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-

Question Number 19704 by Tinkutara last updated on 14/Aug/17

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 ?

Answered by mrW1 last updated on 15/Aug/17

“ 1^{″} means 1, 2, 4, 8, 16, 32 from

right to left .

with three “ 1^{″} we can get

following possible numbers :

32 + 16 + 8

32 + 16 + 4

32 + 16 + 2

32 + 16 + 1 \Rightarrow Σ = 4 \times 32 + 5 \times 16 - 1 = 207

32 + 8 + 4

32 + 8 + 2

32 + 8 + 1 \Rightarrow Σ = 3 \times 32 + 4 \times 8 - 1 = 127

32 + 4 + 2

32 + 4 + 1 \Rightarrow Σ = 2 \times 32 + 3 \times 4 - 1 = 75

32 + 2 + 1 \Rightarrow 35

16 + 8 + 4

16 + 8 + 2

16 + 8 + 1 \Rightarrow Σ = 3 \times 16 + 4 \times 8 - 1 = 79

16 + 4 + 2

16 + 4 + 1 \Rightarrow Σ = 2 \times 16 + 3 \times 4 - 1 = 43

16 + 2 + 1 \Rightarrow 19

8 + 4 + 2

8 + 4 + 1 \Rightarrow Σ = 2 \times 8 + 3 \times 4 - 1 = 27

8 + 2 + 1 \Rightarrow 11

4 + 2 + 1 \Rightarrow 7

Σ = 207 + 127 + 75 + 35 + 79 + 43 + 19 + 27 + 11 + 7

= 630

Commented by Tinkutara last updated on 15/Aug/17

Thank you very much Sir!

Commented by mrW1 last updated on 16/Aug/17

there is an easier way to find the sum

without knowing each number :

2 Σ = (1 + 2 + 4 + 8 + 16 + 32) \times 5 \times 4 = 1260

\Rightarrow Σ = 1260 / 2 = 630

Commented by ajfour last updated on 16/Aug/17

111

1110, and 2 other 4 digit ones

11100, and 5 other 5 digit ones

111000, and 9 other 6 digit ones

111 + Σ 1110 + Σ 11100 + Σ 111000

= 7 + 3 \times 8 + 2 (4 + 2 + 1) +

6 \times 16 + 3 (8 + 4 + 2 + 1) +

10 \times 32 + 4 (16 + 8 + 4 + 2 + 1)

= 7 + 24 + 14 + 96 + 45 + 320 + 124

= 630 .

Commented by Tinkutara last updated on 16/Aug/17

Thank you both mrW 1 and ajfour .