Menu Close

Solve-for-positive-integers-a-3-9b-2-9c-2-2017-where-a-b-c-

Tinku Tara 0 Comments
Pin




Question Number 159057 by HongKing last updated on 12/Nov/21
Solve for positive integers: a^3 + 9b^2 + 9c^2 = 2017 where a ≥ b ≥ c
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}: \\ $$$$\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{9b}^{\mathrm{2}} \:+\:\mathrm{9c}^{\mathrm{2}} \:=\:\mathrm{2017} \\ $$$$\mathrm{where}\:\:\mathrm{a}\:\geqslant\:\mathrm{b}\:\geqslant\:\mathrm{c} \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 12/Nov/21
(a,b,c)=(10,8,7)
$$\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)=\left(\mathrm{10},\mathrm{8},\mathrm{7}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 12/Nov/21
a^3 + 9b^2 + 9c^2 = 2017 where a ≥ b ≥ c ∧ a,b,c∈Z^+ b^2 +c^2 =((2017−a^3 )/9) ⇒9∣(2017−a^3 ) a≤⌊ ((2017))^(1/3) ⌋=12 a=12: 9∤(2017−12^3 ) (rejected) We can observe that only 9∣2017−10^3 9∣2017−7^3 9∣2017−4^3 9∣2017−1^3 So possible values for a are: 10,7,4,1 a=10 b^2 +c^2 =((2017−a^3 )/9)=((2017−10^3 )/9)=113 Note that b & c can have values upto a Here b,c≤10,we can easily find 113=8^2 +7^2 a=7 b^2 +c^2 =((2017−7^3 )/9)=186 maximum value for b & c is 7 and max(b^2 +c^2 )=7^2 +7^2 =98 b^2 +c^2 ≠186 Similar logic shows that for a=4,1 there are no values for b,c under the above conditions. ∨∧∨∧∨∧∨∧∨∧∨∧∨∧∨∧∨∧∨∧∨∧∨∧ _(⌢) ^(⌢^ ) 1st Filter: determinant (((a≤⌊((2017))^(1/3) ⌋=12))) 2nd Filter: determinant (((9∣(2017−a^3 )))) 3rd Filter: determinant (((b,c≤a ∧ b^2 +c^2 =(2017−a^3 )/9)))
$$\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{9b}^{\mathrm{2}} \:+\:\mathrm{9c}^{\mathrm{2}} \:=\:\mathrm{2017} \\ $$$$\mathrm{where}\:\:\mathrm{a}\:\geqslant\:\mathrm{b}\:\geqslant\:\mathrm{c}\:\wedge\:\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{Z}^{+} \\ $$$$\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\frac{\mathrm{2017}−\mathrm{a}^{\mathrm{3}} }{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{9}\mid\left(\mathrm{2017}−\mathrm{a}^{\mathrm{3}} \right) \\ $$$$\mathrm{a}\leqslant\lfloor\:\sqrt[{\mathrm{3}}]{\mathrm{2017}}\:\rfloor=\mathrm{12} \\ $$$$\mathrm{a}=\mathrm{12}:\:\mathrm{9}\nmid\left(\mathrm{2017}−\mathrm{12}^{\mathrm{3}} \right)\:\left({rejected}\right) \\ $$$${We}\:{can}\:{observe}\:{that}\:{only}\: \\ $$$$\mathrm{9}\mid\mathrm{2017}−\mathrm{10}^{\mathrm{3}} \\ $$$$\mathrm{9}\mid\mathrm{2017}−\mathrm{7}^{\mathrm{3}} \\ $$$$\mathrm{9}\mid\mathrm{2017}−\mathrm{4}^{\mathrm{3}} \\ $$$$\mathrm{9}\mid\mathrm{2017}−\mathrm{1}^{\mathrm{3}} \\ $$$${So}\:{possible}\:{values}\:{for}\:\mathrm{a}\:\mathrm{are}: \\ $$$$\mathrm{10},\mathrm{7},\mathrm{4},\mathrm{1} \\ $$$$\:\mathrm{a}=\mathrm{10} \\ $$$$\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\frac{\mathrm{2017}−\mathrm{a}^{\mathrm{3}} }{\mathrm{9}}=\frac{\mathrm{2017}−\mathrm{10}^{\mathrm{3}} }{\mathrm{9}}=\mathrm{113} \\ $$$${Note}\:{that}\:\mathrm{b}\:\&\:\mathrm{c}\:{can}\:{have}\:{values}\:{upto}\:\:\mathrm{a} \\ $$$${Here}\:\mathrm{b},\mathrm{c}\leqslant\mathrm{10},{we}\:{can}\:{easily}\:{find} \\ $$$$\:\:\:\mathrm{113}=\mathrm{8}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \\ $$$$\mathrm{a}=\mathrm{7} \\ $$$$\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\frac{\mathrm{2017}−\mathrm{7}^{\mathrm{3}} }{\mathrm{9}}=\mathrm{186} \\ $$$${maximum}\:{value}\:{for}\:{b}\:\&\:{c}\:{is}\:\mathrm{7} \\ $$$${and}\:\:\:{max}\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)=\mathrm{7}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} =\mathrm{98} \\ $$$$\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \neq\mathrm{186} \\ $$$${Similar}\:{logic}\:{shows}\:{that}\:{for}\:\mathrm{a}=\mathrm{4},\mathrm{1} \\ $$$${there}\:{are}\:{no}\:{values}\:{for}\:\mathrm{b},\mathrm{c}\:{under} \\ $$$${the}\:{above}\:{conditions}. \\ $$$$\underset{\frown} {\overset{\overset{} {\frown}} {\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\mathrm{1st}\:\mathrm{Filter}:\:\begin{array}{|c|}{\mathrm{a}\leqslant\lfloor\sqrt[{\mathrm{3}}]{\mathrm{2017}}\:\rfloor=\mathrm{12}}\\\hline\end{array} \\ $$$$\:\mathrm{2nd}\:\mathrm{Filter}:\:\begin{array}{|c|}{\mathrm{9}\mid\left(\mathrm{2017}−\mathrm{a}^{\mathrm{3}} \right)}\\\hline\end{array} \\ $$$$\:\mathrm{3rd}\:\mathrm{Filter}: \\ $$$$\:\begin{array}{|c|}{\mathrm{b},\mathrm{c}\leqslant\mathrm{a}\:\:\wedge\:\:\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\left(\mathrm{2017}−\mathrm{a}^{\mathrm{3}} \right)/\mathrm{9}}\\\hline\end{array} \\ $$
Commented by HongKing last updated on 14/Nov/21
very nice dear Ser thank you
$$\mathrm{very}\:\mathrm{nice}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you} \\ $$

Pin

Leave a Reply

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