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Question Number 112318 by Aina Samuel Temidayo last updated on 07/Sep/20
A triangle ABC has the following properties BC=1, AB=AC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.
$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}\:\mathrm{BC}=\mathrm{1},\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AC}}\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{a}\:\mathrm{median}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{its}/\mathrm{their} \\ $$$$\mathrm{side}−\mathrm{lengths}\:\mathrm{and}\:\mathrm{angles}. \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Solution please?
$$\mathrm{Solution}\:\mathrm{please}? \\ $$
Commented by mr W last updated on 07/Sep/20
the only possibility: equilateral triangle AB=BC=CA=1
$${the}\:{only}\:{possibility}: \\ $$$${equilateral}\:{triangle} \\ $$$${AB}={BC}={CA}=\mathrm{1} \\ $$
Answered by mr W last updated on 07/Sep/20
Commented by mr W last updated on 07/Sep/20
((CD)/(BC))=((sin β)/(sin ∠BDC)) ((DA)/(BA))=((sin β)/(sin ∠BDA))=((sin β)/(sin ∠BDC))=((CD)/(BC)) DA=CD as given ⇒BC=BA ⇒BC=BA=CA ⇒ΔABC is equilateral
$$\frac{{CD}}{{BC}}=\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\angle{BDC}} \\ $$$$\frac{{DA}}{{BA}}=\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\angle{BDA}}=\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\angle{BDC}}=\frac{{CD}}{{BC}} \\ $$$${DA}={CD}\:{as}\:{given} \\ $$$$\Rightarrow{BC}={BA} \\ $$$$\Rightarrow{BC}={BA}={CA} \\ $$$$\Rightarrow\Delta{ABC}\:{is}\:{equilateral} \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$$$ \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Can you also help with this please?
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{also}\:\mathrm{help}\:\mathrm{with}\:\mathrm{this}\:\mathrm{please}? \\ $$
Commented by 1549442205PVT last updated on 08/Sep/20
The length of the median from the vertex A equal to m_a = (1/2)(√(2(b^2 +c^2 )−a^2 )) The length of the bisector of the angle ABC^(�) equal to l_b =(√(ac−((b^2 ac)/((a+c)^2 )))) The length of the altitude from the vertex C equal to h_C =(2/c)(√(p(p−a)(p−b)(p−c))) From the hypothesis we get m_a =l_b =h_c ⇔(1/2)(√(2(b^2 +c^2 )−a^2 )) = (√(ac))((√(1−(b^2 /((a+c)^2 )))) )=(2/c)(√(p(p−a)(p−b)(p−c)))
$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{median}\:\mathrm{from}\:\mathrm{the}\: \\ $$$$\mathrm{vertex}\:\mathrm{A}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{m}_{\mathrm{a}} =\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)−\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\widehat {\mathrm{ABC}}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{l}_{\mathrm{b}} =\sqrt{\mathrm{ac}−\frac{\mathrm{b}^{\mathrm{2}} \mathrm{ac}}{\left(\mathrm{a}+\mathrm{c}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{altitude}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{vertex}\:\mathrm{C}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{h}_{\mathrm{C}} =\frac{\mathrm{2}}{\mathrm{c}}\sqrt{\mathrm{p}\left(\mathrm{p}−\mathrm{a}\right)\left(\mathrm{p}−\mathrm{b}\right)\left(\mathrm{p}−\mathrm{c}\right)} \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{m}_{\mathrm{a}} =\mathrm{l}_{\mathrm{b}} =\mathrm{h}_{\mathrm{c}} \Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)−\mathrm{a}^{\mathrm{2}} }\:= \\ $$$$\sqrt{\mathrm{ac}}\left(\sqrt{\mathrm{1}−\frac{\mathrm{b}^{\mathrm{2}} }{\left(\mathrm{a}+\mathrm{c}\right)^{\mathrm{2}} }}\:\right)=\frac{\mathrm{2}}{\mathrm{c}}\sqrt{\mathrm{p}\left(\mathrm{p}−\mathrm{a}\right)\left(\mathrm{p}−\mathrm{b}\right)\left(\mathrm{p}−\mathrm{c}\right)} \\ $$
Commented by Aina Samuel Temidayo last updated on 08/Sep/20
Is it complete?
