Question

Find the number of natural number between 101 and 999 which are divisible by 2 or 5

Answer 1

Answer:

Factors of 2 which have multiple over 101= 51 to 494

Factors of 5 which have multiples over 101 = 21 to 199

So,number of  numbers which are divisible by 2 = 494 - 51 = 443+2 (we have to include 51 and 494 also. So +2) = 445

Number of numbers divisible by 5 = 199 - 21 = 178+2 = 180

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Answer 2

Answer:

Step-by-step explanation:

Solution :-

Formula to be used :-

a(n) = a + (n - 1)d

The sequence goes like this,

110, 120, 130, ........, 990

Since, they have a common difference of 10, they form an A.P.

Here,

a = 110, a(n) = 990 and d = 10

a(n) = a + (n - 1)d

⇒ 990 = 110 + (n - 1) × 10

⇒ 990 - 110 = (n - 1) × 10

⇒ 880 = (n - 1) × 10

⇒ 880/10 = n - 1

⇒ n - 1 = 88

⇒ n = 88 + 1

⇒ n = 89

Hence, there are 89 terms between 101 and 999 which are divisible by both 2 and 5.