The correct option is C 22
→ Every 3rd number has one power of 3 in it.
So, n1=[503]=16 numbers have one 3 in them.
→ Every 9th number has 2 powers of 3, one of which has been counted earlier.
So, n2=[509]=5 numbers have a second 3 in them.
→ Every 27th number has 3 powers of 3, 2 of which has been counted.
So, n3=[5027]=1
So, the Total no. of powers, r in 50!
= [503]+[509]+[5027]+ ________
= 16+5+1
r=22