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Question Number 19574 by Tinkutara last updated on 12/Aug/17
Carol was given three numbers and was asked to add the largest of the three to the product of the other two. Instead, she multiplied the largest with the sum of the other two, but still got the right answer. What is the sum of the three numbers?
$$\mathrm{Carol}\:\mathrm{was}\:\mathrm{given}\:\mathrm{three}\:\mathrm{numbers}\:\mathrm{and} \\ $$$$\mathrm{was}\:\mathrm{asked}\:\mathrm{to}\:\mathrm{add}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{three}\:\mathrm{to}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{two}. \\ $$$$\mathrm{Instead},\:\mathrm{she}\:\mathrm{multiplied}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{two},\:\mathrm{but}\:\mathrm{still}\:\mathrm{got} \\ $$$$\mathrm{the}\:\mathrm{right}\:\mathrm{answer}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{three}\:\mathrm{numbers}? \\ $$
Commented by Tinkutara last updated on 13/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 12/Aug/17
positive integers ?
$$\mathrm{positive}\:\mathrm{integers}\:? \\ $$
Commented by Tinkutara last updated on 12/Aug/17
This was the exact question but it was cancelled. So I think they should be positive integers and we have to find the minimum sum.
$$\mathrm{This}\:\mathrm{was}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{question}\:\mathrm{but}\:\mathrm{it}\:\mathrm{was} \\ $$$$\mathrm{cancelled}.\:\mathrm{So}\:\mathrm{I}\:\mathrm{think}\:\mathrm{they}\:\mathrm{should}\:\mathrm{be} \\ $$$$\mathrm{positive}\:\mathrm{integers}\:\mathrm{and}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{sum}. \\ $$
Commented by RasheedSindhi last updated on 12/Aug/17
Many triplets are there having given propefty: 2,1,0;3,1,0;4,1,0 etc 2+1×0=2×(1+0) 3+1×0=3×(1+0) 4+1×0=4×(1+0) So there′s no unique sum.
$$\mathrm{Many}\:\mathrm{triplets}\:\mathrm{are}\:\mathrm{there}\:\mathrm{having} \\ $$$$\mathrm{given}\:\mathrm{propefty}: \\ $$$$\mathrm{2},\mathrm{1},\mathrm{0};\mathrm{3},\mathrm{1},\mathrm{0};\mathrm{4},\mathrm{1},\mathrm{0}\:\mathrm{etc} \\ $$$$\mathrm{2}+\mathrm{1}×\mathrm{0}=\mathrm{2}×\left(\mathrm{1}+\mathrm{0}\right) \\ $$$$\mathrm{3}+\mathrm{1}×\mathrm{0}=\mathrm{3}×\left(\mathrm{1}+\mathrm{0}\right) \\ $$$$\mathrm{4}+\mathrm{1}×\mathrm{0}=\mathrm{4}×\left(\mathrm{1}+\mathrm{0}\right) \\ $$$$\mathrm{So}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{unique}\:\mathrm{sum}. \\ $$
Commented by ajfour last updated on 13/Aug/17
let 0<a<b<c c(a+b)=c+ab ⇒ c=((ab)/(a+b−1)) = (b/(1+(((b−1)/a)))) if c>b then 1+(((b−1)/a))<1 or ((b−1)/a)<0 ⇒ b<a+1 but b>a ⇒ a<b<a+1 how can this be possible for positive integers ?.. therefore let a=b=n so c+ab=c(a+b) ⇒ c=(n^2 /(2n−1)) if a=b=n=1 ⇒ c=1 minimum sum =3 .
$$\mathrm{let}\:\mathrm{0}<\mathrm{a}<\mathrm{b}<\mathrm{c} \\ $$$$\:\:\:\mathrm{c}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{c}+\mathrm{ab} \\ $$$$\Rightarrow\:\:\:\:\mathrm{c}=\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}−\mathrm{1}}\:=\:\frac{\mathrm{b}}{\mathrm{1}+\left(\frac{\mathrm{b}−\mathrm{1}}{\mathrm{a}}\right)} \\ $$$$\mathrm{if}\:\mathrm{c}>\mathrm{b}\:\:\:\mathrm{then}\:\:\:\mathrm{1}+\left(\frac{\mathrm{b}−\mathrm{1}}{\mathrm{a}}\right)<\mathrm{1} \\ $$$$\mathrm{or}\:\:\:\:\frac{\mathrm{b}−\mathrm{1}}{\mathrm{a}}<\mathrm{0}\:\:\:\Rightarrow\:\:\mathrm{b}<\mathrm{a}+\mathrm{1} \\ $$$$\:\:\:\mathrm{but}\:\mathrm{b}>\mathrm{a}\:\:\:\Rightarrow\:\:\:\:\mathrm{a}<\mathrm{b}<\mathrm{a}+\mathrm{1} \\ $$$$\mathrm{how}\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{possible}\:\mathrm{for}\: \\ $$$$\mathrm{positive}\:\mathrm{integers}\:?.. \\ $$$$\mathrm{therefore}\:\mathrm{let}\:\:\mathrm{a}=\mathrm{b}=\mathrm{n} \\ $$$$\mathrm{so}\:\:\:\:\mathrm{c}+\mathrm{ab}=\mathrm{c}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{c}=\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{2n}−\mathrm{1}} \\ $$$$\mathrm{if}\:\:\:\mathrm{a}=\mathrm{b}=\mathrm{n}=\mathrm{1}\:\:\:\Rightarrow\:\:\mathrm{c}=\mathrm{1} \\ $$$$\mathrm{minimum}\:\mathrm{sum}\:=\mathrm{3}\:. \\ $$

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