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Find-n-N-n-100-4-2-n-n-3-Note-that-S-denotes-the-cardinality-or-number-of-elements-of-a-set-S-




Question Number 112011 by Aina Samuel Temidayo last updated on 05/Sep/20
Find ∣{n∈N∣n≤100, 4!∣2^n −n^3 }∣.  (Note that ∣S∣ denotes the cardinality  or number of elements of a set,S).
$$\mathrm{Find}\:\mid\left\{\mathrm{n}\in\mathbb{N}\mid\mathrm{n}\leqslant\mathrm{100},\:\mathrm{4}!\mid\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} \right\}\mid. \\ $$$$\left(\mathrm{Note}\:\mathrm{that}\:\mid\mathrm{S}\mid\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{cardinality}\right. \\ $$$$\left.\mathrm{or}\:\mathrm{number}\:\mathrm{of}\:\mathrm{elements}\:\mathrm{of}\:\mathrm{a}\:\mathrm{set},\mathrm{S}\right). \\ $$
Answered by Rasheed.Sindhi last updated on 07/Sep/20
n may be odd or even.  ^▶ n=2k+1       2^n −n^3 =2^(2k+1) −(2k+1)^3        =2.4^k −(8k^3 +1+3(2k)(2k+1))       =2.4^k −8k^3 −1−12k^2 −6k       =2.4^k −8k^3 −12k^2 −6k−1      =an odd number(not divisible                                       by  4)  ^▶ n=2k         2^n −n^3 =2^(2k) −(2k)^3                         =4^k −8k^3                         =4(4^(k−1) −2k^3 )  ^▶ Hence only for n∈E        4 ∣ 2^n −n^3   ^▶ Even numbers upto 100 is 50  This proves that  ∣{n∈N∣n≤100, 4!∣2^n −n^3 }∣=50  Recall that I considerd      N={1,2,3,...}   ( If we consider     N={0,1,2,3,...}      Then the answer is 51)
$${n}\:{may}\:{be}\:{odd}\:{or}\:{even}. \\ $$$$\:^{\blacktriangleright} {n}=\mathrm{2}{k}+\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} =\mathrm{2}^{\mathrm{2}{k}+\mathrm{1}} −\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:=\mathrm{2}.\mathrm{4}^{{k}} −\left(\mathrm{8}{k}^{\mathrm{3}} +\mathrm{1}+\mathrm{3}\left(\mathrm{2}{k}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\right) \\ $$$$\:\:\:\:\:=\mathrm{2}.\mathrm{4}^{{k}} −\mathrm{8}{k}^{\mathrm{3}} −\mathrm{1}−\mathrm{12}{k}^{\mathrm{2}} −\mathrm{6}{k} \\ $$$$\:\:\:\:\:=\mathrm{2}.\mathrm{4}^{{k}} −\mathrm{8}{k}^{\mathrm{3}} −\mathrm{12}{k}^{\mathrm{2}} −\mathrm{6}{k}−\mathrm{1} \\ $$$$\:\:\:\:={an}\:{odd}\:{number}\left({not}\:{divisible}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{by}\:\:\mathrm{4}\right) \\ $$$$\:^{\blacktriangleright} {n}=\mathrm{2}{k} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} =\mathrm{2}^{\mathrm{2}{k}} −\left(\mathrm{2}{k}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}^{{k}} −\mathrm{8}{k}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\left(\mathrm{4}^{{k}−\mathrm{1}} −\mathrm{2}{k}^{\mathrm{3}} \right) \\ $$$$\:^{\blacktriangleright} \mathcal{H}{ence}\:{only}\:{for}\:{n}\in\mathbb{E} \\ $$$$\:\:\:\:\:\:\mathrm{4}\:\mid\:\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} \\ $$$$\:^{\blacktriangleright} \mathcal{E}{ven}\:{numbers}\:{upto}\:\mathrm{100}\:{is}\:\mathrm{50} \\ $$$${This}\:{proves}\:{that} \\ $$$$\mid\left\{\mathrm{n}\in\mathbb{N}\mid\mathrm{n}\leqslant\mathrm{100},\:\mathrm{4}!\mid\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} \right\}\mid=\mathrm{50} \\ $$$${Recall}\:{that}\:{I}\:{considerd}\: \\ $$$$\:\:\:\mathbb{N}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},…\right\} \\ $$$$\:\left(\:{If}\:{we}\:{consider}\right. \\ $$$$\:\:\:\mathbb{N}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},…\right\} \\ $$$$\left.\:\:\:\:{Then}\:{the}\:{answer}\:{is}\:\mathrm{51}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 07/Sep/20
Sorry I can′t do it.However  with the help of calculator, the  numbers are 4,10,16,....,6n−2  So the number of these numbers  upto 100 is 17
$${Sorry}\:{I}\:{can}'{t}\:{do}\:{it}.{However} \\ $$$${with}\:{the}\:{help}\:{of}\:{calculator},\:{the} \\ $$$${numbers}\:{are}\:\mathrm{4},\mathrm{10},\mathrm{16},….,\mathrm{6}{n}−\mathrm{2} \\ $$$${So}\:{the}\:{number}\:{of}\:{these}\:{numbers} \\ $$$${upto}\:\mathrm{100}\:{is}\:\mathrm{17} \\ $$
Commented by kaivan.ahmadi last updated on 05/Sep/20
hi dear rasheed  the number in question is 4!=4×3×2×1
$${hi}\:{dear}\:{rasheed} \\ $$$${the}\:{number}\:{in}\:{question}\:{is}\:\mathrm{4}!=\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1} \\ $$
Commented by Rasheed.Sindhi last updated on 05/Sep/20
Misreading sir misreading!  I′ll review my answer.  ThanX  dear kaivan!
