Find-the-product-of-101-10001-100000001-1000-01-where-the-last-factor-has-2-7-1-zeros-between-the-ones-Find-the-number-of-ones-in-the-product-

Question Number 18663 by Tinkutara last updated on 26/Jul/17

Find the product of 101 \times 10001 \times

100000001 \times \dots \times (1000 \dots 01) where the

last factor has 2^{7} - 1 zeros between the

ones . Find the number of ones in the

product .

Commented by diofanto last updated on 27/Jul/17

(10^{2} + 1) \times (10^{2^{2}} + 1) \times (10^{2^{3}} + 1) \times \dots \times (10^{2^{7}} + 1)

every even number ⩽ 2^{8} - 2 can be represented by

a sum of elements from the set {2, 2^{2}, 2^{3}, \dots, 2^{7}} .

The product is hence :

10^{0} + 10^{2} + 10^{4} + \dots + 10^{2^{8} - 2}

= 1010101 \dots 101 with 2^{7} ones .

Note that, indeed, the product of 7 sums

with 2 terms each should have 2^{7} terms .

Commented by Tinkutara last updated on 28/Jul/17

Thanks Sir!

Answered by prakash jain last updated on 27/Jul/17

N = (10^{2} + 1) (10^{2^{2}} + 1) \dots (10^{2^{n}} + 1)

(10^{2} - 1) N = (10^{2} - 1) (10^{2} + 1) . . (10^{2^{n}} + 1)

= 10^{2^{n + 1}} - 1

N = \frac{10^{2^{n + 1}} - 1}{10^{2} - 1}

N = 10^{0} + 10^{2} + \dots + 10^{2^{n}}

Commented by Tinkutara last updated on 28/Jul/17

Thanks Sir!