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if-x-is-a-selected-number-of-the-number-from-20-99-then-what-is-probalility-x-3-x-is-divided-by-12-

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Question Number 98773 by john santu last updated on 16/Jun/20
if x is a selected number of the number from 20−99, then what is probalility x^3 −x is divided by 12?
$$\mathrm{if}\:\mathrm{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{selected}\:\mathrm{number}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{from}\:\mathrm{20}−\mathrm{99},\:\mathrm{then}\:\mathrm{what} \\ $$$$\mathrm{is}\:\mathrm{probalility}\:\mathrm{x}^{\mathrm{3}} −\mathrm{x}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\mathrm{12}?\: \\ $$
Answered by mr W last updated on 16/Jun/20
x^3 −x=(x−1)x(x+1) x−1=12k ⇒x=12k+1, k=2..8 x=12k ⇒x=12k, k=2..8 x+1=12k ⇒x=12k−1, k=2..8 3×7=21 possible values for x ⇒p=((21)/(80))
$${x}^{\mathrm{3}} −{x}=\left({x}−\mathrm{1}\right){x}\left({x}+\mathrm{1}\right) \\ $$$${x}−\mathrm{1}=\mathrm{12}{k}\:\Rightarrow{x}=\mathrm{12}{k}+\mathrm{1},\:{k}=\mathrm{2}..\mathrm{8} \\ $$$${x}=\mathrm{12}{k}\:\Rightarrow{x}=\mathrm{12}{k},\:{k}=\mathrm{2}..\mathrm{8} \\ $$$${x}+\mathrm{1}=\mathrm{12}{k}\:\Rightarrow{x}=\mathrm{12}{k}−\mathrm{1},\:{k}=\mathrm{2}..\mathrm{8} \\ $$$$\mathrm{3}×\mathrm{7}=\mathrm{21}\:{possible}\:{values}\:{for}\:{x} \\ $$$$\Rightarrow{p}=\frac{\mathrm{21}}{\mathrm{80}} \\ $$
Commented by mr W last updated on 16/Jun/20
you are right sir! p=((60)/(80))=(3/4)
$${you}\:{are}\:{right}\:{sir}! \\ $$$${p}=\frac{\mathrm{60}}{\mathrm{80}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 16/Jun/20
case 1: x−1 is even, i.e. x is odd, i.e. x=21, 23, ..., 99 ⇒ 40 numbers case 2: x is divisible by 4, i.e. x=20, 24, ..., 96 ⇒ 20 numbers ⇒totally 60 values for x.
$${case}\:\mathrm{1}:\:{x}−\mathrm{1}\:{is}\:{even},\:{i}.{e}.\:{x}\:{is}\:{odd}, \\ $$$${i}.{e}.\:{x}=\mathrm{21},\:\mathrm{23},\:…,\:\mathrm{99}\:\Rightarrow\:\mathrm{40}\:{numbers} \\ $$$${case}\:\mathrm{2}:\:{x}\:{is}\:{divisible}\:{by}\:\mathrm{4}, \\ $$$${i}.{e}.\:{x}=\mathrm{20},\:\mathrm{24},\:…,\:\mathrm{96}\:\Rightarrow\:\mathrm{20}\:{numbers} \\ $$$$\Rightarrow{totally}\:\mathrm{60}\:{values}\:{for}\:{x}. \\ $$
Commented by john santu last updated on 16/Jun/20
yes. thank you friend
$$\mathrm{yes}.\:\mathrm{thank}\:\mathrm{you}\:\mathrm{friend}\: \\ $$
Commented by MJS last updated on 16/Jun/20
x^3 −x is always divisible by 3 thus we only need to check which numbers are divisible by 4
$${x}^{\mathrm{3}} −{x}\:\mathrm{is}\:\mathrm{always}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3} \\ $$$$\mathrm{thus}\:\mathrm{we}\:\mathrm{only}\:\mathrm{need}\:\mathrm{to}\:\mathrm{check}\:\mathrm{which}\:\mathrm{numbers} \\ $$$$\mathrm{are}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{4} \\ $$
