How many positive integers are not divisible by 2, 3, or 5 up to 120?

[evinda]

Hi! (Smile)

I am looking at this exercise:

How many positive integers,that are not greater that $120$, do not get divided by $2,3 \text{ and } 5$?

I thought to write $120$ as a product of prime numbers ($120=2^3 \cdot 3 \cdot 5^2$),and then find the number of multiples of $2,3,5$ and subtract their sum from $120$. (Blush)

Is it right? And...how can I find the number of multiples of $2,3 \text{ and } 5$ ?

 

[chisigma]

Hi! (Smile)

I am looking at this exercise:

How many positive integers,that are not greater that $120$, do not get divided by $2,3 \text{ and } 5$?

I thought to write $120$ as a product of prime numbers ($120=2^3 \cdot 3 \cdot 5^2$),and then find the number of multiples of $2,3,5$ and subtract their sum from $120$. (Blush)

Is it right? And...how can I find the number of multiples of $2,3 \text{ and } 5$ ?

In this case the Sieve of Heratosthenes is a relatively comfortable method...

Kind regards

$\chi$ $\sigma$

 

[evinda]

In this case the Sieve of Heratosthenes is a relatively comfortable method...

Kind regards

$\chi$ $\sigma$

But...is my idea right or could we also do it in an other way?

 

[I like Serena]

Hi! (Smile)

I am looking at this exercise:

How many positive integers,that are not greater that $120$, do not get divided by $2,3 \text{ and } 5$?

I thought to write $120$ as a product of prime numbers ($120=2^3 \cdot 3 \cdot 5^2$),and then find the number of multiples of $2,3,5$ and subtract their sum from $120$. (Blush)

Is it right?

Yep! That is right! (Sun)

And...how can I find the number of multiples of $2,3 \text{ and } 5$ ?

Well, in these number the factor 2 occurs 0 up to 3 times, the factor 3 occurs 0 to 1 times, etcetera... (Thinking)

 

[evinda]

Yep! That is right! (Sun)
Well, in these number the factor 2 occurs 0 up to 3 times, the factor 3 occurs 0 to 1 times, etcetera... (Thinking)

So,there are $4$ multiples of $2$, $2$ of $3$ and $3$ of $5$? (Thinking)

 

[soroban]

Hello, evinda!

How many integers from 1 to 120 are not divisible by 2, 3 or 5?

Let's find the number of integers which are divisible by 2, 3, or 5.

Every second integer is divisible by 2.
.. $$2,4,6,8,10,12,14,16,18\text{ . . .}$$
Hence: .$$n(2) \:=\:\tfrac{120}{2} \:=\: 60$$ multiples of 2.

Every third integer is divisible by 3.
. . $$3,6,9,12,15,18,\text{ . . .}$$
Hence: .$$n(3) \:=\:\tfrac{120}{3} \:=\:40$$ multiples of 3.

But both these lists contain 6, 12, 18, . . .
. . The mutiples of 6 are counted twice.
Hence: .$$n(2,3) \:=\:\tfrac{120}{6} \:=\:20$$ multiples of 6.Every fifth integer is divisible by 5.
. . $$5, 10, 15, 20, 25, 30,\text{ . . .}$$
Hence: .$$n(5) \:=\:\tfrac{120}{5} \:=\:24$$ multiples of 5.

But we counted the multiples of 10, and 15 twice.
There are: .$$\tfrac{120}{10} = 12$$ multiples of 10, $$n(2,5) = 12$$
. . . . and: .$$\tfrac{120}{15} = 8$$ multiples of 15, $$n(3,5) = 8$$

Also, there are: .$$\tfrac{120}{30} = 4$$ multiples of 2, 3 and 5.

$$\begin{array}{cccc}\text{Formula:}\\ n(A \cup B \cup C) &=& n(A) + n(B) + n(C) \\ && -n(A\cap B) -n(B\cap C) - n(A\cap C) \\ && + n(A\cap B\cap C) \end{array}$$

$$\begin{array}{ccccc}n(2 \cup 3 \cup 5) &=& n(2) + n(3) + n(5) \\ && -n(2\cap3) - n(3\cap5) - n(2\cap 5) \\ && + n(2\cap3\cap5)\end{array}$$

$$n(2\cup3\cup5)\:=\:60 + 40 + 24 - 20 - 8 - 12 + 4 \:=\:88$$There are: .$$120-88 \,=\,32$$ numbers not divisible by 2, 3 or 5.

 

[I like Serena]

So,there are $4$ multiples of $2$, $2$ of $3$ and $3$ of $5$? (Thinking)

Wait! (Wait)
I was counting the wrong numbers.

Anyway, Soroban gave a proper method, which also matches chisigma's suggestion. (Emo)