Question

He sum of three consecutive natural numbers each divisible by 3 is 72. what is the largest among them?

Answer 1

let 1st natural no be x
then 2nd be x+3
3rd be x+6
a/q
x+ x+ 3 x+ 6=72
3x+9=72
3x= 72-9=63
x= 63/3=21
so, 1st no = 21
2nd no = 21+3=24
3rd no = 21+6=27

Answer 2

let three consecutive numbers be
x, x+3,x+6
sum= x+x+3+x+6=72
3x+9=72
3x=72-9
3x=63
x=63/3 x=21
therefore x= 21
x+3 x+6 =21+3 ,=21+6
=24 ,=27