Answered by MJS last updated on 16/Jun/20
n=x^3 −x=x(x−1)(x+1) ⇒ 3∣n ∀x∈Z ⇒ { ((x=2k+1 ⇒ 12∣n)),((x=2n ⇒ { ((x=4n ⇒ 12∣n)),((x=4n+2 ⇒ 12∤n)) :})) :} ⇒ for 20≤x≤99 we have 20 numbers for which 12∤n ⇒ searched probability is (3/4)
$${n}={x}^{\mathrm{3}} −{x}={x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right) \\ $$$$\Rightarrow\:\mathrm{3}\mid{n}\:\forall{x}\in\mathbb{Z} \\ $$$$\Rightarrow\:\begin{cases}{{x}=\mathrm{2}{k}+\mathrm{1}\:\Rightarrow\:\mathrm{12}\mid{n}}\\{{x}=\mathrm{2}{n}\:\Rightarrow\:\begin{cases}{{x}=\mathrm{4}{n}\:\Rightarrow\:\mathrm{12}\mid{n}}\\{{x}=\mathrm{4}{n}+\mathrm{2}\:\Rightarrow\:\mathrm{12}\nmid{n}}\end{cases}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{for}\:\mathrm{20}\leqslant{x}\leqslant\mathrm{99}\:\mathrm{we}\:\mathrm{have}\:\mathrm{20}\:\mathrm{numbers}\:\mathrm{for} \\ $$$$\mathrm{which}\:\mathrm{12}\nmid{n} \\ $$$$\Rightarrow\:\mathrm{searched}\:\mathrm{probability}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by john santu last updated on 16/Jun/20
why the method mr W not same sir. i got too((21)/(80))
$$\mathrm{why}\:\mathrm{the}\:\mathrm{method}\:\mathrm{mr}\:\mathrm{W}\:\mathrm{not}\:\mathrm{same}\:\mathrm{sir}. \\ $$$$\mathrm{i}\:\mathrm{got}\:\mathrm{too}\frac{\mathrm{21}}{\mathrm{80}} \\ $$
Commented by MJS last updated on 16/Jun/20
n=x(x−1)(x+1) x n 20 12×665 21 12×770 22 10626 23 12×1012 24 12×1150 25 12×1300 26 17550 27 12×1638 28 12×1827 29 12×2030 ...
$${n}={x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right) \\ $$$${x}\:\:\:\:\:{n} \\ $$$$\mathrm{20}\:\:\:\mathrm{12}×\mathrm{665} \\ $$$$\mathrm{21}\:\:\:\mathrm{12}×\mathrm{770} \\ $$$$\mathrm{22}\:\:\:\mathrm{10626} \\ $$$$\mathrm{23}\:\:\:\mathrm{12}×\mathrm{1012} \\ $$$$\mathrm{24}\:\:\:\mathrm{12}×\mathrm{1150} \\ $$$$\mathrm{25}\:\:\:\mathrm{12}×\mathrm{1300} \\ $$$$\mathrm{26}\:\:\:\mathrm{17550} \\ $$$$\mathrm{27}\:\:\:\mathrm{12}×\mathrm{1638} \\ $$$$\mathrm{28}\:\:\:\mathrm{12}×\mathrm{1827} \\ $$$$\mathrm{29}\:\:\:\mathrm{12}×\mathrm{2030} \\ $$$$… \\ $$
Commented by bemath last updated on 16/Jun/20
for x =22 , 26 , ... not satisfy sir
$$\mathrm{for}\:\mathrm{x}\:=\mathrm{22}\:,\:\mathrm{26}\:,\:…\:\mathrm{not}\:\mathrm{satisfy}\:\mathrm{sir} \\ $$
Commented by bemath last updated on 16/Jun/20
30 30×29×31 ∤ 12 34 34×33×35 ∤12 38 38×37×39 ∤ 12 correct sir?
$$\mathrm{30}\:\:\:\mathrm{30}×\mathrm{29}×\mathrm{31}\:\nmid\:\mathrm{12} \\ $$$$\mathrm{34}\:\:\:\:\mathrm{34}×\mathrm{33}×\mathrm{35}\:\nmid\mathrm{12} \\ $$$$\mathrm{38}\:\:\:\:\mathrm{38}×\mathrm{37}×\mathrm{39}\:\nmid\:\mathrm{12} \\ $$$$\mathrm{correct}\:\mathrm{sir}? \\ $$
Commented by MJS last updated on 16/Jun/20
yes
$${yes} \\ $$
Commented by john santu last updated on 16/Jun/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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