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Question Number 110887 by I want to learn more last updated on 31/Aug/20

Commented by Aina Samuel Temidayo last updated on 31/Aug/20

$$\mathrm{\exists }100\mathrm{a}+10\mathrm{b}+\mathrm{c}\in (100,999)\mathrm{s}.\mathrm{t}.$$$$100\mathrm{a}+10\mathrm{b}+\mathrm{c}={\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3$$$$100\mathrm{a}+10\mathrm{b}+\mathrm{c}=$$$$(\mathrm{a}+\mathrm{b}+\mathrm{c})({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca})$$$$+3\mathrm{abc}=10(10\mathrm{a}+\mathrm{b})+\mathrm{c}$$$$$$$$\mathrm{Satisfying}\mathrm{these}\mathrm{conditions},\mathrm{the}$$$$\mathrm{numbers}\mathrm{are}.$$$$$$$$153$$$$370$$$$371$$$$407$$$$$$$$\mathrm{They}\mathrm{are}\mathrm{called}\mathrm{ARMSTRONG}$$$$\mathrm{numbers}.$$

Commented by I want to learn more last updated on 31/Aug/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}$$

Answered by Rasheed.Sindhi last updated on 02/Sep/20

$$100h+10t+u=h^3+t^3+u^3$$$$(h^3-100h)+(t^3-10t)+(u^3-u)=0$$$$h(h-10)(h+10)$$$$+t(t^2-10)$$$$+u(u-1)(u+1)=0..\mathbb{A}$$$${}^{\bullet }\mathcal{F}oru=0,1:$$$$h(h-10)(h+10)+t(t^2-10)=0$$$$h(10-h)(10+h)=t(t^2-10)$$$$h(10-h)(10+h)>0$$$$\Rightarrow t(t^2-10)>0$$$$\Rightarrow t>3$$$$u=0,1\Rightarrow t>3$$$$h^3-100h+t(t^2-10)$$$$verifyingvariousvaluesoft$$$$(>3)andofcourseupto9$$$$lett=7andh=3satisfytheequation:$$$$h(100-h^2)=7(7^2-10)$$$$h^3-100h+273=0$$$$h=3$$$$Hencetworequirednumbers$$$$are:370\mathrm{\&}371$$$${}^{\bullet }\mathcal{F}oru>1thisapproachis$$$$difficult.$$

Commented by I want to learn more last updated on 31/Aug/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

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