Question

find the natural number between 102 and 999 which are divisible by both 2 and 5.
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Answer 1

Given :-

• First term, a = 110
• Last term, l = 990
• Common Difference, d = 10

To Find :-

• The number of numbers which are divisible by both 2 and 5.

Solution :-

Let us find the number of numbers between 102 & 999 which is divisible by both 2 & 5,

According to the question :-

=> aₙ = a + (n - 1)d

=> 110 + (n - 1)10 = 990

=> 110 + 10n - 10 = 990

=> 10n = 990 - 100

=> 10n = 890

=> n = 89

Hence,

• There are 89 numbers between 102 and 999.

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Answer 2

Step-by-step explanation:

The list of numbers between 101 and 999 that are divisible by 2 and 5 are:

110,120,130,...990

The numbers are in A.P, with first term, a=110, common difference, d=10

Last term, a

n

=990

We know that, a

n

=a+(n−1)d

990=110+(n−1)10

⇒990–110=10n–10

⇒880+10=10n

⇒890=10n

⇒n=89

Therefore, the number of terms between 101 and 999 that are divisible by 2 and 5 are 89.