$$\mathrm{Is}\:\mathrm{it}\:\mathrm{complete}? \\ $$
Commented by 1549442205PVT last updated on 11/Sep/20
You only need give c one some value e.x,c=6 and solve equation to find a,b It can have some of triangles satisfying that conditions
$$\mathrm{You}\:\mathrm{only}\:\mathrm{need}\:\:\mathrm{give}\:\mathrm{c}\:\mathrm{one}\:\mathrm{some}\:\mathrm{value}\: \\ $$$$\mathrm{e}.\mathrm{x},\mathrm{c}=\mathrm{6}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{equation}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a},\mathrm{b} \\ $$$$\mathrm{It}\:\mathrm{can}\:\mathrm{have}\:\mathrm{some}\:\:\mathrm{of}\:\mathrm{triangles} \\ $$$$\mathrm{satisfying}\:\mathrm{that}\:\mathrm{conditions} \\ $$
Commented by Aina Samuel Temidayo last updated on 10/Sep/20
I don′t understand.
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}. \\ $$
Answered by 1549442205PVT last updated on 07/Sep/20
From the hypothesis AB=AC we infer the triangle ABC is isosceles at the vertex A.From the hypothesis the bisector of the angle B^(�) is also the median we infer the triangle ABC is isosceles at the vertex B⇒BA=BC.Therefore we have AB=AC=BC=1 (In course of secondary school level proved the result that:if one triangle has the bisector belong to any side that is also the median(or altitude) then that triangle is isosceles)
$$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{AB}=\mathrm{AC}\:\mathrm{we}\:\mathrm{infer} \\ $$$$\mathrm{the}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{is}\:\mathrm{isosceles}\:\mathrm{at}\:\mathrm{the}\: \\ $$$$\mathrm{vertex}\:\mathrm{A}.\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{the}\:\mathrm{bisector}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{angle}\:\widehat {\mathrm{B}}\:\mathrm{is}\:\mathrm{also}\:\mathrm{the}\:\mathrm{median}\:\mathrm{we}\: \\ $$$$\mathrm{infer}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{is}\:\mathrm{isosceles}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{vertex}\:\mathrm{B}\Rightarrow\mathrm{BA}=\mathrm{BC}.\mathrm{Therefore}\:\mathrm{we} \\ $$$$\mathrm{have}\:\mathrm{AB}=\mathrm{AC}=\mathrm{BC}=\mathrm{1} \\ $$$$\left(\mathrm{In}\:\mathrm{course}\:\mathrm{of}\:\mathrm{secondary}\:\mathrm{school}\:\mathrm{level}\right. \\ $$$$\mathrm{proved}\:\mathrm{the}\:\mathrm{result}\:\:\mathrm{that}:\mathrm{if}\:\mathrm{one}\:\mathrm{triangle} \\ $$$$\mathrm{has}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{belong}\:\mathrm{to}\:\mathrm{any}\:\mathrm{side} \\ $$$$\mathrm{that}\:\:\mathrm{is}\:\mathrm{also}\:\mathrm{the}\:\mathrm{median}\left(\mathrm{or}\:\mathrm{altitude}\right) \\ $$$$\left.\mathrm{then}\:\mathrm{that}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{isosceles}\right) \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Please help @1549442205PVT, @mr W
$$\mathrm{Please}\:\mathrm{help}\:@\mathrm{1549442205PVT},\:@\mathrm{mr} \\ $$$$\mathrm{W} \\ $$
