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 50987 by naka3546 last updated on 23/Dec/18

(√(x + y + 9))  +  (√(x − y + 8))  =  33^2   x > y  x, y  ∈  Z^+   (x^y  + y^x )  mod  (1000)  =  ?

$$\sqrt{{x}\:+\:{y}\:+\:\mathrm{9}}\:\:+\:\:\sqrt{{x}\:−\:{y}\:+\:\mathrm{8}}\:\:=\:\:\mathrm{33}^{\mathrm{2}} \\ $$ $${x}\:>\:{y} \\ $$ $${x},\:{y}\:\:\in\:\:\mathbb{Z}^{+} \\ $$ $$\left({x}^{{y}} \:+\:{y}^{{x}} \right)\:\:{mod}\:\:\left(\mathrm{1000}\right)\:\:=\:\:? \\ $$

Commented byRasheed.Sindhi last updated on 23/Dec/18

No unique answer.  For example:  x=320,y=247       (320^(247) +247^(320) )mod 1000=401                        AND  x=336,y=280       (336^(280) +280^(336) )mod 1000=176

$${No}\:{unique}\:{answer}. \\ $$ $${For}\:{example}: \\ $$ $${x}=\mathrm{320},{y}=\mathrm{247} \\ $$ $$\:\:\:\:\:\left(\mathrm{320}^{\mathrm{247}} +\mathrm{247}^{\mathrm{320}} \right){mod}\:\mathrm{1000}=\mathrm{401} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{AND} \\ $$ $${x}=\mathrm{336},{y}=\mathrm{280} \\ $$ $$\:\:\:\:\:\left(\mathrm{336}^{\mathrm{280}} +\mathrm{280}^{\mathrm{336}} \right){mod}\:\mathrm{1000}=\mathrm{176} \\ $$

Commented bynaka3546 last updated on 23/Dec/18

sorry,  I  have  edited  my  post.

$${sorry},\:\:{I}\:\:{have}\:\:{edited}\:\:{my}\:\:{post}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com