Question 576746
Since a and b are factors of 96, we can let


*[tex \LARGE a = 2^m3^n]


*[tex \LARGE b = 2^p3^q]


Also note that 96 = (2^5)(3^1) and 4 = (2^2)(3^0). From the given information we can conclude that:


at least one of m or p is 5
at least one of n or q is 1
at least one of m or p is 2
at least one of n or q is 0


The first two statements come from the fact that the LCM of two numbers a and b is computed by taking the highest exponent for each prime factor of a and b. Similar for GCF. However, we have two known values for m and p, and two known values for n and q. We can use casework to find all possible pairs {a,b}.


m = 5, n = 1 --> p = 2, q = 0 --> a = 96, b = 4
m = 5, n = 0 --> p = 2, q = 1 --> a = 32, b = 12
m = 2, n = 1 --> p = 5, q = 0 --> a = 12, b = 32
m = 2, n = 0 --> p = 5, q = 1 --> a = 4, b = 96


Therefore the only ordered pairs {a,b} are {96,4}, {32,12}, {12,32}, and {4,96}.