Question

Find three largest consecutive natural numbers such that the sum of one-third of first, one-fourth of second and one-fifth of third is atmost 25.

Answer 1

x/3 + (x+1)/4 + (x+2)/5 = 25
.: 20x+15x+ 15+ 12x+24= 1500
.: 42x+39=1500
.:42x= 1461
.:6x=208.7
:.x=34.78

Now find the numbers which are closer to x and which satisfy the given condition. Hope you find this useful.
Thanks.

Answer 2

Answer:

McCracken

$22252x {x}^{.22.}$