# Integer Part of a number

1. Oct 15, 2011

### mtayab1994

1. The problem statement, all variables and given/known data

2*[x]-3*[3x]+7=0

2. Relevant equations

solve for x when [x] is the integer part of the number.

3. The attempt at a solution

To solve it i removed the brackets and got x=1 and then i found the integer part of 1 is 1.

Is it correct? If not is there a concept to solving it?

2. Oct 15, 2011

### phyzguy

What if x = 1.07? Will this be a solution to the equation?

3. Oct 15, 2011

### mtayab1994

what do u mean by what if x=1.07

The integer part of 1.07 is 1 right? so that can be a solution as well and i also thought about have it as x= [1,2[

4. Oct 15, 2011

### HallsofIvy

Don't forget that $[3x]\ne 3[x]$. For example, if 1.5, then [x]= 1 but 3x= 4.5 so [3x]= 4, not 3. With x= 1.5, 2[x]- 3[3x]+ 7= 2(1)- 3(4)+ 7= 2- 12+ 7= -3, not 0.

Last edited by a moderator: Oct 15, 2011
5. Oct 15, 2011

### mtayab1994

yea thanx for the tip but i'm getting no where with this one at all.

6. Oct 15, 2011

### phyzguy

What are the largest and smallest values of x that will be solutions to the equation? The solution interval is not [1,2], like you said earlier, since, as HallsofIvy showed, 1.5 is not a solution.

7. Oct 15, 2011

### Dickfore

You need to use general intervals. Remember that:
$$[3 x] = N \in \mathbb{Z}, \; N \le 3 x < N + 1 \Leftrightarrow \frac{N}{3} \le x < \frac{N + 1}{3}$$

So, consider the following 3 families of intervals:
First family:

$$N = 3 k, k \le x < k + \frac{1}{3}, \; \mathbb{Z}$$
Then you have:
$$[x] = k, [3 x] = N = 3 k$$
$$2 k - 3 (3 k) + 7 = 0$$

Second family:
$$N = 3 k + 1, k + \frac{1}{3} \le x < k + \frac{2}{3}$$
Then, you have:
$$[x] = k, [3 x] = N = 3 k + 1$$
$$2 k - 3 ( 3 k + 1) + 7 = 0$$

Third family:
$$N = 3 k + 2, k + \frac{2}{3} \le x < k + 1$$
Then you have:
$$[x] = k, [3 x] = N = 3 k + 2$$
$$2 k - 3 (3 k + 2) + 7 = 0$$

Thus, you get three cases that exhaust the whole set of real numbers and for each case you get an equation. The solution to those equations has to be an integer!

8. Oct 15, 2011

### phyzguy

Isn't it true that only the first family has a solution?

9. Oct 15, 2011

### Dickfore

It seems that way, but I wanted the op to check that themselves.