Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 6831 by Tawakalitu. last updated on 30/Jul/16

Find the 100th digits of this sum   1 + 50 + 50^2  + ..... + 50^(999)

$${Find}\:{the}\:\mathrm{100}{th}\:{digits}\:{of}\:{this}\:{sum}\: \\ $$$$\mathrm{1}\:+\:\mathrm{50}\:+\:\mathrm{50}^{\mathrm{2}} \:+\:.....\:+\:\mathrm{50}^{\mathrm{999}} \\ $$

Commented by prakash jain last updated on 31/Jul/16

How did you get  5^(1000) ×10^(1000) −1=(5^(1000) −1)×(999...9999)?

$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get} \\ $$$$\mathrm{5}^{\mathrm{1000}} ×\mathrm{10}^{\mathrm{1000}} −\mathrm{1}=\left(\mathrm{5}^{\mathrm{1000}} −\mathrm{1}\right)×\left(\mathrm{999}...\mathrm{9999}\right)? \\ $$

Commented by Yozzii last updated on 30/Jul/16

Σ_(i=0) ^(999) 50^i =((50^(1000) −1)/(50−1))=((50^(1000) −1)/(49))  =((5^(1000) ×10^(1000) −1)/(49))  Now, n×10^m =n000...000  (m zeroes) (n∈N)  ⇒n10^m −1=(n−1)999...999 (m nines)    ∴ 5^(1000) ×10^(1000) =(5^(1000) )000...000 (1000 zeroes trailing)  ⇒5^(1000) ×10^(1000) −1=(5^(1000) −1)999...999 (1000 nines trailing)  (continue)

$$\underset{{i}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\mathrm{50}^{{i}} =\frac{\mathrm{50}^{\mathrm{1000}} −\mathrm{1}}{\mathrm{50}−\mathrm{1}}=\frac{\mathrm{50}^{\mathrm{1000}} −\mathrm{1}}{\mathrm{49}} \\ $$$$=\frac{\mathrm{5}^{\mathrm{1000}} ×\mathrm{10}^{\mathrm{1000}} −\mathrm{1}}{\mathrm{49}} \\ $$$${Now},\:{n}×\mathrm{10}^{{m}} ={n}\mathrm{000}...\mathrm{000}\:\:\left({m}\:{zeroes}\right)\:\left({n}\in\mathbb{N}\right) \\ $$$$\Rightarrow{n}\mathrm{10}^{{m}} −\mathrm{1}=\left({n}−\mathrm{1}\right)\mathrm{999}...\mathrm{999}\:\left({m}\:{nines}\right) \\ $$$$ \\ $$$$\therefore\:\mathrm{5}^{\mathrm{1000}} ×\mathrm{10}^{\mathrm{1000}} =\left(\mathrm{5}^{\mathrm{1000}} \right)\mathrm{000}...\mathrm{000}\:\left(\mathrm{1000}\:{zeroes}\:{trailing}\right) \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{1000}} ×\mathrm{10}^{\mathrm{1000}} −\mathrm{1}=\left(\mathrm{5}^{\mathrm{1000}} −\mathrm{1}\right)\mathrm{999}...\mathrm{999}\:\left(\mathrm{1000}\:{nines}\:{trailing}\right) \\ $$$$\left({continue}\right) \\ $$

Commented by Yozzii last updated on 31/Jul/16

I was trying to write the number  in decimal form.  e.g 5^2 ×10^3 =25000  ⇒5^2 ×10^3 −1=24999=(5^2 −1)999  (in decimal  representation, not a product)

$${I}\:{was}\:{trying}\:{to}\:{write}\:{the}\:{number} \\ $$$${in}\:{decimal}\:{form}. \\ $$$${e}.{g}\:\mathrm{5}^{\mathrm{2}} ×\mathrm{10}^{\mathrm{3}} =\mathrm{25000} \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{2}} ×\mathrm{10}^{\mathrm{3}} −\mathrm{1}=\mathrm{24999}=\left(\mathrm{5}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{999}\:\:\left({in}\:{decimal}\right. \\ $$$$\left.{representation},\:{not}\:{a}\:{product}\right) \\ $$

Commented by Tawakalitu. last updated on 31/Jul/16

Thanks sir

$${Thanks}\:{sir} \\ $$

Commented by Tawakalitu. last updated on 31/Jul/16

Thanks sir

$${Thanks}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com