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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a subset of the natural number sequence (positive integers starting from 1) we could say that $\frac12$ of the numbers in the set are divisible by 2.

e.g if the set were ${[1,2,3,4,5,6,7]}$ we could say that $3\frac12$ of the numbers in it are divisible by 2.

If we now wanted to work out how many numbers are divisible by 3, we could work it out as $\frac73 = 2\frac13$ and we know this is correct because if we look at the set we can see that the numbers 3 and 6 are the 2 numbers that are divisble by 3.

If we wanted to work out how many numbers are divisible by 2 OR 3. At first glace I thought I could add up the 2 fractions and then subtract the overlap.

enter image description here

This would then equate to $\frac12+\frac13-(\frac12*\frac13) = \frac23$

This makes sense $\frac23$ of all natural numbers are divisible by $2$ or $3$. So if we wanted to see how many numbers were divisible by $2$ OR $3$ in the set ${1,2,3,4,5,6,7,8}$ we could say $8 * \frac23 = 5\frac13$ and this makse sense because the $5$ numbers divisible by $2$ or $3$ are ${2,3,4,6,8}$

This is where I get confused, when I test this against the number 10 for example I get $10 * \frac23 = 6\frac23$ BUT there are $7$ numbers under $10$ that are divisible by both $2$ or $3$, so I was expecting the whole number component to be $7$

Please help me understand. Is it possible to create such a fraction that would tell me the number of elements in the set that are divisible by 2 or 3?

Thanks in advance

share|cite|improve this question
up vote 2 down vote accepted

Of the numbers $1$ to $n$, $\lfloor n/m \rfloor$ (i.e. the greatest integer $\le n/m$) are divisible by $m$. Now $x$ is divisible by both $m_1$ and $m_2$ if and only if $x$ is divisible by $\text{lcm}(m_1, m_2)$ (the least common multiple of $m_1$ and $m_2$: this is $m_1 m_2$ if $m_1$ and $m_2$ have no common factor $> 1$). So of the numbers $1$ to $n$, $\lfloor n/m_1 \rfloor + \lfloor n/m_2 \rfloor - \lfloor n/\text{lcm}(m_1, m_2) \rfloor$ are divisible by $m_1$ or $m_2$. For example, of the numbers $1$ to $7$, there are $$\lfloor 7/2 \rfloor + \lfloor 7/3 \rfloor - \lfloor 7/6 \rfloor = 3 + 2 - 1 = 4$$ divisible by $2$ or $3$.

Since $\dfrac{n}{m} - 1 < \left\lfloor \dfrac{n}{m} \right\rfloor \le \dfrac{n}{m}$, $$ \eqalign{\dfrac{n}{m_1} + \dfrac{n}{m_2} - \dfrac{n}{\text{lcm}(m_1,m_2)} - 2 &< \left\lfloor \dfrac{n}{m_1} \right\rfloor + \left\lfloor \dfrac{n}{m_2} \right\rfloor - \left\lfloor \dfrac{n}{\text{lcm}(m_1, m_2)}\right\rfloor \cr &< \dfrac{n}{m_1} + \dfrac{n}{m_2} - \dfrac{n}{\text{lcm}(m_1,m_2)} + 1\cr}$$

share|cite|improve this answer

In general, no, even if the sets are contiguous as you have here. But most sets are not contiguous numbers. For instance, how many elements of the set $S=\{1,5,7,8,11,13,15,17,\}$ are divisible by 2 or 3? For a contiguous set, the larger the number of elements in the set, the closer the 2/3 approximation gets to the real number of elements, but it won't usually be exact unless the number of elements in the set is divisible by both 2 and 3. In fact, your estimates for 7 and 8 were off by the same amount from the real answer.

share|cite|improve this answer

Your Answer


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.