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Question Number 110565 by Aina Samuel Temidayo last updated on 29/Aug/20

Let n∈N. Using the formula lcm(a,b) = ((ab)/(gcd(a,b))) and lcm(a,b,c) =lcm(lcm(a,b),c), find all the possible value of ((6•lcm(n,n+1,n+2,n+3))/(n(n+1)(n+2)(n+3)))

$$\mathrm{Let}\:\mathrm{n}\in\mathbb{N}.\:\mathrm{Using}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{lcm}\left(\mathrm{a},\mathrm{b}\right) \\ $$$$=\:\frac{\mathrm{ab}}{\mathrm{gcd}\left(\mathrm{a},\mathrm{b}\right)}\:\mathrm{and}\:\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right) \\ $$$$=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b}\right),\mathrm{c}\right),\:\mathrm{find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{possible} \\ $$$$\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{6}\bullet\mathrm{lcm}\left(\mathrm{n},\mathrm{n}+\mathrm{1},\mathrm{n}+\mathrm{2},\mathrm{n}+\mathrm{3}\right)}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)} \\ $$

Commented by kaivan.ahmadi last updated on 29/Aug/20

let m=((6.lcm(n,n+1,n+2,n+3))/(n(n+1)(n+2)(n+3))) lcm(n,n+1)=n(n+1)= lcm(n+2,n+3)=(n+2)(n+3) if n=2k⇒lcm(n(n+1),(n+2)(n+3))= lcm(2k(2k+1),2(k+1)(2k+3))=2k(k+1)(2k+1)(2k+3) ⇒m=((12k(k+1)(2k+1)(2k+3))/(4k(2k+1)(k+1)(2k+3)))=3 if n=2k+1⇒lcm(n(n+1),(n+2)(n+3))= lcm(2(2k+1)(k+1),2(2k+3)(k+2))= 2(k+1)(k+2)(2k+1)(2k+3) ⇒m=((12(k+1)(k+2)(2k+1)(2k+3))/(4(2k+1)(k+1)(2k+3)(k+2)))=3 so m=3

$${let}\:{m}=\frac{\mathrm{6}.{lcm}\left({n},{n}+\mathrm{1},{n}+\mathrm{2},{n}+\mathrm{3}\right)}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$${lcm}\left({n},{n}+\mathrm{1}\right)={n}\left({n}+\mathrm{1}\right)= \\ $$$${lcm}\left({n}+\mathrm{2},{n}+\mathrm{3}\right)=\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right) \\ $$$${if}\:{n}=\mathrm{2}{k}\Rightarrow{lcm}\left({n}\left({n}+\mathrm{1}\right),\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)\right)= \\ $$$${lcm}\left(\mathrm{2}{k}\left(\mathrm{2}{k}+\mathrm{1}\right),\mathrm{2}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)\right)=\mathrm{2}{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right) \\ $$$$\Rightarrow{m}=\frac{\mathrm{12}{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)}{\mathrm{4}{k}\left(\mathrm{2}{k}+\mathrm{1}\right)\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)}=\mathrm{3} \\ $$$${if}\:{n}=\mathrm{2}{k}+\mathrm{1}\Rightarrow{lcm}\left({n}\left({n}+\mathrm{1}\right),\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)\right)= \\ $$$${lcm}\left(\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)\left({k}+\mathrm{1}\right),\mathrm{2}\left(\mathrm{2}{k}+\mathrm{3}\right)\left({k}+\mathrm{2}\right)\right)= \\ $$$$\mathrm{2}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right) \\ $$$$\Rightarrow{m}=\frac{\mathrm{12}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)}{\mathrm{4}\left(\mathrm{2}{k}+\mathrm{1}\right)\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)\left({k}+\mathrm{2}\right)}=\mathrm{3} \\ $$$${so}\:{m}=\mathrm{3} \\ $$$$ \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Thanks

$$\mathrm{Thanks} \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

If anyone has any other solution. Please post it.

$$\mathrm{If}\:\mathrm{anyone}\:\mathrm{has}\:\mathrm{any}\:\mathrm{other}\:\mathrm{solution}. \\ $$$$\mathrm{Please}\:\mathrm{post}\:\mathrm{it}. \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Why does lcm(n,n+1,n+2,n+3) = lcm((n(n+1),(n+2)(n+3))? @kaivan.ahmadi

$$\mathrm{Why}\:\mathrm{does}\:\mathrm{lcm}\left(\mathrm{n},\mathrm{n}+\mathrm{1},\mathrm{n}+\mathrm{2},\mathrm{n}+\mathrm{3}\right)\:= \\ $$$$\mathrm{lcm}\left(\left(\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right),\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)\right)?\right. \\ $$$$@\mathrm{kaivan}.\mathrm{ahmadi} \\ $$

