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 $7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot 1$.)

CorbellaB.15 Mar 19, 2019

#1**0 **

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 :)

Guest Sep 7, 2019