Menu Close

Question-88050




Question Number 88050 by Sahil vampire last updated on 08/Apr/20
Commented by mr W last updated on 08/Apr/20
i found there is only one such number:  588 2353  588^2 +2353^2 =588 2353
$${i}\:{found}\:{there}\:{is}\:{only}\:{one}\:{such}\:{number}: \\ $$$$\mathrm{588}\:\mathrm{2353} \\ $$$$\mathrm{588}^{\mathrm{2}} +\mathrm{2353}^{\mathrm{2}} =\mathrm{588}\:\mathrm{2353} \\ $$
Commented by Sahil vampire last updated on 08/Apr/20
are you using some kind of matlB software/for it
$${are}\:{you}\:{using}\:{some}\:{kind}\:{of}\:{matlB}\:{software}/{for}\:{it} \\ $$
Commented by mr W last updated on 08/Apr/20
no. what is matlB?
$${no}.\:{what}\:{is}\:{matlB}? \\ $$
Commented by Sahil vampire last updated on 08/Apr/20
it is software.  tell na how you calculate the value. please so the math behin  behind it
$${it}\:{is}\:{software}. \\ $$$${tell}\:{na}\:{how}\:{you}\:{calculate}\:{the}\:{value}.\:{please}\:{so}\:{the}\:{math}\:{behin} \\ $$$${behind}\:{it} \\ $$
Commented by mr W last updated on 08/Apr/20
x(10000−x)=y(y−1)  100≤x≤999  1000≤y<3000  in this range i got the only solution  x=588, y=2353  with help of Grapher.
$${x}\left(\mathrm{10000}−{x}\right)={y}\left({y}−\mathrm{1}\right) \\ $$$$\mathrm{100}\leqslant{x}\leqslant\mathrm{999} \\ $$$$\mathrm{1000}\leqslant{y}<\mathrm{3000} \\ $$$${in}\:{this}\:{range}\:{i}\:{got}\:{the}\:{only}\:{solution} \\ $$$${x}=\mathrm{588},\:{y}=\mathrm{2353} \\ $$$${with}\:{help}\:{of}\:{Grapher}. \\ $$
Commented by Sahil vampire last updated on 08/Apr/20
oh youusefacwbook
$${oh}\:{youusefacwbook} \\ $$
Answered by Rasheed.Sindhi last updated on 08/Apr/20
Let the number is abcdefg.  (10^2 a+10b+c)^2 +(10^3 d+10^2 e+10f+g)^2       =10^6 a+10^5 b+10^4 c+10^3 d+10^2 e+10f+g  The term containing 10^6  in lhs:                         is equal to  The term containing 10^6  in rhs:                              a       WRONG IDEA
$${Let}\:{the}\:{number}\:{is}\:\mathrm{abcdefg}. \\ $$$$\left(\mathrm{10}^{\mathrm{2}} \mathrm{a}+\mathrm{10b}+\mathrm{c}\right)^{\mathrm{2}} +\left(\mathrm{10}^{\mathrm{3}} \mathrm{d}+\mathrm{10}^{\mathrm{2}} \mathrm{e}+\mathrm{10f}+\mathrm{g}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{10}^{\mathrm{6}} \mathrm{a}+\mathrm{10}^{\mathrm{5}} \mathrm{b}+\mathrm{10}^{\mathrm{4}} \mathrm{c}+\mathrm{10}^{\mathrm{3}} \mathrm{d}+\mathrm{10}^{\mathrm{2}} \mathrm{e}+\mathrm{10f}+\mathrm{g} \\ $$$${The}\:{term}\:{containing}\:\mathrm{10}^{\mathrm{6}} \:{in}\:{lhs}: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{is}\:{equal}\:{to} \\ $$$${The}\:{term}\:{containing}\:\mathrm{10}^{\mathrm{6}} \:{in}\:{rhs}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a} \\ $$$$\:\:\:\:\:\mathbb{WRONG}\:\mathbb{IDEA} \\ $$
Commented by mr W last updated on 08/Apr/20
say the number is abcdefg.  say x=abc  say y=defg  100≤x≤999  1000≤y≤9999  given: x^2 +y^2 =10000x+y  ⇒x(10000−x)=y(y−1)  ......
$${say}\:{the}\:{number}\:{is}\:\mathrm{abcdefg}. \\ $$$${say}\:{x}=\mathrm{abc} \\ $$$${say}\:{y}=\mathrm{defg} \\ $$$$\mathrm{100}\leqslant{x}\leqslant\mathrm{999} \\ $$$$\mathrm{1000}\leqslant{y}\leqslant\mathrm{9999} \\ $$$${given}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{10000}{x}+{y} \\ $$$$\Rightarrow{x}\left(\mathrm{10000}−{x}\right)={y}\left({y}−\mathrm{1}\right) \\ $$$$…… \\ $$

Leave a Reply

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