By 14.

@Lars Ninh , thanks for the problem. I discover (without proof) a general formula to solve this type of problem.

For a range of consecutive positive integers starting with $a$ and ending with $l$ , then the number of multiples of $m < l$ is given by:

$n(m) = \left \lfloor \frac lm \right \rfloor - \left \lceil \frac am \right \rceil + 1$

where $\lfloor \cdot \rfloor$ denotes the floor function and $\lceil \cdot \rceil$ denotes the ceiling function .

For $a=1000$ , $l = 9999$ , and $m = 14$ , we have $n(14) = \left \lfloor \dfrac {9999}{14} \right \rfloor - \left \lceil \dfrac {1000}{14} \right \rceil + 1 = 714 - 72 + 1 = \boxed {643}$

• Call the number can be divided by 14 is 14k

  1000 ≤ 14k ≤ 9999

  $\frac{1000}{14}$ ≤ k ≤ $\frac{9999}{14}$

72 ≤ k ≤ 714

So k = (714 -72) +1 = 643