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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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