Commented by 1549442205PVT last updated on 08/Sep/20
2)Given x,y,z∈N;yz+x is prime, (yz+x)∣(zx+y),(yz+x)∣(xy+z)(∗) We need find all possible values of P=(((xy+z)((zx+y))/((yz+x)^2 )) i)Case z=0 then P=((xy^2 )/x^2 )=(y^2 /x)=m^2 where y=mx, x∈P(prime) ii)Case y=0 then P=((xz^2 )/x^2 )=(z^2 /x)=n^2 where z=nx,x∈P(prime) iii)Case y,z≠0 From the hypothesis we have zx+y=m(yz+x),xy+z=n(yz+x) where m,n∈N^∗ .Adding two equalities we get x(y+z)+y+z=(m+n)(yz+x) ⇔(x+1)(y+z)=(m+n)(yz+x) Since yz+x is prime we infer that (yz+x)∣(x+1) or (yz+x)∣(y+z) a) If (yz+x)∣(x+1)then yz+x≤x+1 ⇒yz≤1⇒y=z=1(since y,z∈N,≠0) Then P=(((x+1)^2 )/((x+1)^2 ))=1(Here we see that has infinite x so that x+1 is prime) b)If (yz+x)∣(y+z)then yz+x≤y+z ⇔(y−1)(z−1)≤1−x(1) Since y−1≥0,z−1≥0,1−x≤0 ,we see that the equality (1)ocurrs if and only if x=y=z=1.Then P=((2.2)/2^2 )=1 Combining all cases we have P=m^2 for z=0,y=mx,x∈P P=n^(2 ) for y=0,z=nx,x∈P or P=1 when y=z=1, x∈P or x=y=z=1 3)Find ∣{n∈N∣n≤100,4!∣2^n −n^3 }∣ We need find n∈N∣n≤100 and A=2^n −n^3 ⋮4!=24 If n is odd then 2^n −n^3 is odd number ,so A isn′t divisible by 24.Hence,n must be an even number ,infer n=2k⇒A=2^(2k) −(2k)^3 =4^k −8k^3 (1) We need find k so that A⋮24 We need must have 0≤A≤100,so k=5 because when k=4 we get A=4^4 −8.4^3 <0.When k=6 we get A=4^6 −8.6^3 =2368 when k=5 we get A=4^5 −8.5^3 = 1024−8.125=24⋮4! Thus,has unique n=10 satisfying the condition of our problem:n∈N, n≤100,2^n −n^3 ⋮4!.Therefore ∣{n∈N∣n≤100,2^n −n^3 ⋮4!}∣=1
$$\left.\mathrm{2}\right)\mathrm{Given}\:\mathrm{x},\mathrm{y},\mathrm{z}\in\mathrm{N};\mathrm{yz}+\mathrm{x}\:\mathrm{is}\:\mathrm{prime}, \\ $$$$\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{zx}+\mathrm{y}\right),\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{xy}+\mathrm{z}\right)\left(\ast\right) \\ $$$$\mathrm{We}\:\mathrm{need}\:\mathrm{find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of} \\ $$$$\mathrm{P}=\frac{\left(\mathrm{xy}+\mathrm{z}\right)\left(\left(\mathrm{zx}+\mathrm{y}\right)\right.}{\left(\mathrm{yz}+\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{i}\right)\mathrm{Case}\:\:\mathrm{z}=\mathrm{0}\:\mathrm{then}\:\mathrm{P}=\frac{\mathrm{xy}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}}=\mathrm{m}^{\mathrm{2}} \\ $$$$\mathrm{where}\:\mathrm{y}=\mathrm{mx},\:\:\mathrm{x}\in\mathscr{P}\left(\mathrm{prime}\right) \\ $$$$\left.\mathrm{ii}\right)\mathrm{Case}\:\mathrm{y}=\mathrm{0}\:\mathrm{then}\:\mathrm{P}=\frac{\mathrm{xz}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{x}}=\mathrm{n}^{\mathrm{2}} \\ $$$$\mathrm{where}\:\:\mathrm{z}=\mathrm{nx},\mathrm{x}\in\mathscr{P}\left(\mathrm{prime}\right) \\ $$$$\left.