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Question Number 19193 by Tinkutara last updated on 06/Aug/17
The sum of two positive integers is 52  and their LCM is 168. Find the numbers.
$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{is}\:\mathrm{52} \\ $$$$\mathrm{and}\:\mathrm{their}\:\mathrm{LCM}\:\mathrm{is}\:\mathrm{168}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{numbers}. \\ $$
Answered by RasheedSindhi last updated on 06/Aug/17
Let the numbers are x and y  and their HCF is h  x+y=52.......(i)  xy=168h......(ii)       [ ∵ product of the numbers=LCM×HCF ]  (i) & (ii):  x+((168h)/x)=52               x^2 −52x+168h=0        x=((52±(√(52^2 −4(168h))))/2)           =((52±4(√(169−42h)))/2)        x=26±2(√(169−42h))  Since x∈Z^+  , 169−42h is +ve perfect square      and also h>0 ⇒h=4         x=26±2(√(169−42×4))             =28,24         y=52−28,52−24=24,28  Hence the required numbers are           24  and  28
$$\mathrm{Let}\:\mathrm{the}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$$$\mathrm{and}\:\mathrm{their}\:\mathrm{HCF}\:\mathrm{is}\:\mathrm{h} \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{52}…….\left(\mathrm{i}\right) \\ $$$$\mathrm{xy}=\mathrm{168h}……\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\left[\:\because\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}=\mathrm{LCM}×\mathrm{HCF}\:\right] \\ $$$$\left(\mathrm{i}\right)\:\&\:\left(\mathrm{ii}\right):\:\:\mathrm{x}+\frac{\mathrm{168h}}{\mathrm{x}}=\mathrm{52} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{52x}+\mathrm{168h}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{x}=\frac{\mathrm{52}\pm\sqrt{\mathrm{52}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{168h}\right)}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{52}\pm\mathrm{4}\sqrt{\mathrm{169}−\mathrm{42h}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{x}=\mathrm{26}\pm\mathrm{2}\sqrt{\mathrm{169}−\mathrm{42h}} \\ $$$$\mathrm{Since}\:\mathrm{x}\in\mathbb{Z}^{+} \:,\:\mathrm{169}−\mathrm{42h}\:\mathrm{is}\:+\mathrm{ve}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\:\:\:\:\mathrm{and}\:\mathrm{also}\:\mathrm{h}>\mathrm{0}\:\Rightarrow\mathrm{h}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}=\mathrm{26}\pm\mathrm{2}\sqrt{\mathrm{169}−\mathrm{42}×\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{28},\mathrm{24} \\ $$$$\:\:\:\:\:\:\:\mathrm{y}=\mathrm{52}−\mathrm{28},\mathrm{52}−\mathrm{24}=\mathrm{24},\mathrm{28} \\ $$$$\mathrm{Hence}\:\mathrm{the}\:\mathrm{required}\:\mathrm{numbers}\:\mathrm{are} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{24}\:\:\mathrm{and}\:\:\mathrm{28} \\ $$$$ \\ $$
Commented by Tinkutara last updated on 06/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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