Anyone remember this tough math question or one like it?

<p>My daughter was telling me about a tough math question on the SAT in which she was clueless about how to solve it.</p>

<p>It was something about 'how many prime numbers' but it gave information such as P and P*</p>

<p>Has anyone had a question like this and can point me in the right direction for getting my daughter some help?</p>

<p>MM</p>

<p>I believe it was something like this:
n* is equal to the sum of all of its possible prime factors. For example 15* = 3 + 5 = 8. For how many values, if p is prime, is p* = (100p)*?
a) None
b) One
c) Two
d) Three
e) More than three</p>

<p>Here’s an excerpt from an email I sent to a student when she emailed me about it:</p>

<p>You’re right, it’s none. Let’s plug in some values to see what’s going on.</p>

<p>2* = 2
3* = 3
5* = 5
7* = 7
etc.</p>

<p>Ok, got it. Now what are the prime factors of 100?</p>

<p>I’m getting 2 and 5. (full prime factorization is 2<em>2</em>5*5)</p>

<p>So any time you’re multiplying a prime number by 100, you’re adding those to the prime factorization.</p>

<p>Find the full prime factorizations of a few “100p”'s just to be sure.</p>

<p>700 = 2x2x5x5x7, so 700* is 2+5+7
1100 = 2x2x5x5x11, so 1100* is 2+5+11
1300 = 2x2x5x5x13, so 1300* is 2+5+13</p>

<p>See the pattern? Every (100p)* is going to be 2+5+p. So you’ll never have 2+5+p = p.</p>

<p>By the way, 2 is the only even prime integer. 1 isn’t prime.
0 is neither negative nor positive integer.</p>

<p>What you have to realize is that the question is asking you to factor p and 100p and add up the numbers you get. The factorization of p is p, but the factorization of 100p=2<em>2</em>5<em>5</em>p. 2+5+p=p will never happen, as PWNtheSAT explained.</p>