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Question-110887

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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
∃ 100a+10b+c ∈ (100,999) s.t. 100a+10b+c = a^3 +b^3 +c^3 100a+10b+c = (a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) +3abc=10(10a+b) +c Satisfying these conditions, the numbers are. 153 370 371 407 They are called ARMSTRONG numbers.
$$\exists\:\mathrm{100a}+\mathrm{10b}+\mathrm{c}\:\in\:\left(\mathrm{100},\mathrm{999}\right)\:\mathrm{s}.\mathrm{t}. \\ $$$$\mathrm{100a}+\mathrm{10b}+\mathrm{c}\:=\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \\ $$$$\mathrm{100a}+\mathrm{10b}+\mathrm{c}\:= \\ $$$$\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{ab}−\mathrm{bc}−\mathrm{ca}\right) \\ $$$$+\mathrm{3abc}=\mathrm{10}\left(\mathrm{10a}+\mathrm{b}\right)\:+\mathrm{c} \\ $$$$ \\ $$$$\mathrm{Satisfying}\:\mathrm{these}\:\mathrm{conditions},\:\mathrm{the} \\ $$$$\mathrm{numbers}\:\mathrm{are}. \\ $$$$ \\ $$$$\mathrm{153} \\ $$$$\mathrm{370} \\ $$$$\mathrm{371} \\ $$$$\mathrm{407} \\ $$$$ \\ $$$$\mathrm{They}\:\mathrm{are}\:\mathrm{called}\:\mathrm{ARMSTRONG} \\ $$$$\mathrm{numbers}. \\ $$
Commented by I want to learn more last updated on 31/Aug/20
Thanks sir, i appreciate
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate} \\ $$
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..A ^• For u=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 ⇒t(t^2 −10)>0 ⇒t>3 u=0,1⇒t>3 h^3 −100h+t(t^2 −10) verifying various values of t (>3) and of course upto 9 let t=7 and h=3 satisfy the equation: h(100−h^2 )=7(7^2 −10) h^3 −100h+273=0 h=3 Hence two required numbers are:370 & 371 ^• For u>1 this approach is difficult.
$$\mathrm{100}{h}+\mathrm{10}{t}+{u}={h}^{\mathrm{3}} +{t}^{\mathrm{3}} +{u}^{\mathrm{3}} \\ $$$$\left({h}^{\mathrm{3}} −\mathrm{100}{h}\right)+\left({t}^{\mathrm{3}} −\mathrm{10}{t}\right)+\left({u}^{\mathrm{3}} −{u}\right)=\mathrm{0} \\ $$$$\:\:\:{h}\left({h}−\mathrm{10}\right)\left({h}+\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{t}\left({t}^{\mathrm{2}} −\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{u}\left({u}−\mathrm{1}\right)\left({u}+\mathrm{1}\right)=\mathrm{0}..\mathbb{A} \\ $$$$\:^{\bullet} \mathcal{F}{or}\:{u}=\mathrm{0},\mathrm{1}: \\ $$$${h}\left({h}−\mathrm{10}\right)\left({h}+\mathrm{10}\right)+{t}\left({t}^{\mathrm{2}} −\mathrm{10}\right)=\mathrm{0} \\ $$$${h}\left(\mathrm{10}−{h}\right)\left(\mathrm{10}+{h}\right)={t}\left({t}^{\mathrm{2}} −\mathrm{10}\right) \\ $$$${h}\left(\mathrm{10}−{h}\right)\left(\mathrm{10}+{h}\right)>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{t}\left({t}^{\mathrm{2}} −\mathrm{10}\right)>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{t}>\mathrm{3} \\ $$$$\:\:\:{u}=\mathrm{0},\mathrm{1}\Rightarrow{t}>\mathrm{3} \\ $$$${h}^{\mathrm{3}} −\mathrm{100}{h}+{t}\left({t}^{\mathrm{2}} −\mathrm{10}\right) \\ $$$${verifying}\:{various}\:{values}\:{of}\:{t} \\ $$$$\left(>\mathrm{3}\right)\:{and}\:{of}\:{course}\:{upto}\:\mathrm{9} \\ $$$${let}\:{t}=\mathrm{7}\:{and}\:{h}=\mathrm{3}\:{satisfy}\:{the}\:{equation}: \\ $$$${h}\left(\mathrm{100}−{h}^{\mathrm{2}} \right)=\mathrm{7}\left(\mathrm{7}^{\mathrm{2}} −\mathrm{10}\right) \\ $$$${h}^{\mathrm{3}} −\mathrm{100}{h}+\mathrm{273}=\mathrm{0} \\ $$$${h}=\mathrm{3} \\ $$$${Hence}\:{two}\:{required}\:{numbers} \\ $$$${are}:\mathrm{370}\:\&\:\mathrm{371} \\ $$$$\:^{\bullet} \mathcal{F}{or}\:{u}>\mathrm{1}\:{this}\:{approach}\:{is} \\ $$$${difficult}. \\ $$
Commented by I want to learn more last updated on 31/Aug/20
Thanks sir, i appreciate.
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$

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