• Nov 16th 2010, 10:35 AM
Nforce
We have two sets:

A.... numbers that are divisible by 15
B.... numbers that are divisible by 33

We are looking just numbers between 1 and 10^6.

We need to find:

$
A \cap B^c
$

so we need to calculate this

$
[\frac{10^6}{15}(1 - \frac{1}{11})]
$

I understand $\frac{10^6}{15}$, it's logical.

But how do we get;
$
(1 - \frac{1}{11})
$
• Nov 16th 2010, 11:01 AM
Plato
It is really difficult to interpret what you have posted.
I take it that A is the set of multiples of 15 from 1 to $10^6$.
B is the set of multiples of 33 from 1 to $10^6$.

The question want to know how elements are in the set $A\cap B^c$.
That is the number of integers from 1 to $10^6$ which are multiples of 15 and not multiples of 33.
That is $\left\lfloor {\frac{{10^6 }}{{15}}} \right\rfloor - \left\lfloor {\frac{{10^6 }}{{165}}} \right\rfloor$.

If that is not what you mean, then what do you mean?
• Nov 16th 2010, 12:25 PM
emakarov
In fact, $\left\lfloor {\frac{{10^6 }}{{15}}} \right\rfloor - \left\lfloor {\frac{{10^6 }}{{165}}} \right\rfloor =\left\lfloor\frac{10^6}{15}(1 - \frac{1}{11})\right\rfloor$. As the OP said, $10^6/15$ part makes sense. To exclude multiple of 33 from the set of multiples of 15, we in fact need to exclude multiples of 165, the least common multiple of 15 and 33. So, from 11 numbers 1 * 15, 2 * 15, ..., 11 * 15, we leave the first 10, and we do similarly for other segments $(11k+i)\cdot15$, $i=1,\dots,11$, $k\in\mathbb{N}$.
• Nov 16th 2010, 12:41 PM
Nforce
Yes you understood me correctly.

I have an old book about logic and sets and it's difficult to understand the solutions.

What about if we need to find just $B^c$

so it's (universal set) - (elements in B),

$10^6 - (\frac{10^6}{33})$
• Nov 16th 2010, 12:45 PM
Plato
Quote:

Originally Posted by Nforce
What about if we need to find just $B^c$
so it's (universal set) - (elements in B),
$10^6 - (\frac{10^6}{33})$

It is important to use the floor function here.
$10^6 - \left\lfloor {\frac{{10^6 }}{{33}}} \right\rfloor$