What is the sum (in base 10) of all the natural numbers less than 64 which have exactly three ones in their base 2 representation?
The numbers will have at most $6$ digits in their binary representation and let ﬁrst digits can be $0$ , hence all numbers can be made six digit numbers. The number of numbers with exactly $3$ ones $= 6C3= 6·5·4 6 = 20$. Each number has $3$ ones and $3$ zeros. So totally out of $20·6$ digits $20·3$ one’s are there. So there are exactly $20*3/6$ = $10$ one's in each place.
I don't know how I can proceed for here!