Question in proofs homework in our "sets" unit. I'm not sure if I need to use unions/intersects. Just confused as how to begin to solve this question. Help please!
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$\begingroup$Since $3$ and $4$ are coprime integers we need to find the number of integer divisible by $12 = 3 \cdot 4$. And the number of integer between $1$ and $100$ that are divisible by $12$ is:
$$\left\lfloor\frac{100}{12}\right\rfloor = 8$$
$\endgroup$ $\begingroup$If a number is divisible by both $3$ and $4$, then it is divisible by $3\times 4 = 12$. Why?
Try dividing $100$ by $12$ and rounding down: keeping only the integer result, not the remainder. To see how this works, we can simply take multiples of $12$:
$$12,\;24,\;36,\;\ldots,\;96$$ (the next multiple of $12$ takes us over $100$). So between $1$ and $96$, there are $$\dfrac{96}{12}= 8$$ multiples of twelve.
$\endgroup$ $\begingroup$Hint: For any integers $a$ and $b$ such that $\gcd(a,b)=1$, we have for any integer $n$ that $$a\mid n\;\;\mathsf{\text{and}}\;\;b\mid n\iff ab\mid n.$$ How many integers between $1$ and $100$ are multiples of $12$?
$\endgroup$ 1 $\begingroup$A number is divisible by $3$ and $4$ if and only if it's divisible by $12$ since $3$ and $4$ are coprime.
Now we have $$100:12=8+\frac13$$ so there's $8$ number multiple of $12$ between $1$ and $100$.
$\endgroup$ $\begingroup$EDIT: I read your question wrong. The following answer explains how to get the number of integers between $1$ and $100$ that are divisible by $3$ OR $4$.
To get the number of integers between $1$ and $100$ that are divisible by $3$, let $3k \leq 100$ and solve for $k$, noting that $k$ has to be an integer (hint: use the floor function). Then similarly find the number of integers less than or equal to $100$ that are divisible by $4$.
Now you can add those two numbers together - however, you would be counting all the integers that are divisible by both $3$ and $4$ twice. Therefore, to set things right, you need to subtract from that sum the number of integers that are divisible by both $3$ and $4$, so that these integers will effectively have been counted once (this is called the inclusion-exclusion principle).
The least common multiple of $3$ and $4$ is $12$, so the integers that are divisible by both are exactly those that are divisible by $12$.
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