Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. Please answer soon...The one who answers first will be marked as brainliest...*PINKY PROMISE*
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Answered by
2
the total number of numbers divisible by both 2 and 5 between 101 and 999 are divisible a of 10
= 89
= 89
LoveJapan:
where is the whole process dude? =_=
u told answer so i gave
plz mark as brainliest
i will give u process in comments
sorry dude...comment section isn't the place for explaining process...so...better luck next time!
Answered by
9
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.
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.
hope it helped
s mark as brainliest
*pls
i took more time because i was showing the whole process too...
you deserved it dude!
aww... thanks
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