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Question-59316




Question Number 59316 by mr W last updated on 07/May/19
Answered by MJS last updated on 07/May/19
a+b+c=25 ∧ a, b, c ∈N^★  ⇒ 1≤a, b, c≤23  S={n∈N^★ ∣1≤n≤23 ∧ mod(360, n)=0}=  ={1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20}  triples (a, b, c) with a, b, c ∈S ∧ a+b+c=25 (let a≥b≥c at first)  (20, 4, 1) =80  (20, 3, 2)=120  (18, 6, 1)=108  (18, 5, 2)=180  (18, 4, 3)=216  (15, 9, 1)=135  (15, 8, 2)=240  (15, 6, 4)=360 •  (15, 5, 5)=375  (12, 12, 1)=144  (12, 10, 3)=360 •  (12, 9, 4)=432  (12, 8, 5)=480  (10, 10, 5)=500  (10, 9, 6)=540  (9, 8, 8)=576  ⇒ 12 possible triples  (3, 10, 12) and its permutations  (4, 6, 15) and its permutations
$${a}+{b}+{c}=\mathrm{25}\:\wedge\:{a},\:{b},\:{c}\:\in\mathbb{N}^{\bigstar} \:\Rightarrow\:\mathrm{1}\leqslant{a},\:{b},\:{c}\leqslant\mathrm{23} \\ $$$$\mathbb{S}=\left\{{n}\in\mathbb{N}^{\bigstar} \mid\mathrm{1}\leqslant{n}\leqslant\mathrm{23}\:\wedge\:\mathrm{mod}\left(\mathrm{360},\:{n}\right)=\mathrm{0}\right\}= \\ $$$$=\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{8},\:\mathrm{9},\:\mathrm{10},\:\mathrm{12},\:\mathrm{15},\:\mathrm{18},\:\mathrm{20}\right\} \\ $$$$\mathrm{triples}\:\left({a},\:{b},\:{c}\right)\:\mathrm{with}\:{a},\:{b},\:{c}\:\in\mathbb{S}\:\wedge\:{a}+{b}+{c}=\mathrm{25}\:\left(\mathrm{let}\:{a}\geqslant{b}\geqslant{c}\:\mathrm{at}\:\mathrm{first}\right) \\ $$$$\left(\mathrm{20},\:\mathrm{4},\:\mathrm{1}\right)\:=\mathrm{80} \\ $$$$\left(\mathrm{20},\:\mathrm{3},\:\mathrm{2}\right)=\mathrm{120} \\ $$$$\left(\mathrm{18},\:\mathrm{6},\:\mathrm{1}\right)=\mathrm{108} \\ $$$$\left(\mathrm{18},\:\mathrm{5},\:\mathrm{2}\right)=\mathrm{180} \\ $$$$\left(\mathrm{18},\:\mathrm{4},\:\mathrm{3}\right)=\mathrm{216} \\ $$$$\left(\mathrm{15},\:\mathrm{9},\:\mathrm{1}\right)=\mathrm{135} \\ $$$$\left(\mathrm{15},\:\mathrm{8},\:\mathrm{2}\right)=\mathrm{240} \\ $$$$\left(\mathrm{15},\:\mathrm{6},\:\mathrm{4}\right)=\mathrm{360}\:\bullet \\ $$$$\left(\mathrm{15},\:\mathrm{5},\:\mathrm{5}\right)=\mathrm{375} \\ $$$$\left(\mathrm{12},\:\mathrm{12},\:\mathrm{1}\right)=\mathrm{144} \\ $$$$\left(\mathrm{12},\:\mathrm{10},\:\mathrm{3}\right)=\mathrm{360}\:\bullet \\ $$$$\left(\mathrm{12},\:\mathrm{9},\:\mathrm{4}\right)=\mathrm{432} \\ $$$$\left(\mathrm{12},\:\mathrm{8},\:\mathrm{5}\right)=\mathrm{480} \\ $$$$\left(\mathrm{10},\:\mathrm{10},\:\mathrm{5}\right)=\mathrm{500} \\ $$$$\left(\mathrm{10},\:\mathrm{9},\:\mathrm{6}\right)=\mathrm{540} \\ $$$$\left(\mathrm{9},\:\mathrm{8},\:\mathrm{8}\right)=\mathrm{576} \\ $$$$\Rightarrow\:\mathrm{12}\:\mathrm{possible}\:\mathrm{triples} \\ $$$$\left(\mathrm{3},\:\mathrm{10},\:\mathrm{12}\right)\:\mathrm{and}\:\mathrm{its}\:\mathrm{permutations} \\ $$$$\left(\mathrm{4},\:\mathrm{6},\:\mathrm{15}\right)\:\mathrm{and}\:\mathrm{its}\:\mathrm{permutations} \\ $$