\mathrm{iii}\right)\mathrm{Case}\:\mathrm{y},\mathrm{z}\neq\mathrm{0} \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{zx}+\mathrm{y}=\mathrm{m}\left(\mathrm{yz}+\mathrm{x}\right),\mathrm{xy}+\mathrm{z}=\mathrm{n}\left(\mathrm{yz}+\mathrm{x}\right) \\ $$$$\mathrm{where}\:\mathrm{m},\mathrm{n}\in\mathrm{N}^{\ast} .\mathrm{Adding}\:\mathrm{two}\:\mathrm{equalities} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{x}\left(\mathrm{y}+\mathrm{z}\right)+\mathrm{y}+\mathrm{z}=\left(\mathrm{m}+\mathrm{n}\right)\left(\mathrm{yz}+\mathrm{x}\right) \\ $$$$\Leftrightarrow\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{z}\right)=\left(\mathrm{m}+\mathrm{n}\right)\left(\mathrm{yz}+\mathrm{x}\right) \\ $$$$\mathrm{Since}\:\mathrm{yz}+\mathrm{x}\:\mathrm{is}\:\mathrm{prime}\:\mathrm{we}\:\mathrm{infer}\:\mathrm{that} \\ $$$$\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{x}+\mathrm{1}\right)\:\mathrm{or}\:\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{y}+\mathrm{z}\right) \\ $$$$\left.\mathrm{a}\right)\:\mathrm{If}\:\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{x}+\mathrm{1}\right)\mathrm{then}\:\mathrm{yz}+\mathrm{x}\leqslant\mathrm{x}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{yz}\leqslant\mathrm{1}\Rightarrow\mathrm{y}=\mathrm{z}=\mathrm{1}\left(\mathrm{since}\:\mathrm{y},\mathrm{z}\in\mathrm{N},\neq\mathrm{0}\right) \\ $$$$\mathrm{Then}\:\mathrm{P}=\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1}\left(\mathrm{Here}\:\mathrm{we}\:\mathrm{see}\:\mathrm{that}\right. \\ $$$$\left.\mathrm{has}\:\mathrm{infinite}\:\mathrm{x}\:\mathrm{so}\:\mathrm{that}\:\mathrm{x}+\mathrm{1}\:\mathrm{is}\:\mathrm{prime}\right) \\ $$$$\left.\mathrm{b}\right)\mathrm{If}\:\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{y}+\mathrm{z}\right)\mathrm{then}\:\mathrm{yz}+\mathrm{x}\leqslant\mathrm{y}+\mathrm{z} \\ $$$$\Leftrightarrow\left(\mathrm{y}−\mathrm{1}\right)\left(\mathrm{z}−\mathrm{1}\right)\leqslant\mathrm{1}−\mathrm{x}\left(\mathrm{1}\right) \\ $$$$\mathrm{Since}\:\mathrm{y}−\mathrm{1}\geqslant\mathrm{0},\mathrm{z}−\mathrm{1}\geqslant\mathrm{0},\mathrm{1}−\mathrm{x}\leqslant\mathrm{0}\:,\mathrm{we}\:\mathrm{see} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equality}\:\left(\mathrm{1}\right)\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only} \\ $$$$\mathrm{if}\:\mathrm{x}=\mathrm{y}=\mathrm{z}=\mathrm{1}.\mathrm{Then} \\ $$$$\mathrm{P}=\frac{\mathrm{2}.\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\mathrm{Combining}\:\mathrm{all}\:\mathrm{cases}\:\mathrm{we}\:\mathrm{have} \\ $$$$\boldsymbol{\mathrm{P}}=\boldsymbol{\mathrm{m}}^{\mathrm{2}} \:\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{z}}=\mathrm{0},\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{mx}},\boldsymbol{\mathrm{x}}\in\mathscr{P} \\ $$$$\boldsymbol{\mathrm{P}}=\boldsymbol{\mathrm{n}}^{\mathrm{2}\:} \boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{y}}=\mathrm{0},\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{nx}},\boldsymbol{\mathrm{x}}\in\mathscr{P} \\ $$$$\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{P}}=\mathrm{1}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{z}}=\mathrm{1},\:\boldsymbol{\mathrm{x}}\in\mathscr{P} \\ $$$$\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\left.