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.

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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).

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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\$.