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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-

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Question Number 18663 by Tinkutara last updated on 26/Jul/17
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.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{101}\:×\:\mathrm{10001}\:× \\ $$$$\mathrm{100000001}\:×\:…\:×\:\left(\mathrm{1000}…\mathrm{01}\right)\:\mathrm{where}\:\mathrm{the} \\ $$$$\mathrm{last}\:\mathrm{factor}\:\mathrm{has}\:\mathrm{2}^{\mathrm{7}} \:−\:\mathrm{1}\:\mathrm{zeros}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{ones}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ones}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{product}. \\ $$
Commented by diofanto last updated on 27/Jul/17
(10^2 + 1)×(10^2^2 + 1)×(10^2^3 + 1)×...×(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 ,...,2^7 }. The product is hence: 10^0 + 10^2 + 10^4 + ... + 10^(2^8 −2) = 1010101...101 with 2^7 ones. Note that, indeed, the product of 7 sums with 2 terms each should have 2^7 terms.
$$\left(\mathrm{10}^{\mathrm{2}} \:+\:\mathrm{1}\right)×\left(\mathrm{10}^{\mathrm{2}^{\mathrm{2}} } \:+\:\mathrm{1}\right)×\left(\mathrm{10}^{\mathrm{2}^{\mathrm{3}} } +\:\mathrm{1}\right)×…×\left(\mathrm{10}^{\mathrm{2}^{\mathrm{7}} } +\mathrm{1}\right) \\ $$$$\mathrm{every}\:\mathrm{even}\:\mathrm{number}\:\leqslant\:\mathrm{2}^{\mathrm{8}} −\mathrm{2}\:\mathrm{can}\:\mathrm{be}\:\mathrm{represented}\:\mathrm{by} \\ $$$$\mathrm{a}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{elements}\:\mathrm{from}\:\mathrm{the}\:\mathrm{set}\:\left\{\mathrm{2},\mathrm{2}^{\mathrm{2}} ,\mathrm{2}^{\mathrm{3}} ,…,\mathrm{2}^{\mathrm{7}} \right\}. \\ $$$$\mathrm{The}\:\mathrm{product}\:\mathrm{is}\:\mathrm{hence}: \\ $$$$\mathrm{10}^{\mathrm{0}} \:+\:\mathrm{10}^{\mathrm{2}} \:+\:\mathrm{10}^{\mathrm{4}} \:+\:…\:+\:\mathrm{10}^{\mathrm{2}^{\mathrm{8}} −\mathrm{2}} \: \\ $$$$=\:\mathrm{1010101}…\mathrm{101}\:\mathrm{with}\:\mathrm{2}^{\mathrm{7}} \:\mathrm{ones}. \\ $$$$\mathrm{Note}\:\mathrm{that},\:\mathrm{indeed},\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{7}\:\mathrm{sums} \\ $$$$\mathrm{with}\:\mathrm{2}\:\mathrm{terms}\:\mathrm{each}\:\mathrm{should}\:\mathrm{have}\:\mathrm{2}^{\mathrm{7}} \:\mathrm{terms}. \\ $$
Commented by Tinkutara last updated on 28/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by prakash jain last updated on 27/Jul/17
N=(10^2 +1)(10^2^2 +1)...(10^2^n +1) (10^2 −1)N=(10^2 −1)(10^2 +1)..(10^2^n +1) =10^2^(n+1) −1 N=((10^2^(n+1) −1)/(10^2 −1)) N=10^0 +10^2 +...+10^2^n
$$\mathrm{N}=\left(\mathrm{10}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{10}^{\mathrm{2}^{\mathrm{2}} } +\mathrm{1}\right)…\left(\mathrm{10}^{\mathrm{2}^{{n}} } +\mathrm{1}\right) \\ $$$$\left(\mathrm{10}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{N}=\left(\mathrm{10}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{10}^{\mathrm{2}} +\mathrm{1}\right)..\left(\mathrm{10}^{\mathrm{2}^{{n}} } +\mathrm{1}\right) \\ $$$$=\mathrm{10}^{\mathrm{2}^{{n}+\mathrm{1}} } −\mathrm{1} \\ $$$$\mathrm{N}=\frac{\mathrm{10}^{\mathrm{2}^{{n}+\mathrm{1}} } −\mathrm{1}}{\mathrm{10}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{N}=\mathrm{10}^{\mathrm{0}} +\mathrm{10}^{\mathrm{2}} +…+\mathrm{10}^{\mathrm{2}^{{n}} } \\ $$
Commented by Tinkutara last updated on 28/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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