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Determine the number of positive integers $(\leq 720)$ which are not divisible by any of $2,3$ or $5.$
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Numbers divisible by $5 = \left\lfloor\frac{720}{5}\right\rfloor = 144.$

Numbers divisible by $2 = \left\lfloor\frac{720}{2}\right\rfloor = 360.$

Numbers divisible by $3 = \left\lfloor\frac{720}{3}\right\rfloor = 240.$

Numbers divisible by either $5,2$ or $3 = 144 + 360 + 240-$ Numbers divisible by $2,3-$ Numbers divisible by $2,5-$  Numbers divisible by $3,5+$ Numbers divisible by $2,3,5.$

Since all $2,3,5$ are prime numbers,

• Numbers divisible by $2,3=$ Numbers divisible by $2\times 3 = \left\lfloor\frac{720}{6}\right\rfloor = 120$
• Numbers divisible by $2,5=$ Numbers divisible by $2 \times 5 = \left\lfloor\frac{720}{10}\right\rfloor = 72$
• Numbers divisible by $3,5=$ Numbers divisible by $3 \times 5 = \left\lfloor\frac{720}{15}\right\rfloor = 48$
• Numbers divisible by $2,3,5=$ Numbers divisible by $2 \times 3 \times 5 = \left\lfloor\frac{720}{30}\right\rfloor = 24$

So,

Numbers divisible by either $5,2$ or $3 = 144+360 + 240 - 120 - 72 -48 + 24 = 528.$

So, numbers which are not divisible by any of $2,3$ or $5 = 720 -528 = 192.$

by Veteran (431k points)
edited by
0
Shouldn't we use floor function in order to find the no. of multiples, i.e.

shouldn't Numbers divisible by $5 = \left \lfloor {\frac{720}{5}} \right \rfloor = 144$, not that it affects the final result as $720$ is a multiple of $5$ itself.

However, in situations where it isn't multiple, e.g. we try to find no. of positive integers ($\leq 10$) divisible by $3$, we use $\left \lfloor {\frac{10}{3}} \right \rfloor = 3$, which is correct as the positive integers ($\leq 10$) divisible by $3$ are $3$, $6$ and $9$.
+1
Thanks. Yes, I was thinking floor only but typed ceil. Corrected now. 👍
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Numbers divisible by 2,3= Numbers divisible by 2×3 ----> How this is true?
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@belliot

actually the numbers divisible by both a,b=numbers divisible by Least common multiple of(a,b).

for example,numbers divisible by 2={2,4,6,8,10,12,14,16,18,20,22,24,26,28......}

and numbers divisible by 4={4,8,12,16,20,24,28....}.

Now the numbers divisible by both 2,4

=intersection of {2,4,6,8,10,12,14,16,18,20,22,24,26,28......} and {4,8,12,16,20,24,28....}.

={4,8,12,16,20..}

=Numbers divisible by 4

=numbers divisible by l.c.m of 2,4

But if the numbers a,b are co prime, their l.c.m is nothing but their product i.e., a*b.

Here 2,3 are co prime so their l.c.m is 2*3=6. same thing applies for 2,5 and 3,5.

@Arjun sir ,i think it is better to add this point.

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@Arjun

One small correction. The peach colored circle in the venn diagram needs to have 48 instead of 44.

An idea of probability can be used here. Let $p, ~ q,~ r$ be the probability of having a number divisible by $2, ~3, ~5$ respectively.

If we notice from the set of natural numbers $\mathbb{N}=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,\cdots,\infty\}$, we observe that any one of every $2$ consecutive numbers is divisible by $2$. Again, any one of every $3$ consecutive numbers is divisible by $3$. The same goes for $5$. Therefore $$p = \frac{1}{2}\\q = \frac{1}{3}\\r = \frac{1}{5}$$

Now the complement of these probabilities are $p', ~ q', ~ r'$ meaning that the probabilities of having a number NOT divisible by $2,~3,~5$ respectively. Therefore, $$p' =1- \frac{1}{2}=\frac{1}{2}\\q' =1- \frac{1}{3}=\frac{2}{3}\\r' =1- \frac{1}{5}=\frac{4}{5}$$

So the probability of having a number NOT divisible by 2 or 3 or 5 is $\mathrm{P}= p'\times q'\times r'$. [It can be thought of De Morgan's law, $(p \vee q \vee r)'=p' \wedge q' \wedge r'$ ]

$$\therefore \mathrm{P}=p' \times q' \times r' =\frac{1}{2} \times \frac{2}{3} \times \frac{4}{5}= \frac{4}{15}$$

The number of positive integers out of the first 720 positive integers NOT divisible by 2 or 3 or 5 is $\left \lfloor 720 \mathrm{P}\right \rfloor=\left \lfloor 720\times \frac{4}{15} \right \rfloor=192$.

Note: $\left \lfloor x \right \rfloor$ means the greatest integer value less than or equal to x.

So the answer is $192$.

by Active (3.5k points)
edited
+2
good approach.