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The least common multiple of two positive integers is $7!$, and their greatest common divisor is $9$. If one of the integers is $315$, then what is the other? (Note that $7!$ means $7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot 1$.)

 Mar 19, 2019
edited by CorbellaB.15  Mar 19, 2019
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The identity lcm[a,b]*gcd(a,b)=ab holds for all positive integers a and b.

 

Let a=315 in this identity, and let b be the number we are looking for. Thus

 

\(7! \cdot 9=315 \cdot b\)

so

 

\(b = \frac{7!\cdot 9}{315} = \frac{7!\cdot 9}{35\cdot 9} = \frac{7!}{35} = \frac{{7}\cdot 6\cdot {5}\cdot 4\cdot 3\cdot 2\cdot 1}{{35}} = 6\cdot4\cdot3\cdot2 = \boxed{144}.\)

 

Hope this helps :)

 Sep 7, 2019

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