Question

How many integers between 1 to 300 that are divisible by at least one of 3, 5, 7?

Answer 1

3,5,7,11,13,17,19,22,23,25,29,31,37............... 300

Answer 2

Answer:

Total 161 integers between 1 to 300 are divisible by at least one of 3, 5, 7 .

Step-by-step explanation:

We can easily find integers between 1 to 300 that are divisible by atleast one of  3, 5 and 7 by using concepts like arithematic progression and sets.

Finding the number of integers between 1 to 300 divisible by 3:

3, 6, 9, 12, ...., 297

total number of terms:

$a_{n}=a+(n-1)d$

Given, $a_{n}=297, a=3, d=3$

Putting the values,

$297=3+(n-1)3\\294=(n-1)3\\n-1=98\\n=99$

Finding the number of integers between 1 to 300 divisible by 5:

5, 10, 15, ...., 295

total number of terms:

$a_{n}=a+(n-1)d$

Given, $a_{n}=295, a=5, d=5$

Putting the values,

$295=5+(n-1)5\\290=(n-1)5\\n-1=58\\n=59$

Finding the number of integers between 1 to 300 divisible by 7:

7, 14, 21, ...., 294

total number of terms:

$a_{n}=a+(n-1)d$

Given, $a_{n}=294, a=7, d=7$

Putting the values,

$294=7+(n-1)7\\287=(n-1)7\\n-1=41\\n=42$

Finding the number of integers between 1 to 300 divisible by 15:

15, 30, 45, ...., 285

total number of terms:

$a_{n}=a+(n-1)d$

Given, $a_{n}=285, a=15, d=15$

Putting the values,

$285=15+(n-1)15\\270=(n-1)15\\n-1=18\\n=19$

Finding the number of integers between 1 to 300 divisible by 35:

35, 70, ...., 280

total number of terms:

$a_{n}=a+(n-1)d$

Given, $a_{n}=280, a=35, d=35$

Putting the values,

$280=35+(n-1)35\\245=(n-1)35\\n-1=7\\n=8$

Finding the number of integers between 1 to 300 divisible by 21:

21, 42, 63, ...., 294

total number of terms:

$a_{n}=a+(n-1)d$

Given, $a_{n}=294, a=21, d=21$

Putting the values,

$294=21+(n-1)21\\273=(n-1)21\\n-1=13\\n=14$

Finding the number of integers between 1 to 300 divisible by 105:

105, 210

$n=2$

Total integers between 1 to 300 that are divisible by at least one of 3, 5, 7 are:

n(A∪B∪C)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A∩C)+n(A∩B∩C)

Given,

n(A) is number of terms divisible by 3

n(B) is number of terms divisible by 5

n(C) is number of terms divisible by 7

n(A∩B) is number of terms divisible by 15

n(B∩C) is number of terms divisible by 35

n(A∩C) is number of terms divisible by 21

n(A∩B∩C) is number of terms divisible by 105

Putting the values,

n(A∪B∪C) = $99+59+42-19-8-14+2$ = $161$