How many positive integers less than 1000 are divisible by both 7 and 11?

Solution:

Given, the number is 1000

We have to find the number of positive integers less than 1000 that are divisible by both 7 and 11.

Using arithmetic progression,

aₙ = a + (n - 1)d

Where, a is the first term

n is the total number of terms

d is the common difference

The last number less 1000 than divisible by 7 = 994

So, aₙ = 994

a = 7

d = 7

994 = 7 + (n - 1)(7)

994 = 7 + 7n - 7

7n = 994

n = 994/7

n = 142

The last number less 1000 divisible by 11 = 990

So, aₙ = 990

a = 11

d = 11

990 = 11 + (n - 1)(11)

990 = 11 + 11n - 11

11n = 990

n = 990/11

n = 90

The last number less than 1000 divisible by 77 = 924

So, aₙ = 924

a = 77

d = 77

924 = 77 + (n - 1)(77)

924 = 77 + 77n - 77

77n = 924

n = 924/77

n = 12

Total number of positive integers that are divisible by both 7 and 11 = 12

Therefore, the required number of positive integers is 12.

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How many positive integers less than 1000 are divisible by both 7 and 11?

Summary:

12 positive integers less than 1000 are divisible by both 7 and 11.