Question Number 62449 by Tawa1 last updated on 21/Jun/19
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{digit}\:\mathrm{in}\:\:\:\:\mathrm{2}^{\mathrm{50}} \\ $$
Commented by Tawa1 last updated on 21/Jun/19
$$\mathrm{And}\:\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{a}\:\mathrm{general}\:\mathrm{nth}\:\mathrm{number}\:\mathrm{of}\:\mathrm{term}\:\mathrm{in}\:\mathrm{any}\:\mathrm{number}\:\mathrm{and}\:\:\mathrm{powers} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Jun/19
![By experimenting below we can see that first three powers of 2 are 1-digit next three are 2-digit,after that next three are 3-digit numbers... [(2^m ,N,d),(2^1 ,2,1),(2^2 ,4,1),(2^3 ,8,1),(2^4 ,(16),2),(2^5 ,(32),2),(2^6 ,(64),2),(2^7 ,(128),3),(2^8 ,(256),3),(2^9 ,(512),3),(2^(10) ,(1024),4),((...),(...),(...)) ] We observe that: 2^(3n) have n digit 2^(3n−1) ,2^(3n−2) also have n digit Now, 2^(51) =2^(3×17) have 17 digits 2^(50) =2^(3×17−1) also have 17 digits](Q62461.png)
$${By}\:{experimenting}\:{below}\:{we}\:{can}\:{see} \\ $$$${that}\:{first}\:{three}\:{powers}\:{of}\:\mathrm{2}\:{are}\:\mathrm{1}-{digit} \\ $$$${next}\:{three}\:{are}\:\mathrm{2}-{digit},{after}\:{that} \\ $$$${next}\:{three}\:{are}\:\mathrm{3}-{digit}\:{numbers}... \\ $$$$ \\ $$$$\begin{bmatrix}{\mathrm{2}^{{m}} }&{{N}}&{{d}}\\{\mathrm{2}^{\mathrm{1}} }&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}^{\mathrm{2}} }&{\mathrm{4}}&{\mathrm{1}}\\{\mathrm{2}^{\mathrm{3}} }&{\mathrm{8}}&{\mathrm{1}}\\{\mathrm{2}^{\mathrm{4}} }&{\mathrm{16}}&{\mathrm{2}}\\{\mathrm{2}^{\mathrm{5}} }&{\mathrm{32}}&{\mathrm{2}}\\{\mathrm{2}^{\mathrm{6}} }&{\mathrm{64}}&{\mathrm{2}}\\{\mathrm{2}^{\mathrm{7}} }&{\mathrm{128}}&{\mathrm{3}}\\{\mathrm{2}^{\mathrm{8}} }&{\mathrm{256}}&{\mathrm{3}}\\{\mathrm{2}^{\mathrm{9}} }&{\mathrm{512}}&{\mathrm{3}}\\{\mathrm{2}^{\mathrm{10}} }&{\mathrm{1024}}&{\mathrm{4}}\\{...}&{...}&{...}\end{bmatrix} \\ $$$${We}\:{observe}\:{that}: \\ $$$$\:\:\:\mathrm{2}^{\mathrm{3}{n}} {have}\:{n}\:{digit} \\ $$$$\:\:\mathrm{2}^{\mathrm{3}{n}−\mathrm{1}} ,\mathrm{2}^{\mathrm{3}{n}−\mathrm{2}} \:{also}\:{have}\:{n}\:{digit} \\ $$$${Now}, \\ $$$$\:\:\:\mathrm{2}^{\mathrm{51}} =\mathrm{2}^{\mathrm{3}×\mathrm{17}} {have}\:\mathrm{17}\:{digits} \\ $$$$\:\:\:\mathrm{2}^{\mathrm{50}} =\mathrm{2}^{\mathrm{3}×\mathrm{17}−\mathrm{1}} {also}\:{have}\:\mathrm{17}\:{digits} \\ $$
Commented by mr W last updated on 21/Jun/19
