retagged by
4,293 views

2 Answers

Best answer
24 votes
24 votes

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 $3 = 144 + 360 + 240$
$\qquad \qquad -$ Numbers divisible by $2$ and $3$
$\qquad \qquad-$ Numbers divisible by $2$ and $5$
$\qquad \qquad-$  Numbers divisible by $3$ and $5$
$\qquad \qquad+$ Numbers divisible by $2,3$ and $5.$

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

  • Numbers divisible by $2$ and $3=$ Numbers divisible by $(2\times 3) = \left\lfloor\frac{720}{6}\right\rfloor = 120$
  • Numbers divisible by $2$ and $5=$ Numbers divisible by $(2 \times 5) = \left\lfloor\frac{720}{10}\right\rfloor = 72$
  • Numbers divisible by $3$ and $5=$ Numbers divisible by $(3 \times 5) = \left\lfloor\frac{720}{15}\right\rfloor = 48$
  • Numbers divisible by $2,3$ and $5=$ 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 $5 = 720 -528 = 192.$

Correct Answer: 192.

edited by
21 votes
21 votes

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[ 720 \mathrm{P} \right]=\left[ 720\times \frac{4}{15} \right]=192$.

 

So the answer is $192$.

edited by
Answer:

Related questions

24 votes
24 votes
6 answers
1
Kathleen asked Oct 8, 2014
16,183 views
The number of elements in the power set $P(S)$ of the set $S=\{\{\emptyset\}, 1, \{2, 3\}\}$ is:$2$$4$$8$None of the above
6 votes
6 votes
2 answers
2
Kathleen asked Oct 8, 2014
1,796 views
Determine the number of divisors of $600.$
11 votes
11 votes
3 answers
3
Kathleen asked Oct 8, 2014
1,489 views
Prove using mathematical induction for $n \geq 5, 2^n n^2$
27 votes
27 votes
5 answers
4
Kathleen asked Oct 8, 2014
6,465 views
Let $G_1$ and $G_2$ be subgroups of a group $G$.Show that $G_1 \cap G_2$ is also a subgroup of $G$.Is $G_1 \cup G_2$ always a subgroup of $G$?.