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

$$\mathrm{Find}\mathrm{the}\mathrm{product}\mathrm{of}101\times 10001\times$$$$100000001\times ...\times (1000...01)\mathrm{where}\mathrm{the}$$$$\mathrm{last}\mathrm{factor}\mathrm{has}{2}^7-1\mathrm{zeros}\mathrm{between}\mathrm{the}$$$$\mathrm{ones}.\mathrm{Find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ones}\mathrm{in}\mathrm{the}$$$$\mathrm{product}.$$

Commented by diofanto last updated on 27/Jul/17

$$({10}^2+1)\times ({10}^{2^2}+1)\times ({10}^{2^3}+1)\times ...\times ({10}^{2^7}+1)$$$$\mathrm{every}\mathrm{even}\mathrm{number}⩽2^8-2\mathrm{can}\mathrm{be}\mathrm{represented}\mathrm{by}$$$$\mathrm{a}\mathrm{sum}\mathrm{of}\mathrm{elements}\mathrm{from}\mathrm{the}\mathrm{set}\{2,2^2,2^3,...,2^7\}.$$$$\mathrm{The}\mathrm{product}\mathrm{is}\mathrm{hence}:$$$${10}^0+{10}^2+{10}^4+...+{10}^{2^8-2}$$$$=1010101...101\mathrm{with}{2}^7\mathrm{ones}.$$$$\mathrm{Note}\mathrm{that},\mathrm{indeed},\mathrm{the}\mathrm{product}\mathrm{of}7\mathrm{sums}$$$$\mathrm{with}2\mathrm{terms}\mathrm{each}\mathrm{should}\mathrm{have}{2}^7\mathrm{terms}.$$

Commented by Tinkutara last updated on 28/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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

$$\mathrm{N}=({10}^2+1)({10}^{2^2}+1)...({10}^{2^n}+1)$$$$({10}^2-1)\mathrm{N}=({10}^2-1)({10}^2+1)..({10}^{2^n}+1)$$$$={10}^{2^{n+1}}-1$$$$\mathrm{N}=\frac{{10}^{2^{n+1}}-1}{{10}^2-1}$$$$\mathrm{N}={10}^0+{10}^2+...+{10}^{2^n}$$

Commented by Tinkutara last updated on 28/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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