Answered by MJS last updated on 08/May/19
other method  a+b+c=25 ⇒ c=25−a−b  abc=360 ⇏ c=((360)/(ab))  25−a−b=((369)/(ab))  b^2 +(a−25)b+((360)/a)=0  ⇒ b=((25−a)/2)±((√(a^3 −50a^2 +625a−1440))/(2(√a)))  ⇒ a^3 −50a^2 +625a−1440≥0 ∧ a≥0  ⇒ 3≤a≤15 [∨ a≥32 but 1≤a, b, c≤23]  now try values  of a∈N^★  which lead to b∈N^★   the 3 values a, b^− , b^+  represent triples a, b, c  due to symmetry of both given equations  a=3 ⇒ b^− =10; b^+ =12  a=4 ⇒ b^− =6; b^+ =15  no other solutions  ⇒ 6 permutations of (3, 10, 12)  and 6 permutations of (4, 6, 15)
$$\mathrm{other}\:\mathrm{method} \\ $$$${a}+{b}+{c}=\mathrm{25}\:\Rightarrow\:{c}=\mathrm{25}−{a}−{b} \\ $$$${abc}=\mathrm{360}\:\nRightarrow\:{c}=\frac{\mathrm{360}}{{ab}} \\ $$$$\mathrm{25}−{a}−{b}=\frac{\mathrm{369}}{{ab}} \\ $$$${b}^{\mathrm{2}} +\left({a}−\mathrm{25}\right){b}+\frac{\mathrm{360}}{{a}}=\mathrm{0} \\ $$$$\Rightarrow\:{b}=\frac{\mathrm{25}−{a}}{\mathrm{2}}\pm\frac{\sqrt{{a}^{\mathrm{3}} −\mathrm{50}{a}^{\mathrm{2}} +\mathrm{625}{a}−\mathrm{1440}}}{\mathrm{2}\sqrt{{a}}} \\ $$$$\Rightarrow\:{a}^{\mathrm{3}} −\mathrm{50}{a}^{\mathrm{2}} +\mathrm{625}{a}−\mathrm{1440}\geqslant\mathrm{0}\:\wedge\:{a}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{3}\leqslant{a}\leqslant\mathrm{15}\:\left[\vee\:{a}\geqslant\mathrm{32}\:\mathrm{but}\:\mathrm{1}\leqslant{a},\:{b},\:{c}\leqslant\mathrm{23}\right] \\ $$$$\mathrm{now}\:\mathrm{try}\:\mathrm{values}\:\:\mathrm{of}\:{a}\in\mathbb{N}^{\bigstar} \:\mathrm{which}\:\mathrm{lead}\:\mathrm{to}\:{b}\in\mathbb{N}^{\bigstar} \\ $$$$\mathrm{the}\:\mathrm{3}\:\mathrm{values}\:{a},\:{b}^{−} ,\:{b}^{+} \:\mathrm{represent}\:\mathrm{triples}\:{a},\:{b},\:{c} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{symmetry}\:\mathrm{of}\:\mathrm{both}\:\mathrm{given}\:\mathrm{equations} \\ $$$${a}=\mathrm{3}\:\Rightarrow\:{b}^{−} =\mathrm{10};\:{b}^{+} =\mathrm{12} \\ $$$${a}=\mathrm{4}\:\Rightarrow\:{b}^{−} =\mathrm{6};\:{b}^{+} =\mathrm{15} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{solutions} \\ $$$$\Rightarrow\:\mathrm{6}\:\mathrm{permutations}\:\mathrm{of}\:\left(\mathrm{3},\:\mathrm{10},\:\mathrm{12}\right) \\ $$$$\mathrm{and}\:\mathrm{6}\:\mathrm{permutations}\:\mathrm{of}\:\left(\mathrm{4},\:\mathrm{6},\:\mathrm{15}\right) \\ $$
Commented by mr W last updated on 08/May/19
thank you sir!
$${thank}\:{you}\:{sir}! \\ $$
Commented by malwaan last updated on 08/May/19
how 3≤a≤15 or a≥32 ?
