discrete math

functions pigeonhole principle

How many numbers in the set $\{1,2,3,\dots ,346\}$ are divisible by 5 or 7 ?

Dylan Yoder
2022-04-14
Answered

discrete math

functions pigeonhole principle

How many numbers in the set $\{1,2,3,\dots ,346\}$ are divisible by 5 or 7 ?

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szalbierzfytg

Answered 2022-04-15
Author has **13** answers

In the given problem we have to find how many numbers in the set $\{1,2,3,\dots ,346\}$ are divisible by 5 or 7.

Denote A= $\{1,2,3,\dots ,346\}$

Suppose $A}_{5$ be the subset of A containing those numbers which are divisible by 5 and

$A}_{7$ be the subset of A containing those numbers which are divisible by 7.

Then $A}_{5}\cap {A}_{7$ contain those numbers of A which are divisible by both 5 and 7.

And a number is divisible by both 5 and 7 if it is divisible by their least common multiple and $\text{lcm}(5,7)=35$ .

Therefore a number is divisible by both 5 and 7 if it is divisible by 35.

$A}_{5}\cup {A}_{7$ contain those numbers of A which are divisible by both 5 or 7.

Therefore we have

$n\left({A}_{5}\right)=\left[\frac{346}{5}\right]=69$ where [] is the greatest integer function.

and

$n\left({A}_{7}\right)=\left[\frac{346}{7}\right]=49$

and

$n({A}_{5}\cap {A}_{7})=\left[\frac{346}{35}\right]=9$

Now

$n({A}_{5}\cup {A}_{7})=n\left({A}_{5}\right)+\left({A}_{7}\right)-n({A}_{5}\cap {A}_{7})$

$\Rightarrow n({A}_{5}\cup {A}_{7})=69+49-9$

$\Rightarrow n({A}_{5}\cup {A}_{7})=109$

Hence there are 109 numbers in the set $\{1,2,3,\dots ,346\}$ are divisible by 5 or 7.

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