\mathrm{3}\right)\mathrm{Find}\:\mid\left\{\mathrm{n}\in\mathrm{N}\mid\mathrm{n}\leqslant\mathrm{100},\mathrm{4}!\mid\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} \right\}\mid \\ $$$$\mathrm{We}\:\mathrm{need}\:\mathrm{find}\:\mathrm{n}\in\mathrm{N}\mid\mathrm{n}\leqslant\mathrm{100}\:\mathrm{and} \\ $$$$\mathrm{A}=\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} \vdots\mathrm{4}!=\mathrm{24} \\ $$$$\mathrm{If}\:\mathrm{n}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{then}\:\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} \mathrm{is}\:\mathrm{odd}\:\mathrm{number} \\ $$$$,\mathrm{so}\:\mathrm{A}\:\mathrm{isn}'\mathrm{t}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{24}.\mathrm{Hence},\mathrm{n} \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{an}\:\mathrm{even}\:\mathrm{number}\:,\mathrm{infer} \\ $$$$\mathrm{n}=\mathrm{2k}\Rightarrow\mathrm{A}=\mathrm{2}^{\mathrm{2k}} −\left(\mathrm{2k}\right)^{\mathrm{3}} =\mathrm{4}^{\mathrm{k}} −\mathrm{8k}^{\mathrm{3}} \left(\mathrm{1}\right) \\ $$$$\mathrm{We}\:\mathrm{need}\:\mathrm{find}\:\mathrm{k}\:\mathrm{so}\:\mathrm{that}\:\mathrm{A}\vdots\mathrm{24} \\ $$$$\mathrm{We}\:\mathrm{need}\:\mathrm{must}\:\mathrm{have}\:\mathrm{0}\leqslant\mathrm{A}\leqslant\mathrm{100},\mathrm{so} \\ $$$$\mathrm{k}=\mathrm{5}\:\mathrm{because}\:\mathrm{when}\:\mathrm{k}=\mathrm{4}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{A}=\mathrm{4}^{\mathrm{4}} −\mathrm{8}.\mathrm{4}^{\mathrm{3}} <\mathrm{0}.\mathrm{When}\:\mathrm{k}=\mathrm{6}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{A}=\mathrm{4}^{\mathrm{6}} −\mathrm{8}.\mathrm{6}^{\mathrm{3}} =\mathrm{2368} \\ $$$$\mathrm{when}\:\mathrm{k}=\mathrm{5}\:\mathrm{we}\:\mathrm{get}\:\mathrm{A}=\mathrm{4}^{\mathrm{5}} −\mathrm{8}.\mathrm{5}^{\mathrm{3}} = \\ $$$$\mathrm{1024}−\mathrm{8}.\mathrm{125}=\mathrm{24}\vdots\mathrm{4}! \\ $$$$\mathrm{Thus},\mathrm{has}\:\mathrm{unique}\:\mathrm{n}=\mathrm{10}\:\mathrm{satisfying} \\ $$$$\mathrm{the}\:\mathrm{condition}\:\mathrm{of}\:\mathrm{our}\:\mathrm{problem}:\mathrm{n}\in\mathrm{N}, \\ $$$$\boldsymbol{\mathrm{n}}\leqslant\mathrm{100},\mathrm{2}^{\boldsymbol{\mathrm{n}}} −\boldsymbol{\mathrm{n}}^{\mathrm{3}} \vdots\mathrm{4}!.\mathrm{Therefore} \\ $$$$\mid\left\{\boldsymbol{\mathrm{n}}\in\boldsymbol{\mathrm{N}}\mid\boldsymbol{\mathrm{n}}\leqslant\mathrm{100},\mathrm{2}^{\boldsymbol{\mathrm{n}}} −\boldsymbol{\mathrm{n}}^{\mathrm{3}} \vdots\mathrm{4}!\right\}\mid=\mathrm{1} \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Thanks. Please do the rest.