$${dear}\:{rasheed}\:{sir},\:{the}\:{idea}\:{is}\:{great}! \\ $$$${but}\:{the}\:{number}\:{of}\:\left({decimal}\right)\:{digits} \\ $$$${of}\:\mathrm{2}^{{n}\:} {doesn}'{t}\:{increase}\:{in}\:{a}\:{constant} \\ $$$${step}\:{of}\:\mathrm{3}\:{as}\:{you}\:{described}.\:{you}\:{have} \\ $$$${had}\:{seen}\:{this}\:{if}\:{you}\:{had}\:{a}\:{couple}\: \\ $$$${examples}\:{more},\:{e}.{g}. \\ $$$$\mathrm{2}^{\mathrm{10}} =\mathrm{1024} \\ $$$$\mathrm{2}^{\mathrm{11}} =\mathrm{2048} \\ $$$$\mathrm{2}^{\mathrm{12}} =\mathrm{4096} \\ $$$$\mathrm{2}^{\mathrm{13}} =\mathrm{8192}\:\:\:\:\Rightarrow\:\mathrm{4}\:{numbers}\:{with}\:\mathrm{4}\:{digits}! \\ $$$$\mathrm{2}^{\mathrm{14}} =\mathrm{16384} \\ $$$$\mathrm{2}^{\mathrm{15}} =\mathrm{32768} \\ $$$$\mathrm{2}^{\mathrm{16}} =\mathrm{65536}\:\Rightarrow\:\mathrm{3}\:{numbers}\:{with}\:\mathrm{5}\:{digits} \\ $$$$\mathrm{2}^{\mathrm{17}} =\mathrm{131072} \\ $$$$...... \\ $$$${in}\:{fact}\:\mathrm{2}^{\mathrm{50}} =\mathrm{11258}\:\mathrm{99906}\:\mathrm{84262}\:\mathrm{4} \\ $$$${it}\:{has}\:\mathrm{16}\:{digits},\:{not}\:\mathrm{17}. \\ $$
Commented by Rasheed.Sindhi last updated on 21/Jun/19
$$\boldsymbol{{Sir}}\:{thanks}\:{for}\:{guidance}.{I}\:{learnt} \\ $$$${that}\:{one}\:{should}\:{be}\:{more}\:{careful} \\ $$$${in}\:{handling}\:{numbers}! \\ $$$${Upto}\:\mathrm{2}^{\mathrm{30}} \:{the}\:{rule}\:'{number}\:{of}\:{decimal} \\ $$$${digits}\:{of}\:\mathrm{2}^{\mathrm{3}{n}} ={n}'\:{works}\:{well},{but}\:{at} \\ $$$$\mathrm{2}^{\mathrm{33}} {the}\:{rule}\:{doesn}'{t}\:{work}! \\ $$$$\mathcal{T}{hank}\mathcal{S}\:{again}\:{sir}! \\ $$
Commented by mr W last updated on 21/Jun/19
$${dear}\:{sir},\:{i}\:{appreciate}\:{your}\:{deep}\:{thoughts} \\ $$$${very}\:{much}. \\ $$
Answered by mr W last updated on 22/Jun/19
Commented by Rasheed.Sindhi last updated on 22/Jun/19
$${Nice}\:{Approach}\:{Sir}! \\ $$$${That}'{s}\:\:{according}\:{to}\:{logical}\:{thinking}! \\ $$$${And}\:{that}'{s}\:{deep}\:{thinking}\:{actually}! \\ $$$$\:\left({Whereas}\:{I}\:{thought}\:{in}\:{my}\:{answer}\:{patternically},{in}\:{a}\:{blind}\:{way}.\right) \\ $$$$ \\ $$$${BTW}\:{Sir}, \\ $$$$\:\:\:\:\lfloor{n}\mathrm{log2}\rfloor+\mathrm{1}\overset{?} {=}\lceil{n}\mathrm{log2}\rceil \\ $$$$\left({Recall}\:{that}\:{n}\mathrm{log2}\:{isn}'{t}\:{a}\:{whole}\:{number}\right) \\ $$
Commented by mr W last updated on 22/Jun/19
$${similarly}\:{the}\:{number}\:{of}\:{digits}\:{of}\:\mathrm{3}^{{n}} \\ $$$${is}\:\lfloor{n}\:\mathrm{log}\:\mathrm{3}\rfloor+\mathrm{1}. \\ $$
Commented by mr W last updated on 22/Jun/19
$${thanks}\:{sir}! \\ $$$${you}\:{are}\:{absolutely}\:{correct}\:{with}\: \\ $$$$\lfloor{n}\:\mathrm{log}\:\mathrm{2}\rfloor+\mathrm{1}=\lceil{n}\:\mathrm{log}\:\mathrm{2}\rceil\:{since}\:\mathrm{log}\:\mathrm{2}\:{is} \\ $$$${not}\:{a}\:{rational}\:{number}.\:{actually}\:{i} \\ $$$${also}\:{considered}\:{to}\:{use}\:\lceil{n}\:\mathrm{log}\:\mathrm{2}\rceil\:{instead} \\ $$$${of}\:\lfloor{n}\:\mathrm{log}\:\mathrm{2}\rfloor+\mathrm{1},\:{but}\:{my}\:{personal}\:{taste} \\ $$$${is}\:\:{more}\:{for}\:{floor}\:{function}. \\ $$
Commented by Rasheed.Sindhi last updated on 22/Jun/19
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