How many ordered pairs of positive integers (m,n) satisfy lcm[m,n]=720?


please help i am stuck


i dont think its the right way to just count every solution..

 Jul 15, 2021

you're right, counting each solution would take forever.

Notice that the prime factorization of 720 is \(2^4\cdot3^2\cdot5\). Obviously, both numbers have to be factors of 720, meaning that the powers of the primes in their prime factorization have to be smaller or equal to the powers of the primes of the prime factorization of 720. Also, note that, for each prime, there must be at least one of the 2 numbers, m and n, that have the power of that prime in its prime factorization equal to that of 720; the other number can have the power of that prime in its prime factorization less than equal to that of 720.

Let n be expressed as: \(2^a\cdot3^b\cdot5^c\), and let m be expressed as \(2^x\cdot3^y\cdot5^z\).  Let's do some casework to figure out the cases:

For a and x, these are the ways we can set them equal to:

a=4, x=0

x=4, a=0

a=4, x=1

x=4, a=1

x=4, a=2

a=4, x=2

x=4, a=3

a=4, x=3

a=4, x=4

which is 9 ways in total.

If you were to do a similar process with b, y and c, z, you would get that the number of ways we can set them equal to is 3*2-1=5 and 2*2-1=3, respectively, so the answer is 3*5*9=135

(sorry if this seems confusing; I am bad at explaining things)

 Jul 15, 2021
edited by textot  Jul 15, 2021

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