Commented by floor(10²Eta[1]) last updated on 29/Aug/20

because lcm(a,b,c)=lcm(lcm(a,b),c) the same logic works for 4 numbers: lcm(a,b,c,d)=lcm(lcm(a,b),lcm(c,d)) since n and n+1 are coprimes and n+2 and n+3 also are coprimes so lcm(n, n+1)=n(n+1) and lcm(n+2, n+3)=(n+2)(n+3) so lcm(n, n+1, n+2, n+3)= lcm((n(n+1),(n+2)(n+3))

$$\mathrm{because}\:\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b}\right),\mathrm{c}\right) \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{logic}\:\mathrm{works}\:\mathrm{for}\:\mathrm{4}\:\mathrm{numbers}: \\ $$$$\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b}\right),\mathrm{lcm}\left(\mathrm{c},\mathrm{d}\right)\right) \\ $$$$\mathrm{since}\:\mathrm{n}\:\mathrm{and}\:\mathrm{n}+\mathrm{1}\:\mathrm{are}\:\mathrm{coprimes}\:\mathrm{and} \\ $$$$\mathrm{n}+\mathrm{2}\:\mathrm{and}\:\mathrm{n}+\mathrm{3}\:\mathrm{also}\:\mathrm{are}\:\mathrm{coprimes}\:\mathrm{so} \\ $$$$\mathrm{lcm}\left(\mathrm{n},\:\mathrm{n}+\mathrm{1}\right)=\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\:\mathrm{and} \\ $$$$\mathrm{lcm}\left(\mathrm{n}+\mathrm{2},\:\mathrm{n}+\mathrm{3}\right)=\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right) \\ $$$$\mathrm{so} \\ $$$$\mathrm{lcm}\left(\mathrm{n},\:\mathrm{n}+\mathrm{1},\:\mathrm{n}+\mathrm{2},\:\mathrm{n}+\mathrm{3}\right)= \\ $$$$\mathrm{lcm}\left(\left(\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right),\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)\right)\right. \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

lcm(a,b,c,d)=lcm(lcm(a,b,c),d)=lcm(lcm(lcm(a,b),c),d)

$$\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right),\mathrm{d}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b}\right),\mathrm{c}\right),\mathrm{d}\right) \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

I don′t think lcm(a,b,c,d)=lcm(lcm(a,b),lcm(c,d)) is a good argument.

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think} \\ $$$$\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b}\right),\mathrm{lcm}\left(\mathrm{c},\mathrm{d}\right)\right) \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{good}\:\mathrm{argument}. \\ $$

Commented by floor(10²Eta[1]) last updated on 29/Aug/20

well... it′s true even you think that is not a ′′good argument′′

$$\mathrm{well}...\:\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\mathrm{even}\:\mathrm{you}\:\mathrm{think} \\ $$$$\mathrm{that}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:''\mathrm{good}\:\mathrm{argument}'' \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Can you prove it?

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{prove}\:\mathrm{it}? \\ $$

Commented by floor(10²Eta[1]) last updated on 29/Aug/20

are you serious?? just think about for a second, it′s the same logic of why lcm(a,b,c)=lcm(lcm(a,b),c)

$$\mathrm{are}\:\mathrm{you}\:\mathrm{serious}??\:\mathrm{just}\:\mathrm{think}\:\mathrm{about}\:\mathrm{for}\: \\ $$$$\mathrm{a}\:\mathrm{second},\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{same}\:\mathrm{logic}\:\mathrm{of}\:\mathrm{why} \\ $$$$\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b}\right),\mathrm{c}\right) \\ $$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

lcm(a,b,c,d)=lcm(lcm(a,b,c),d)=lcm(lcm(lcm(a,b),c),d) This is what is correct. (Using that logic).

$$ \\ $$$$\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right),\mathrm{d}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b}\right),\mathrm{c}\right),\mathrm{d}\right) \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{what}\:\mathrm{is}\:\mathrm{correct}.\:\left(\mathrm{Using}\:\mathrm{that}\right. \\ $$$$\left.\mathrm{logic}\right). \\ $$

Commented by floor(10²Eta[1]) last updated on 30/Aug/20

yeah this is also correct

$$\mathrm{yeah}\:\mathrm{this}\:\mathrm{is}\:\mathrm{also}\:\mathrm{correct}\: \\ $$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

The answer is 3 or 1. I′ve been able to solve it.

$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{3}\:\mathrm{or}\:\mathrm{1}.\:\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{able} \\ $$$$\mathrm{to}\:\mathrm{solve}\:\mathrm{it}. \\ $$

Commented by floor(10²Eta[1]) last updated on 30/Aug/20

congratulations

$$\mathrm{congratulations} \\ $$$$ \\ $$

Commented by kaivan.ahmadi last updated on 30/Aug/20

let lcm(a,b)=x ,lcm(c,d)=y⇒ lcm(a,b,c,d)=lcm(lcm(a,b,c),d)= lcm(lcm(lcm(a,b),c),d)=lcm(lcm(x,c),d)= lcm(x,c,d)=lcm(x,lcm(c,d))=lcm(x,y)

$${let}\:{lcm}\left({a},{b}\right)={x}\:,{lcm}\left({c},{d}\right)={y}\Rightarrow \\ $$$${lcm}\left({a},{b},{c},{d}\right)={lcm}\left({lcm}\left({a},{b},{c}\right),{d}\right)= \\ $$$${lcm}\left({lcm}\left({lcm}\left({a},{b}\right),{c}\right),{d}\right)={lcm}\left({lcm}\left({x},{c}\right),{d}\right)= \\ $$$${lcm}\left({x},{c},{d}\right)={lcm}\left({x},{lcm}\left({c},{d}\right)\right)={lcm}\left({x},{y}\right) \\ $$$$ \\ $$

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