If x is an integer and m and n are integers greater than 3, for how many values of x is possible that {2^n 3^m} / x
is an integer?
(Arithmetic and Algebra Session 1 - GregMat)
Minute: 58:00. This will help you with the concept behind it. 
Here’s how I solved it -
Consider choosing numbers for x, m and n
x = 6, m = 4 and n = 4
This implies, \frac{2^{n} 3^{m}}{x} = \frac{2^{4} 3^{4}}{6} , which is an integer.
Here, x = 6 can be written as 2^{1}3^{1}
In x, can raise the power of 2 from 0 to n for the expression to be an integer, since we have 2^{n} in the numerator. If it becomes greater than n, there will be an extra 2 in the denominator and the result is no longer an integer.
Similarly, for the 3 in x, we can raise the power from 0 to m, since we have 3^{m} in the numerator.
0 - n => n + 1 values
0 - m => m + 1 values
Total possible values for x = (n + 1)(m + 1)
Edit:
Since the result can also be negative integers, making the denominator negative we get x = 2(n+1)(m+1)
Thanks @Adnan for that!
I would like to suggest a small correction here. Since they mentioned just the word integer for the value of x, we should also consider the options for negative as well. Therefore making the answer 2(n + 1)(m + 1)
Hey, I do not understand how 2 is multiplied here in the final edit.
The question says the resulting value of the given expression is an integer. So we should consider both positive and negative cases.
Consider the same example as above \frac{2^{3}3^{4}}{6}, where we considered x = 6.
This simplifies to 216.
But, we could consider x = -6 and the result would be -216 which is also an integer.
Since for each positive case, we have a negative case, we multiply the final answer by 2 to get the total number of possible values for the expression.
Ah the magic that the number line holds! Thanks Karishma, I got it now
Thanks😊