Menu Close

Category: Geometry

Let-A-be-a-subset-of-0-1-1997-containing-more-than-1000-elements-Prove-that-A-contains-either-a-power-of-2-or-two-distinct-integers-whose-sum-is-a-power-of-2-

Question Number 10181 by 0942679167 last updated on 29/Jan/17 $${Let}\:{A}\:{be}\:{a}\:{subset}\:{of}\:\left\{\mathrm{0};\mathrm{1};\centerdot\centerdot\centerdot;\mathrm{1997}\right\}\: \\ $$$${containing}\:{more}\:{than}\:\mathrm{1000}\:{elements}. \\ $$$${Prove}\:{that}\:{A}\:{contains}\:{either}\:{a}\:{power} \\ $$$${of}\:\mathrm{2}\:{or}\:{two}\:{distinct}\:{integers}\:{whose}\: \\ $$$${sum}\:{is}\:{a}\:{power}\:{of}\:\mathrm{2}. \\ $$ Terms of Service Privacy Policy…

Question-10166

Question Number 10166 by Joel575 last updated on 28/Jan/17 Commented by Joel575 last updated on 28/Jan/17 $$\mathrm{Two}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{same}\:\mathrm{center}\:\mathrm{point}\:\mathrm{have} \\ $$$$\mathrm{R}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{R}_{\mathrm{2}} \:\mathrm{with}\:\mathrm{R}_{\mathrm{1}} \:<\:\mathrm{R}_{\mathrm{2}} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{long}\:\mathrm{of}\:\mathrm{AB}\:\mathrm{bowstring}\:\mathrm{is}\:\mathrm{10}\:\mathrm{cm}, \\…

x-1-16-1-x-1-8-1-x-1-4-1-x-1-2-1-

Question Number 10005 by konen last updated on 20/Jan/17 $$\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{16}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}\right)=? \\ $$ Answered by mrW1 last updated on 21/Jan/17 $${P}=\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{16}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{8}}}…

a-b-Z-a-3-b-3-19-a-b-

Question Number 9988 by konen last updated on 20/Jan/17 $$\mathrm{a},\mathrm{b}\in\mathrm{Z}^{+} \\ $$$$\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} =\mathrm{19}\:\:\:\Rightarrow\:\mathrm{a}.\mathrm{b}=? \\ $$ Answered by mrW1 last updated on 20/Jan/17 $${since}\:\mathrm{4}^{\mathrm{3}} −\mathrm{3}^{\mathrm{3}}…