floor of $x$, solve the equation:

$$\left[\frac{x+1}3\right]=\frac{x-1}2$$

Choose the correct answer:

a) $(2,7)\cup(9,15)$

b) $(-3,2)\cup[3,4)\cup(6,14)$

c) $(-1,1]\cup[2,3)\cup(5,8)$

d) $(-5,-3)\cup(1,3]\cup[5,7)$

e) $[1,\frac32]\cup(2,4)\cup[5,7)$

f) $[0,2]\cup[4,7]\cup(9,+\infty)$

Can someone please explain how I can solve this type of equation? Thank you very much!

P.S. I'm not sure if $[x]$ is called whole part of $x$ in English, it's not my native language.

share|improve this question
    
This looks really strange to me, the LHS is always an integer, so the RHS has to be an integer two. Hence $x$ has to be an odd integer. So why are all options intervals? –  Simon Markett May 24 '12 at 10:31
    
$$\frac{x-1}{2} \in \mathbb Z$$, so $x$ is an odd number. Seems "e" is the correct answer. Umm, except for $6$! –  Gigili May 24 '12 at 10:32
    
@Grozav Please comment if you would like more help with some notation that you don't fully understand. –  Ronald May 24 '12 at 10:38
add comment

2 Answers 2

up vote 4 down vote accepted

So as $k:=[\frac{x+1}3] \in \mathbb Z$, we have $\frac{x-1}2 =k \in\mathbb Z$ also. So $x = 2k+1$ is an odd integer. We have to solve $[\frac{2k+2}3] = k$, we now consider three cases.

  • $k = 3\ell$ for some integer $\ell$: Then $$ \left[\frac{2k+2}3\right] = \left[\frac{6\ell+2}3\right] = \left[2\ell + \frac 23\right] $$ Now $2\ell$ is an integer, therefore the integer part of this is $\left[2\ell + \frac 23 \right] = 2\ell$. This equals $k = 3\ell$ for $\ell = 0$, which yields $k=0$, hence $x=1$.

  • $k = 3\ell + 1$ for some $\ell \in \mathbb Z$, here $$\left[\frac{2k+2}3\right] = \left[\frac{6\ell+4}3\right] = \left[2\ell + 1 + \frac 13\right] = 2\ell + 1,$$ which equals $k = 3\ell + 1$ for $\ell = 0$, giving $k = 1$, hence $x = 3$.

  • $k = 3\ell + 2$ for some $\ell \in \mathbb Z$, here $$\left[\frac{2k+2}3\right] = \left[\frac{6\ell+6}3\right] = \left[2\ell + 2\right] = 2\ell + 2,$$ which equals $k = 3\ell + 2$ for $\ell = 0$, giving $k = 2$, hence $x = 5$.

The only solutios are there fore $1$, $3$ and $5$, hence it is (e) (intersected with $2\mathbb Z+1$).

share|improve this answer
    
"e" includes $6$ which is not an answer to the equation. –  Gigili May 24 '12 at 10:36
    
Yes. So I wrote "intersected $2\mathbb Z + 1$". At least (e) doesn't contain wrong odd numbers. –  martini May 24 '12 at 10:38
    
Right, didn't notice that part. –  Gigili May 24 '12 at 10:40
    
Thank you very much, this is similar to their explanation, but it contains all the steps into getting the results. –  Grozav Alex Ioan May 24 '12 at 11:19
    
There's one thing I didnt exactly understand. When giving k the values of 3l, 3l+1, and 3l+2, why do we equal the floor equation with 2l, 2l+1, 2l+2? –  Grozav Alex Ioan May 24 '12 at 13:09
show 3 more comments

First of all, notice that $\frac{x-1}{2}$ has to be a whole number (an integer).

Therefore, $x-1$ has to be even, and $x$ has to be an odd integer.

We can use the options to eliminate any number which includes an even number. I guess we are talking only about integers, $\mathbb{Z}$, since any non-integer solutions are definitely out.

For example, the region $(2,7)$ in part (a) includes 4 and 6, so (a) is not correct.

This on its own is not enough to come to a solution, but it should allow us to quickly eliminate wrong options.

In fact, we can go through like this and eliminate all the options that you've given - none of them are right, as it stands.

Perhaps there is a typo in (e)? Since two of the three regions are representing an odd number, but $[5,7)$ incorrectly includes 6. I guess that (e) is supposed to be the correct answer, but maybe it should say $[5,6)$.

share|improve this answer
    
You put too much effort into improving your answer, +1. –  Gigili May 24 '12 at 11:01
    
Thank you for your help, martini's answer contained the practical way into solving the ecuation, however, thank you very much! –  Grozav Alex Ioan May 24 '12 at 11:20
    
@Gigli Haha thanks! Answer first, edit later. –  Ronald May 24 '12 at 11:38
    
@GrozavAlexIoan You are right to choose that answer, but I think using multiple choice options is an important skill. :) –  Ronald May 24 '12 at 11:39
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.