The GCF of three numbers is 6 and their LCM is 900. If two of the numbers are 36 and 60, find the other number.

Saturday, November 10, 2012

The GCF of three numbers is 6 and their LCM is 900. If two of the numbers are 36 and 60, find the other number.

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Denote the unknown number as $\displaystyle{x}.$ Factor the given numbers into primes:

$\displaystyle{36}={{2}}^{{2}}\cdot{{3}}^{{2}},$ $\displaystyle{60}={{2}}^{{2}}\cdot{3}\cdot{5},$ $\displaystyle{6}={2}\cdot{3},$ $\displaystyle{900}={{2}}^{{2}}\cdot{{3}}^{{2}}\cdot{{5}}^{{2}}.$

The $\displaystyle{G}{C}{F}{\left({36},{60}\right)}={{2}}^{{2}}\cdot{3}={12},$ not $\displaystyle{6}.$ Therefore to have $\displaystyle{G}{C}{F}{\left({36},{60},{x}\right)}={6},$ $\displaystyle{x}$ must have $\displaystyle{6}$ as its factor, but not have $\displaystyle{12}$ as its factor.

Further, the $\displaystyle{L}{C}{M}{\left({36},{60}\right)}={{2}}^{{2}}\cdot{{3}}^{{2}}\cdot{5}={180}.$

The given $\displaystyle{L}{C}{M}{\left({36},{60},{x}\right)}={900}={180}\cdot{5},$ so $\displaystyle{x}$ must have one more factor of $\displaystyle{5}$ than $\displaystyle{60},$ i.e. it must have a factor of $\displaystyle{{5}}^{{2}}={25}.$

This way we know that $\displaystyle{x}$ must have $\displaystyle{6}\cdot{25}={150}$ as its factor, so $\displaystyle{x}={150}\cdot{y},$ and $\displaystyle{y}$ must be odd. From the other hand, $\displaystyle{x}$ must be a factor of its multiple $\displaystyle{900},$ so y must be a factor of $\displaystyle\frac{{900}}{{150}}={6},$ and not so many variants remain.

Actually they are $\displaystyle{y}={1},$ $\displaystyle{x}={150}$ and $\displaystyle{y}={3},$ $\displaystyle{x}={450}.$

There are two possible answers: 150 and 450 (not counting -150 and -450).

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Saturday, November 10, 2012