$${Misreading}\:{sir}\:{misreading}! \\ $$$${I}'{ll}\:{review}\:{my}\:{answer}. \\ $$$$\mathcal{T}{han}\mathcal{X}\:\:{dear}\:{kaivan}! \\ $$
Commented by Aina Samuel Temidayo last updated on 06/Sep/20
Please I′m still waiting for the  correction. Thanks.
$$\mathrm{Please}\:\mathrm{I}'\mathrm{m}\:\mathrm{still}\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{correction}.\:\mathrm{Thanks}. \\ $$
Answered by Rasheed.Sindhi last updated on 09/Sep/20
4! ∣ 2^n −n^3   24∣2^n −n^3   2^n ≡n^3 (mod 24)    Let  { ((2^n ≡u(mod 24))),((n^3 ≡v(mod 24)) :}                [(n,(2^n mod 24_((u)) ),(n^3 mod 24_((v)) ),),(1,(         2),(       1),),(2,(        4),(       8),),(3,(        8),(       3),),(4,(      16),(     16),(u=v)),(5,(       8),(      5),),(6,(     16),(      0),),(7,(       8),(     7),),(8,(     16),(     8),),(9,(       8),(     9),),((10),(     16),(   16),(u=v)),((11),(       8),(   11),),((12),(     16),(    0),),((13),(       8),(   13),),((14),(     16),(    8),),((15),(       8),(  15),),((16),(     16),(  16),(u=v)),((17),(       8),(  17),),((18),(     16),(   0),),((19),(       8),(  19),),((20),(     16),(   8),),((21),(      8),(  21),),((22),(    16),(  16),(u=v)),((23),(       8),(  23),),((24),(    16),(   0),) ]  Values of n for which u=v or  2^n ≡n^3 (mod 24) or in other words  4! ∣ 2^n −n^3  are:   4,10,16,...(6n−2)..  This is an AP upto 100 of which  are 17 terms.  ∴ 4! ∣ 2^n −n^3  has 17 solutions       for n∈N ∧ n≤100.
