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Question Number 98342 by Lekhraj last updated on 13/Jun/20

Commented by MJS last updated on 13/Jun/20

$$29$$$$n=29+30k$$

Commented by Lekhraj last updated on 13/Jun/20

$$\mathrm{How}?\mathrm{Explain}$$

Answered by Rasheed.Sindhi last updated on 15/Jun/20

$$1+2^1+2^2+2^3+...+2^n\equiv 0\left(mod77\right)$$$$\Rightarrow 2^{n+1}-1\equiv 0\left(mod77\right)$$$$\Rightarrow 2^{n+1}\equiv 1\left(mod77\right)$$$$\mathrm{Observing}\mathrm{all}\mathrm{powers}\mathrm{of}2$$$$(2,2^2,...,2^{30})\mathrm{we}\mathrm{achieve}\mathrm{that}$$$$2^{30}\mathrm{is}\mathrm{smallest}\mathrm{power}\mathrm{such}\mathrm{that}$$$$2^{30}\equiv 1\left(mod77\right)$$$$\therefore n+1=30\Rightarrow n=29$$

Commented by MJS last updated on 17/Jun/20

$$2^{n+1}-1=77m$$$$2^{n+1}=77m+1\Rightarrow m=2k-1$$$$2^{n+1}=154k-76$$$$2^n=77k-38$$$$\mathrm{we}\mathrm{can}\mathrm{now}\mathrm{try}\mathrm{for}k$$$$\mathrm{anyway}\mathrm{I}\mathrm{think}\mathrm{we}\mathrm{can}\mathrm{only}\mathrm{try}...$$

Commented by Rasheed.Sindhi last updated on 17/Jun/20

$$2^{n+1}=77m+1\Rightarrow m=2k-1$$$$$$

Commented by MJS last updated on 17/Jun/20

$$\mathrm{thank}\mathrm{you},\mathrm{I}\mathrm{corrected}$$

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