How many ordered pairs of positive integers 
satisfy the system
![\begin{align*}\gcd (m^3, n^2) & = 2^2 \cdot 3^2,\\ \text{LCM} [m^2, n^3] & = 2^4 \cdot 3^4 \cdot 5^6,\end{align*}](http://data.artofproblemsolving.com/images/latex/c/a/a/caaf90100fb45a875104cfd0977aabb5d40b25de.gif "\begin{align*}\gcd (m^3, n^2) & = 2^2 \cdot 3^2,\\ \text{LCM} [m^2, n^3] & = 2^4 \cdot 3^4 \cdot 5^6,\end{align*}")
where
and ![\text{LCM}[a, b]](http://data.artofproblemsolving.com/images/latex/6/e/e/6eebfdc0b90298325dca03dabfb75924df0b743f.gif "\text{LCM}[a, b]")
denote the greatest common divisor and least common multiple of 
and 
, respectively?
(A) 0 (B) 1 (C) 2 (3) More than 2
satisfy the systemwhere
and denote the greatest common divisor and least common multiple of and , respectively?(A) 0 (B) 1 (C) 2 (3) More than 2
option (b) 1
ReplyDeletem = 2^2*3^2
n = 2*3*5^2
the solution is incorrect, please try again.
ReplyDeleteanswer is option C
ReplyDelete2 ordered pairs are possible
is it right?
Yeah that's correct
ReplyDeletehmmm well
ReplyDeletem=2^2*3^2
n=2*3
am forgetting the 5 for now...
see that for gcd to be 2^2*3^2 this is only possi when n^2=this value
and 2^4,3^4 to be lcm means m^2 is this
5 can oscillate between m or n
hence 2 such pairs... one m=2^2*3^2*5^3 , n=2*3
and once m=2^2*3^2 , n=2*3*5^2
That's right! 2 solutions
ReplyDelete