Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 26438 by San Sophanethsan069 last updated on 25/Dec/17

Answered by mrW1 last updated on 25/Dec/17

A=999...9 (n digits 9) S(A)=sum of digits of A=9n A=10^n −1 A^2 =(10^n −1)^2 =10^(2n) −2×10^n +1 A^2 =(10^n −2)×10^n +1=999...98000...01 (n−1 digits 9, one digit 8, n−1 digits 0, one digit 1) S(A^2 )=9×(n−1)+8+0×(n−1)+1=9n i.e. A and A^2 have the same sum of digits.

$${A}=\mathrm{999}...\mathrm{9}\:\left({n}\:{digits}\:\mathrm{9}\right) \\ $$$${S}\left({A}\right)={sum}\:{of}\:{digits}\:{of}\:{A}=\mathrm{9}{n} \\ $$$${A}=\mathrm{10}^{{n}} −\mathrm{1} \\ $$$${A}^{\mathrm{2}} =\left(\mathrm{10}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}{n}} −\mathrm{2}×\mathrm{10}^{{n}} +\mathrm{1} \\ $$$${A}^{\mathrm{2}} =\left(\mathrm{10}^{{n}} −\mathrm{2}\right)×\mathrm{10}^{{n}} +\mathrm{1}=\mathrm{999}...\mathrm{98000}...\mathrm{01} \\ $$$$\left({n}−\mathrm{1}\:{digits}\:\mathrm{9},\:{one}\:{digit}\:\mathrm{8},\:{n}−\mathrm{1}\:{digits}\:\mathrm{0},\:{one}\:{digit}\:\mathrm{1}\right) \\ $$$${S}\left({A}^{\mathrm{2}} \right)=\mathrm{9}×\left({n}−\mathrm{1}\right)+\mathrm{8}+\mathrm{0}×\left({n}−\mathrm{1}\right)+\mathrm{1}=\mathrm{9}{n} \\ $$$${i}.{e}.\:{A}\:{and}\:{A}^{\mathrm{2}} \:{have}\:{the}\:{same}\:{sum}\:{of} \\ $$$${digits}. \\ $$

Commented by San Sophanethsan069 last updated on 26/Dec/17

Thank you so much

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com