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Question Number 103606 by bemath last updated on 16/Jul/20
an integer n between 1 and 98 , inclusive is to be chosen at random. what is the probability that n(n+1) will be divisible by 3
$${an}\:{integer}\:{n}\:{between}\:\mathrm{1}\:{and}\:\mathrm{98}\:, \\ $$$${inclusive}\:{is}\:{to}\:{be}\:{chosen}\:{at} \\ $$$${random}.\:{what}\:{is}\:{the}\:{probability} \\ $$$${that}\:{n}\left({n}+\mathrm{1}\right)\:{will}\:{be}\:{divisible}\:{by}\:\mathrm{3} \\ $$
Commented by bemath last updated on 16/Jul/20
thank you all
$${thank}\:{you}\:{all} \\ $$
Answered by OlafThorendsen last updated on 16/Jul/20
1st case : n = 3p, p = 1..32 32 possibilities n(n+1) can be divisible by 3 2nd case : n = 3p+1 then n+1 = 3p+2 n(n+1) = (3p+1)(3p+2) n(n+1) can′t be divisible by 3 3rd case : n = 3p+2, p = 0..32 33 possibilities then n+1 = 3p+3 = 3(p+1) n(n+1) = 3(3p+2)(p+1) n(n+1) can be divisible by 3 Total result : 32+0+33 = 65 probability = ((65)/(98)) ≈ 66,3%
$$\mathrm{1st}\:\mathrm{case}\:: \\ $$$${n}\:=\:\mathrm{3}{p},\:{p}\:=\:\mathrm{1}..\mathrm{32} \\ $$$$\mathrm{32}\:\mathrm{possibilities} \\ $$$${n}\left({n}+\mathrm{1}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3} \\ $$$$\mathrm{2nd}\:\mathrm{case}\:: \\ $$$${n}\:=\:\mathrm{3}{p}+\mathrm{1}\:\mathrm{then}\:{n}+\mathrm{1}\:=\:\mathrm{3}{p}+\mathrm{2} \\ $$$${n}\left({n}+\mathrm{1}\right)\:=\:\left(\mathrm{3}{p}+\mathrm{1}\right)\left(\mathrm{3}{p}+\mathrm{2}\right) \\ $$$${n}\left({n}+\mathrm{1}\right)\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3} \\ $$$$\mathrm{3rd}\:\mathrm{case}\:: \\ $$$${n}\:=\:\mathrm{3}{p}+\mathrm{2},\:{p}\:=\:\mathrm{0}..\mathrm{32} \\ $$$$\mathrm{33}\:\mathrm{possibilities} \\ $$$$\mathrm{then}\:{n}+\mathrm{1}\:=\:\mathrm{3}{p}+\mathrm{3}\:=\:\mathrm{3}\left({p}+\mathrm{1}\right) \\ $$$${n}\left({n}+\mathrm{1}\right)\:=\:\mathrm{3}\left(\mathrm{3}{p}+\mathrm{2}\right)\left({p}+\mathrm{1}\right) \\ $$$${n}\left({n}+\mathrm{1}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3} \\ $$$$\mathrm{Total}\:\mathrm{result}\:: \\ $$$$\mathrm{32}+\mathrm{0}+\mathrm{33}\:=\:\mathrm{65} \\ $$$$\boldsymbol{\mathrm{probability}}\:=\:\frac{\mathrm{65}}{\mathrm{98}}\:\approx\:\mathrm{66},\mathrm{3\%} \\ $$
Answered by floor(10²Eta[1]) last updated on 16/Jul/20
for n(n+1) be divisible by 3, we have 2 cases: 1)n is a multiple of 3 2)n+1 is a multiple of 3 [1, 98]→98 terms 1)AP: (3, 6, 9, ..., 96)→{a_n =3n} 3n=96⇒n=32(the AP has 32 terms) so the probability that n is a multiple of 3 is: P_1 =((32)/(98))=((16)/(49)) 2)AP: (2, 5, 8, ..., 95)→32 terms probability of n+1 multiple of 3: P_2 =((32)/(98))=((16)/(49)) ⇒probability that n(n+1) is divisible by 3: P_1 +P_2 =((32)/(49))≈65%
