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Question Number 50209 by mondodotto@gmail.com last updated on 14/Dec/18
The gcf and lcm of three numbers  are 2 and 1200 respectively.  if the two numbers are 16 and 24,  find the third number
$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{gcf}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{lcm}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{three}}\:\boldsymbol{\mathrm{numbers}} \\ $$$$\boldsymbol{\mathrm{are}}\:\mathrm{2}\:\boldsymbol{\mathrm{and}}\:\mathrm{1200}\:\boldsymbol{\mathrm{respectively}}. \\ $$$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{numbers}}\:\boldsymbol{\mathrm{are}}\:\mathrm{16}\:\boldsymbol{\mathrm{and}}\:\mathrm{24}, \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{third}}\:\boldsymbol{\mathrm{number}} \\ $$
Answered by mr W last updated on 14/Dec/18
let′s say the third number is N  gcf(N,16,24)=2  since 16=2^4  and 24=2^3 ×3  ⇒N=2^1 Πp^i  with p≥3  lcm(N,16,24)=1200  since 1200=2^4 ×3×5^2   ⇒N=2^1 ×5^2 =50 or 2^1 ×3×5^2 =150  i.e. the third number is 50 or 150.
$${let}'{s}\:{say}\:{the}\:{third}\:{number}\:{is}\:{N} \\ $$$${gcf}\left({N},\mathrm{16},\mathrm{24}\right)=\mathrm{2} \\ $$$${since}\:\mathrm{16}=\mathrm{2}^{\mathrm{4}} \:{and}\:\mathrm{24}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3} \\ $$$$\Rightarrow{N}=\mathrm{2}^{\mathrm{1}} \Pi{p}^{{i}} \:{with}\:{p}\geqslant\mathrm{3} \\ $$$${lcm}\left({N},\mathrm{16},\mathrm{24}\right)=\mathrm{1200} \\ $$$${since}\:\mathrm{1200}=\mathrm{2}^{\mathrm{4}} ×\mathrm{3}×\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow{N}=\mathrm{2}^{\mathrm{1}} ×\mathrm{5}^{\mathrm{2}} =\mathrm{50}\:{or}\:\mathrm{2}^{\mathrm{1}} ×\mathrm{3}×\mathrm{5}^{\mathrm{2}} =\mathrm{150} \\ $$$${i}.{e}.\:{the}\:{third}\:{number}\:{is}\:\mathrm{50}\:{or}\:\mathrm{150}. \\ $$
Commented by mr W last updated on 15/Dec/18
MJS sir should check. he is expert for  such questions.
$${MJS}\:{sir}\:{should}\:{check}.\:{he}\:{is}\:{expert}\:{for} \\ $$$${such}\:{questions}. \\ $$
Commented by mondodotto@gmail.com last updated on 15/Dec/18
oky thanks
$$\mathrm{oky}\:\mathrm{thanks} \\ $$

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