<p>If p and q are different prime numbers greater than 7, then how many factors does x have if x=p^4q^6?</p>
<p>Correct answer is 35. Can someone please explain how to get this answer?</p>
<p>You know p and q don’t have common factors. So, p<em>1, p</em>q, p<em>q^2, etc. and q</em>1, q<em>p^2, etc. and q^2</em>p^3 and q^3p^2 etc etc etc are all going to be discrete factors of p^4q^6. Just carry out this process with as many iterations as necessary, and you have the number of factors.</p>
<p>Don’t forget about 1. Let’s create a list with a nice, systematic pattern:</p>
<p>These are the factors: </p>
<p>1, q, q^2, q^3, q^4, q^5, q^6
p, pq, pq^2, pq^3, pq^4, pq^5, pq^6
p^2, …p^2q^6
p^3,…p^3q^6
p^4,…p^4q^6</p>
<p>So there are 7*5 = 35 factors. </p>
<p>Remark, more generally we see that there are (k+1)(r+1) factors of p^k*q^r</p>
<p>This can also be solved by applying the counting principle!</p>
<p>Here’s what I mean: you are trying to build a number that will be a factor of (p^4)(q^6).</p>
<p>Your number can “contain” as factors only p and q, but multiple times each if you like. You have to choose how many factors of each you want. For p, you have 5 choices (0 - 4) and for q you have 7 choices (0 - 6). So overall, you have 5 x 7 = 35 ways to build your number.</p>
<p>I always like seeing new ways to apply this principle. Where is this question from?</p>