Find the natural no.s between 101 and 999 which are divisible by both 2 and 3
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Answered by
1
The number divisible by both 2 and 3 will divisible by 6
102,108---------------------------996
a=102,l=996,d=6
l=a+(n-1)d
996=102+(n-1)6
996-102=(n-1)6
894/6=n-1
149=n-1
149+1=n
n=150
102,108---------------------------996
a=102,l=996,d=6
l=a+(n-1)d
996=102+(n-1)6
996-102=(n-1)6
894/6=n-1
149=n-1
149+1=n
n=150
Answered by
0
n= 150 this is a perfect answer
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