find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
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31
Solution:-
Since the number must be divided by both 2 and 5, then the first term 'a' = 110 and common difference 'd' = 10 and the last term 'l' = 990
Therefore,
The number of terms is,
n = {last term - first term/difference} + 1
= {(990 - 110)/10} + 1
= (880/10) + 1
= 88 + 1
= 89
So, there are 89 numbers which are divisible by both 2 and 5.
Answer.
Since the number must be divided by both 2 and 5, then the first term 'a' = 110 and common difference 'd' = 10 and the last term 'l' = 990
Therefore,
The number of terms is,
n = {last term - first term/difference} + 1
= {(990 - 110)/10} + 1
= (880/10) + 1
= 88 + 1
= 89
So, there are 89 numbers which are divisible by both 2 and 5.
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