Menu Close

Question-48869




Question Number 48869 by behi83417@gmail.com last updated on 29/Nov/18
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18
i can not understand the questions..  let (P/Q)=Intregal part+fractional part  in which location zeros to find..       s
$${i}\:{can}\:{not}\:{understand}\:{the}\:{questions}.. \\ $$$${let}\:\frac{{P}}{{Q}}={Intregal}\:{part}+{fractional}\:{part} \\ $$$${in}\:{which}\:{location}\:{zeros}\:{to}\:{find}.. \\ $$$$ \\ $$$$ \\ $$$$\:{s} \\ $$
Commented by behi83417@gmail.com last updated on 29/Nov/18
P divide by Q.how many zero are  in answer.resault have′t fractional  part.
$${P}\:{divide}\:{by}\:{Q}.{how}\:{many}\:{zero}\:{are} \\ $$$${in}\:{answer}.{resault}\:{have}'{t}\:{fractional} \\ $$$${part}. \\ $$
Commented by Kunal12588 last updated on 30/Nov/18
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Nov/18
2^1 =2  2^2 =4  2^3 =8  2^4 =16  so last digit of 2^4 =6  last digit of 2^(100) =6  last digit of P=2  (6×7=42)  Q=510510     =2×3×5×7×11×13×17  let(P/Q)=D×10^n   last digit of P is 2   and Q=2×3×5×7×11×13×17  so  P is devisible by 2 contained in  Q   after division by two last digit of P is 1  so again by division 3×5×7×11×13×17  last digit is either 3 or 7  so there will be no zeroes in (P/Q)     pls check is it correct
$$\mathrm{2}^{\mathrm{1}} =\mathrm{2} \\ $$$$\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{3}} =\mathrm{8} \\ $$$$\mathrm{2}^{\mathrm{4}} =\mathrm{16} \\ $$$${so}\:{last}\:{digit}\:{of}\:\mathrm{2}^{\mathrm{4}} =\mathrm{6} \\ $$$${last}\:{digit}\:{of}\:\mathrm{2}^{\mathrm{100}} =\mathrm{6} \\ $$$${last}\:{digit}\:{of}\:{P}=\mathrm{2}\:\:\left(\mathrm{6}×\mathrm{7}=\mathrm{42}\right) \\ $$$${Q}=\mathrm{510510} \\ $$$$\:\:\:=\mathrm{2}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{11}×\mathrm{13}×\mathrm{17} \\ $$$${let}\frac{{P}}{{Q}}={D}×\mathrm{10}^{{n}} \\ $$$${last}\:{digit}\:{of}\:{P}\:{is}\:\mathrm{2}\: \\ $$$${and}\:{Q}=\mathrm{2}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{11}×\mathrm{13}×\mathrm{17} \\ $$$${so}\:\:{P}\:{is}\:{devisible}\:{by}\:\mathrm{2}\:{contained}\:{in}\:\:{Q}\: \\ $$$${after}\:{division}\:{by}\:{two}\:{last}\:{digit}\:{of}\:{P}\:{is}\:\mathrm{1} \\ $$$${so}\:{again}\:{by}\:{division}\:\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{11}×\mathrm{13}×\mathrm{17} \\ $$$${last}\:{digit}\:{is}\:{either}\:\mathrm{3}\:{or}\:\mathrm{7} \\ $$$${so}\:{there}\:{will}\:{be}\:{no}\:{zeroes}\:{in}\:\frac{{P}}{{Q}}\: \\ $$$$ \\ $$$${pls}\:{check}\:{is}\:{it}\:{correct} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *