For how many positive values of $n$ are both $\frac n3$ and $3n$ four-digit integers? For how many positive values of $n$ are both $\frac n3$ and $3n$ four-digit integers?
Any help is greatly appreciated. I think the smallest n value is 3000 and the largest n value is 3333. Does this make sense?
 A: Your answer makes sense.
Minimum 4 digit number is $1000$
Maximum 4 digit number is $9999$
$$max = 3n = 9999$$
$$n_{max}=3333$$
$$min=\frac{n}{3}=1000$$
$$n_{min}=3000$$
Keep in mind that n must be divisible by 3. So, the answer would be: $$\frac{3333-3000}{3}+1=112$$
A: 112 values is the number of positive values whose n/3 and n*3 both are 4-digit numbers.
A: The question is asking how many values of n satisfy 


*

*$n \in \mathbb{N}$

*$3n \in [1000, 9999]$ 

*$\frac n3 \in [1000, 9999]$


So find the min value of $n$ satisfying the lower bound:
$$ \frac n3 \ge 1000 $$
And the max value of $n$ satisfying the higher bound:
$$ 3n \le 9999$$
Now only include values in the range $[3000, 3333]$ where $\frac n3$ is an integer (adding 1 since $3000$ is divisible by $3$).
$$ \frac {3333}{3} - \frac {3000}{3} + 1 = 112 $$
A: To be absolutely clear: $n$ is a positive integer, right?
Then for $\frac{n}{3}$ to be a $4$-digit number, you need $2999 < n < 29998$. But you also have to consider that not all those numbers are multiples of $3$.
But for $3n$ to be a $4$-digit number, you need $333 < n < 3334$.
These two ranges overlap, so the answer is simply to count up the multiples of $3$ in the overlap of $2999 < n < 3334$.
A: Some basic thoughts.
1) $\frac n3 < n < 3n$ 
2) If $k$ has four digits then $1000 \le k \le 99999$.
So therefore
$1000 \le \frac n3 < n < 3n \le 9999$
So 
$1000 \le \frac n3 \implies 3000 \le n$.
And $3n \le 9999\implies n \le 3333$.
So $3000 \le n \le 3333$.
A third thing to note is 
3) if $\frac n3 \in \mathbb Z \iff 3|n$ so
the question becomes:  How many multiples of $3$ are there between $3000$ and $3333$ inclusively.
One final basic thought
4)  For any set of consecutive integers every $n$th one of them will be divisible by by $n$.
So between $3000$ and $3333$ inclusively, there are $3334$ integers (don't forget to count $3000$ as the first one and $3001$ as the second one) so there are about $\frac {334}3 = 111\frac 13$ multiples of $3$.  
Of the $333$ integers between $3001$ and $3333$ exactly $\frac 13$ of them, $111$ are divisible by $3$. But $3000$ is also divisible by $3$ so that is $112$ multiples of $3$. 
They are:  $3000, 3003,3006,.......,3327, 3330, 3333$.
