So this isn’t a super important thing as far as how often it appears on the SAT, but it does pop up time and again, so if you’re shooting for perfection (or close to it) you might want to pay attention. Otherwise, you can get by just fine without this little nugget (but you might as well read it, since you’re here anyway).

Do you know what prime factorization is? Basically, the prime factorization of a number is the way you would build that number by multiplying together only prime numbers. To find the prime factorization of a number, divide by 2 if you can. Do that as many times as you can. Once you can’t do that anymore, try dividing by 3 as many times as you can. Then by 5. Then by 7. Then by 11. I think you get the idea.

Let’s try one together, like best friends

What is the prime factorization of 13728?

Whoa. Big number. Lots of people like to make trees when they do this. Let’s do that. Damn I wish you and I were in the same room with a chalkboard right now. This is going to take flippin’ forever.

See how, when I couldn’t divide by 2 anymore, I went to three, and then to 11? I knew I was done when I had two prime numbers, 11 and 13. If I multiplied all those numbers back together, I’d get 13728 again. For serious. Try it:

2 × 2 × 2 × 2 × 2 × 3 × 11 × 13 = 13728

So why do I need to know this?

Because sometimes the SAT asks hard questions (and if you took the October 2011 SAT, you can confirm) about the lowest multiple of two numbers that’s also perfect square. It just so happens that prime factorization is a great way to find a perfect square.

The prime factorization of a perfect square will contain even numbers of each prime number. Look back at the prime factorization of 13728. That’s not a perfect square. There are 5 2s, and one each of 3, 11, and 13. We can use this information to find the lowest multiple of 13728 that’s a perfect square. In order to make a prime number, we’re going to need another 2, another 3, another 11, and another 13. Yikes. That’s gonna be a big number.

2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 11 × 11 × 13 × 13= 11778624

Peep the bold underlines. Those are the factors I’ve added. My new product is huge. It’s also a perfect square. Seriously.

11778624 = 3432

And there are no multiples of 13728 that are less than 11778624 that are perfect squares. Scout’s honor.

What would an SAT question about this look like?

Glad you asked. Try this (no multiple choice — it’s a grid-in):

  1. If p2 is a multiple of both 8 and 35, and p is a positive integer, what is the least possible value of p?

So…yeah. Start by doing a prime factorization of 8 and 35.

8 = 2 × 2 × 2
35 = 5 × 7

Note that you have odd numbers of all three used prime factors. You’re gonna need another 2, another 5, and another 7.

2 × 2 × 2 × 2 × 5 × 5 × 7 × 7 = 19600
p2 = 19600

Confirm that 19600 is a multiple of both 8 and 35 (of course it is):

19600 ÷ 8 = 2450
19600 ÷ 35 = 560

Yes, it worked. So what’s p? Just take the square root of 19600!

19600 = 140
p = 140

Note the tempting false shortcut: just multiply 8 by 35 and square the result. But if you do that, you get 78400 for p2 and 280 for p. That’s not the smallest possible p, as we just showed.

Like I said, you don’t see this often on the SAT, but if you’re shooting for perfection, you’ll want to know this relationship between prime factors and perfect squares.

Comments (6)

This is already incorporated into the Beta, but not as a cohesive chapter. It was a tough call for me, but honestly this only appears once in a blue moon, so I don’t want to misrepresent its importance by devoting a whole chapter to something you’re very unlikely to see. The basic facts are in the “Miscellany” chapter in the beginning, and the Sample Question is a part of the yet-to-be-finished Drill #4. 

I Didint know where to post this question, but can you please explain questions involving range (all possible values) Like this: If d(w)= square rOot of w2+1 for all real values of w, which of the following is NOT a possible value of d(w)?

For questions like this, hit me up on That way more people will get to benefit from the answer. 🙂

The easiest way to get this one, if you have a graphing calculator, is to graph it. It’ll look like this, and you’ll be able to see right away that d(w) will never be 0.

If you don’t have the luxury of a graphing calc, you’re gonna have to do some reasoning. The lowest w^2 can ever be is 0, so the lowest w^2 + 1 can ever be is 1. That means the lowest the square root of (w^2 + 1) can ever be is 1, too.

The March 2015 US SAT had a question relating to lowest number and perfect squares type thing….. it was the last question of the section (I remember you could put either 25 or 49)… i put the latter

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