$$\mathrm{Thanks}.\:\mathrm{Please}\:\mathrm{do}\:\mathrm{the}\:\mathrm{rest}. \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Why did you change the solution for No. 2 number theory? What′s wrong with the first? We weren′t asked to find values for x,y and z.
$$\mathrm{Why}\:\mathrm{did}\:\mathrm{you}\:\mathrm{change}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{for} \\ $$$$\mathrm{No}.\:\mathrm{2}\:\mathrm{number}\:\mathrm{theory}?\:\mathrm{What}'\mathrm{s}\:\mathrm{wrong} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{first}?\:\mathrm{We}\:\mathrm{weren}'\mathrm{t}\:\mathrm{asked}\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{values}\:\mathrm{for}\:\mathrm{x},\mathrm{y}\:\mathrm{and}\:\mathrm{z}. \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
x,y,z are natural numbers. They can′t be 0. I think your first solution is correct. Please send it again.
$$\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{are}\:\mathrm{natural}\:\mathrm{numbers}.\:\mathrm{They}\:\mathrm{can}'\mathrm{t} \\ $$$$\mathrm{be}\:\mathrm{0}.\:\mathrm{I}\:\mathrm{think}\:\mathrm{your}\:\mathrm{first}\:\mathrm{solution}\:\mathrm{is} \\ $$$$\mathrm{correct}.\:\mathrm{Please}\:\mathrm{send}\:\mathrm{it}\:\mathrm{again}. \\ $$
Commented by 1549442205PVT last updated on 07/Sep/20
Ok,excuse me .I mistake.Need note the condition (∗).Set of natural numbers by new definition include zero N={0,1,2,.....},N^∗ ={1,2,3,...}
$$\mathrm{Ok},\mathrm{excuse}\:\mathrm{me}\:.\mathrm{I}\:\mathrm{mistake}.\mathrm{Need}\:\mathrm{note} \\ $$$$\mathrm{the}\:\mathrm{condition}\:\left(\ast\right).\mathrm{Set}\:\mathrm{of}\:\mathrm{natural} \\ $$$$\mathrm{numbers}\:\mathrm{by}\:\mathrm{new}\:\mathrm{definition}\:\mathrm{include} \\ $$$$\mathrm{zero}\:\mathrm{N}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},…..\right\},\mathrm{N}^{\ast} =\left\{\mathrm{1},\mathrm{2},\mathrm{3},…\right\} \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
But you are going against the question.
$$\mathrm{But}\:\mathrm{you}\:\mathrm{are}\:\mathrm{going}\:\mathrm{against}\:\mathrm{the} \\ $$$$\mathrm{question}. \\ $$
Commented by 1549442205PVT last updated on 08/Sep/20
I answered correctly requirements of question
$$\mathrm{I}\:\mathrm{answered}\:\:\mathrm{correctly}\:\mathrm{requirements}\:\mathrm{of} \\ $$$$\mathrm{question}\: \\ $$
Commented by Aina Samuel Temidayo last updated on 08/Sep/20
Answer to no. 3 is 17
$$\mathrm{Answer}\:\mathrm{to}\:\mathrm{no}.\:\mathrm{3}\:\mathrm{is}\:\mathrm{17} \\ $$

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