$${how}\:\mathrm{3}\leqslant{a}\leqslant\mathrm{15}\:{or}\:{a}\geqslant\mathrm{32}\:? \\ $$
Commented by MJS last updated on 08/May/19
a^3 −50a^2 +625a−1440≥0  we must find the zeros  a=t+((50)/3)  t^3 −((625)/3)t−((7630)/(27))=0  D=(1/(27))p^3 +(1/4)q^2 =(1/(27))(−((625)/3))^3 +(1/4)(−((7630)/(27)))^2 =  =−((944800)/3)<0 ⇒ trigonometric solution  a_1 =t_1 +((50)/3)=((50)/3)(1−sin ((1/3)arcsin ((763)/(3125))))≈15.2979  a_2 =t_2 +((50)/3)=((50)/3)(1+sin ((π/3)+(1/3)arcsin ((763)/(3125))))≈31.7360  a_3 =t_3 +((50)/3)=((50)/3)(1−cos ((π/6)+(1/3)arcsin ((763)/(3125))))≈2.96604  (d/da)[a^3 −50a^2 +625a−1440]=3a^2 −100a+625       <0 at a=a_1        >0 at a=a_2 ∧a=a_3   ⇒ a^3 −50a^2 +625a−1440≥0 for a_3 ≤a≤a_1  and a≥a_3   we need a∈N^★  ⇒ ⌈a_3 ⌉≤a≤⌊a_1 ⌋ ∨ a≥⌈a_3 ⌉ ⇒  ⇒ 3≤a≤15 ∨ a≥32
$${a}^{\mathrm{3}} −\mathrm{50}{a}^{\mathrm{2}} +\mathrm{625}{a}−\mathrm{1440}\geqslant\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{find}\:\mathrm{the}\:\mathrm{zeros} \\ $$$${a}={t}+\frac{\mathrm{50}}{\mathrm{3}} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{625}}{\mathrm{3}}{t}−\frac{\mathrm{7630}}{\mathrm{27}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{1}}{\mathrm{27}}{p}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{4}}{q}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{27}}\left(−\frac{\mathrm{625}}{\mathrm{3}}\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\mathrm{7630}}{\mathrm{27}}\right)^{\mathrm{2}} = \\ $$$$=−\frac{\mathrm{944800}}{\mathrm{3}}<\mathrm{0}\:\Rightarrow\:\mathrm{trigonometric}\:\mathrm{solution} \\ $$$${a}_{\mathrm{1}} ={t}_{\mathrm{1}} +\frac{\mathrm{50}}{\mathrm{3}}=\frac{\mathrm{50}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{763}}{\mathrm{3125}}\right)\right)\approx\mathrm{15}.\mathrm{2979} \\ $$$${a}_{\mathrm{2}} ={t}_{\mathrm{2}} +\frac{\mathrm{50}}{\mathrm{3}}=\frac{\mathrm{50}}{\mathrm{3}}\left(\mathrm{1}+\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{763}}{\mathrm{3125}}\right)\right)\approx\mathrm{31}.\mathrm{7360} \\ $$$${a}_{\mathrm{3}} ={t}_{\mathrm{3}} +\frac{\mathrm{50}}{\mathrm{3}}=\frac{\mathrm{50}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{763}}{\mathrm{3125}}\right)\right)\approx\mathrm{2}.\mathrm{96604} \\ $$$$\frac{{d}}{{da}}\left[{a}^{\mathrm{3}} −\mathrm{50}{a}^{\mathrm{2}} +\mathrm{625}{a}−\mathrm{1440}\right]=\mathrm{3}{a}^{\mathrm{2}} −\mathrm{100}{a}+\mathrm{625} \\ $$$$\:\:\:\:\:<\mathrm{0}\:\mathrm{at}\:{a}={a}_{\mathrm{1}} \\ $$$$\:\:\:\:\:>\mathrm{0}\:\mathrm{at}\:{a}={a}_{\mathrm{2}} \wedge{a}={a}_{\mathrm{3}} \\ $$$$\Rightarrow\:{a}^{\mathrm{3}} −\mathrm{50}{a}^{\mathrm{2}} +\mathrm{625}{a}−\mathrm{1440}\geqslant\mathrm{0}\:\mathrm{for}\:{a}_{\mathrm{3}} \leqslant{a}\leqslant{a}_{\mathrm{1}} \:\mathrm{and}\:{a}\geqslant{a}_{\mathrm{3}} \\ $$$$\mathrm{we}\:\mathrm{need}\:{a}\in\mathbb{N}^{\bigstar} \:\Rightarrow\:\lceil{a}_{\mathrm{3}} \rceil\leqslant{a}\leqslant\lfloor{a}_{\mathrm{1}} \rfloor\:\vee\:{a}\geqslant\lceil{a}_{\mathrm{3}} \rceil\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{3}\leqslant{a}\leqslant\mathrm{15}\:\vee\:{a}\geqslant\mathrm{32} \\ $$
Commented by malwaan last updated on 09/May/19
thank you very much sir
$${thank}\:{you}\:{very}\:{much}\:{sir} \\ $$

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