question in proofs homework in our "sets" unit. I"m not sure if I need to use unions/intersects. Just confused as just how to begin to resolve this question.Help please!



Since $3$ and $4$ room coprime integers we require to uncover the number of integer divisible by $12 = 3 cdot 4$. And the variety of integer in between $1$ and also $100$ that room divisible by $12$ is:

$$leftlfloorfrac10012 ight floor = 8$$


If a number is divisible by both $3$ and also $4$, climate it is divisible by $3 imes 4 = 12$. Why?

Try dividing $100$ by $12$ and rounding down: keeping only the essence result, no the remainder. Come see how this works, we have the right to simply take it multiples that $12$:

$$12,;24,;36,;ldots,;96$$ (the next multiple of $12$ takes united state over $100$). So in between $1$ and also $96$, there room $$dfrac9612= 8$$ multiples of twelve.


Hint: For any kind of integers $a$ and $b$ such that $gcd(a,b)=1$, we have for any type of integer $n$ that$$amid n;;eer-selection.comsf extand;;bmid niff abmid n.$$How numerous integers between $1$ and also $100$ room multiples of $12$?


A number is divisible through $3$ and also $4$ if and also only if it"s divisible through $12$ because $3$ and $4$ room coprime.

You are watching: How many numbers between 1 and 100 are divisible by 3

Now we have $$100:12=8+frac13$$so there"s $8$ number many of $12$ in between $1$ and $100$.

EDIT: I read your concern wrong. The adhering to answer explains how to get the number of integers in between $1$ and $100$ that are divisible through $3$ OR $4$.

To acquire the variety of integers in between $1$ and also $100$ that space divisible by $3$, permit $3k leq 100$ and also solve because that $k$, noting the $k$ needs to be an creature (hint: usage the floor function). Then an in similar way find the variety of integers much less than or same to $100$ that space divisible by $4$.

Now girlfriend can add those 2 numbers together - however, you would be counting all the integers that space divisible through both $3$ and also $4$ twice. Therefore, to collection things right, you need to subtract indigenous that amount the number of integers that room divisible by both $3$ and also $4$, so the these integers will successfully have been counted once (this is referred to as the inclusion-exclusion principle).

See more: 1936 Buffalo Nickel F Mint Mark, 1936 Year Us Buffalo Nickels (1913

The least typical multiple of $3$ and also $4$ is $12$, therefore the integers that room divisible by both are specifically those that space divisible by $12$.