$$\mathrm{4}!\:\mid\:\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} \\ $$$$\mathrm{24}\mid\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} \\ $$$$\mathrm{2}^{{n}} \equiv{n}^{\mathrm{3}} \left({mod}\:\mathrm{24}\right) \\ $$$$ \\ $$$${Let}\:\begin{cases}{\mathrm{2}^{{n}} \equiv{u}\left({mod}\:\mathrm{24}\right)}\\{{n}^{\mathrm{3}} \equiv{v}\left({mod}\:\mathrm{24}\right.}\end{cases} \\ $$$$\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\begin{bmatrix}{{n}}&{\underset{\left({u}\right)} {\mathrm{2}^{{n}} {mod}\:\mathrm{24}}}&{\underset{\left({v}\right)} {{n}^{\mathrm{3}} {mod}\:\mathrm{24}}}&{}\\{\mathrm{1}}&{\:\:\:\:\:\:\:\:\:\mathrm{2}}&{\:\:\:\:\:\:\:\mathrm{1}}&{}\\{\mathrm{2}}&{\:\:\:\:\:\:\:\:\mathrm{4}}&{\:\:\:\:\:\:\:\mathrm{8}}&{}\\{\mathrm{3}}&{\:\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\:\:\:\:\:\mathrm{3}}&{}\\{\mathrm{4}}&{\:\:\:\:\:\:\mathrm{16}}&{\:\:\:\:\:\mathrm{16}}&{{u}={v}}\\{\mathrm{5}}&{\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\:\:\:\:\mathrm{5}}&{}\\{\mathrm{6}}&{\:\:\:\:\:\mathrm{16}}&{\:\:\:\:\:\:\mathrm{0}}&{}\\{\mathrm{7}}&{\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\:\:\:\mathrm{7}}&{}\\{\mathrm{8}}&{\:\:\:\:\:\mathrm{16}}&{\:\:\:\:\:\mathrm{8}}&{}\\{\mathrm{9}}&{\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\:\:\:\mathrm{9}}&{}\\{\mathrm{10}}&{\:\:\:\:\:\mathrm{16}}&{\:\:\:\mathrm{16}}&{{u}={v}}\\{\mathrm{11}}&{\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\:\mathrm{11}}&{}\\{\mathrm{12}}&{\:\:\:\:\:\mathrm{16}}&{\:\:\:\:\mathrm{0}}&{}\\{\mathrm{13}}&{\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\:\mathrm{13}}&{}\\{\mathrm{14}}&{\:\:\:\:\:\mathrm{16}}&{\:\:\:\:\mathrm{8}}&{}\\{\mathrm{15}}&{\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\mathrm{15}}&{}\\{\mathrm{16}}&{\:\:\:\:\:\mathrm{16}}&{\:\:\mathrm{16}}&{{u}={v}}\\{\mathrm{17}}&{\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\mathrm{17}}&{}\\{\mathrm{18}}&{\:\:\:\:\:\mathrm{16}}&{\:\:\:\mathrm{0}}&{}\\{\mathrm{19}}&{\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\mathrm{19}}&{}\\{\mathrm{20}}&{\:\:\:\:\:\mathrm{16}}&{\:\:\:\mathrm{8}}&{}\\{\mathrm{21}}&{\:\:\:\:\:\:\mathrm{8}}&{\:\:\mathrm{21}}&{}\\{\mathrm{22}}&{\:\:\:\:\mathrm{16}}&{\:\:\mathrm{16}}&{{u}={v}}\\{\mathrm{23}}&{\:\:\:\:\:\:\:\mathrm{8}}&{\:\:\mathrm{23}}&{}\\{\mathrm{24}}&{\:\:\:\:\mathrm{16}}&{\:\:\:\mathrm{0}}&{}\end{bmatrix} \\ $$$${Values}\:{of}\:{n}\:{for}\:{which}\:{u}={v}\:{or} \\ $$$$\mathrm{2}^{{n}} \equiv{n}^{\mathrm{3}} \left({mod}\:\mathrm{24}\right)\:{or}\:{in}\:{other}\:{words} \\ $$$$\mathrm{4}!\:\mid\:\mathrm{2}^{{n}} −{n}^{\mathrm{3}} \:{are}: \\ $$$$\:\mathrm{4},\mathrm{10},\mathrm{16},…\left(\mathrm{6}{n}−\mathrm{2}\right).. \\ $$$$\mathcal{T}{his}\:{is}\:{an}\:{AP}\:{upto}\:\mathrm{100}\:{of}\:{which} \\ $$$${are}\:\mathrm{17}\:{terms}. \\ $$$$\therefore\:\mathrm{4}!\:\mid\:\mathrm{2}^{\boldsymbol{{n}}} −\boldsymbol{{n}}^{\mathrm{3}} \:\boldsymbol{{has}}\:\mathrm{17}\:\boldsymbol{{solutions}} \\ $$$$\:\:\:\:\:\boldsymbol{{for}}\:\boldsymbol{{n}}\in\mathbb{N}\:\wedge\:\boldsymbol{{n}}\leqslant\mathrm{100}. \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Can you please complete it?
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{complete}\:\mathrm{it}? \\ $$
Commented by Rasheed.Sindhi last updated on 09/Sep/20
The Answer is now complete.  Sorry for late.
$$\mathcal{T}{he}\:\mathcal{A}{nswer}\:{is}\:{now}\:\boldsymbol{{complete}}. \\ $$$$\mathcal{S}{orry}\:{for}\:{late}. \\ $$
Commented by Aina Samuel Temidayo last updated on 09/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$

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