$$\mathrm{for}\:\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\:\mathrm{be}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3},\:\mathrm{we}\:\mathrm{have}\:\mathrm{2} \\ $$$$\mathrm{cases}: \\ $$$$\left.\mathrm{1}\right)\mathrm{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$$$\left.\mathrm{2}\right)\mathrm{n}+\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$$$\left[\mathrm{1},\:\mathrm{98}\right]\rightarrow\mathrm{98}\:\mathrm{terms} \\ $$$$\left.\mathrm{1}\right)\mathrm{AP}:\:\left(\mathrm{3},\:\mathrm{6},\:\mathrm{9},\:…,\:\mathrm{96}\right)\rightarrow\left\{\mathrm{a}_{\mathrm{n}} =\mathrm{3n}\right\} \\ $$$$\mathrm{3n}=\mathrm{96}\Rightarrow\mathrm{n}=\mathrm{32}\left(\mathrm{the}\:\mathrm{AP}\:\mathrm{has}\:\mathrm{32}\:\mathrm{terms}\right) \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of} \\ $$$$\mathrm{3}\:\mathrm{is}:\:\mathrm{P}_{\mathrm{1}} =\frac{\mathrm{32}}{\mathrm{98}}=\frac{\mathrm{16}}{\mathrm{49}} \\ $$$$\left.\mathrm{2}\right)\mathrm{AP}:\:\left(\mathrm{2},\:\mathrm{5},\:\mathrm{8},\:…,\:\mathrm{95}\right)\rightarrow\mathrm{32}\:\mathrm{terms} \\ $$$$\mathrm{probability}\:\mathrm{of}\:\mathrm{n}+\mathrm{1}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}: \\ $$$$\mathrm{P}_{\mathrm{2}} =\frac{\mathrm{32}}{\mathrm{98}}=\frac{\mathrm{16}}{\mathrm{49}} \\ $$$$\Rightarrow\mathrm{probability}\:\mathrm{that}\:\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by}\:\mathrm{3}: \\ $$$$\mathrm{P}_{\mathrm{1}} +\mathrm{P}_{\mathrm{2}} =\frac{\mathrm{32}}{\mathrm{49}}\approx\mathrm{65\%} \\ $$
Commented by OlafThorendsen last updated on 16/Jul/20
Why n = 98 cannot be chosen in 2) ?
$$\mathrm{Why}\:{n}\:=\:\mathrm{98}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{chosen}\:\mathrm{in} \\ $$$$\left.\mathrm{2}\right)\:? \\ $$
Commented by floor(10²Eta[1]) last updated on 16/Jul/20
yeah you′re right i miss that :/
$$\mathrm{yeah}\:\mathrm{you}'\mathrm{re}\:\mathrm{right}\:\mathrm{i}\:\mathrm{miss}\:\mathrm{that}\::/ \\ $$
Answered by bobhans last updated on 16/Jul/20
note that n(n+1) is not divisible by 3 precisely when n ≡ 1 (mod 3) ; that is n has remainder 1 upon division by 3. from 1 to 98 inclusive there are 33 such values of n. hence the desired probability equals to 1−((33)/(98)) = ((65)/(98)) . [ p(A) = 1−p(A^c ) ] □
$${note}\:{that}\:{n}\left({n}+\mathrm{1}\right)\:{is}\:{not}\:{divisible}\:{by}\:\mathrm{3} \\ $$$${precisely}\:{when}\:{n}\:\equiv\:\mathrm{1}\:\left({mod}\:\mathrm{3}\right)\:;\:{that}\:{is}\:{n} \\ $$$${has}\:{remainder}\:\mathrm{1}\:{upon}\:{division}\:{by}\:\mathrm{3}.\:{from} \\ $$$$\mathrm{1}\:{to}\:\mathrm{98}\:{inclusive}\:{there}\:{are}\:\mathrm{33}\:{such}\:{values} \\ $$$${of}\:{n}.\:{hence}\:{the}\:{desired}\:{probability}\:{equals} \\ $$$${to}\:\mathrm{1}−\frac{\mathrm{33}}{\mathrm{98}}\:=\:\frac{\mathrm{65}}{\mathrm{98}}\:.\: \\ $$$$\left[\:{p}\left({A}\right)\:=\:\mathrm{1}−{p}\left({A}^{{c}} \right)\:\right]\:\square \\ $$

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