Question

How many integers between 1 to 567 which are divisible by 3 or 5 or 7?

Answer 1

308 integers between 1 and 567 which are divisible by 3 or 5 or 7.

How many integers between 1 and 567 which are divisible by 3 or 5 or 7.

Divisible by 3 = n(3) :

first term = 3 , last term = 564

so, no of terms = [last term - first term]/3 + 1

= [564 - 3]/3 + 1

= 187 + 1 = 188

Divisible by 5 = n(5) :

first term = 5 , last term = 565

so, no of terms = [last term - first term]/5 + 1

= [565 - 5]/5 + 1

= 113 + 1 = 114

Divisible by 7 = n(7) :

first term = 7 , last term = 560

so, no of terms = [last term - first term]/7 + 1

= [560 - 7]/7 + 1

= 80

Divisible by 3 and 5 = n(3 n 5) :

first term = 15 , last term = 555

so, no of terms = [last term - first term]/15 + 1

= [555 - 15]/15 + 1

= 37

Divisible by 5 and 7 = n(5 n 7) :

first term = 35 , last term = 560

so, no of terms = [last term - first term]/35 + 1

= [560 - 35]/35 + 1

= 16

Divisible by 7 and 3 = n(7 n 3) :

first term = 21 , last term = 546

so, no of terms = [last term - first term]/21 + 1

= [546 - 21]/21 + 1

= 26

divisible by 3 , 5 and 7 = n(3 n 5 n 7) :

first term = 105 , last term = 525

so, no of terms = [last term - first term]/105 + 1

= [525 - 105]/105 + 1

= 5

use formula,

n(A U B U C) = n(A) + n(B) + n(C) - n(A n B) - n(B n C) - n(C n A) + n(A n B n C)

∴ n(3 U 5 U 7) = n(3) + n(5) + n(7) - n(3 n 5) - n(5 n 7) - n(7 n 3) + n(3 n 5 n 7)

= 188 + 114 + 80 - 37 - 16 - 26 + 5

= 308

Therefore 308 integers between 1 and 567 which are divisible by 3 or 5 or 7.