# Perfect Square Factors

December 24, 2007

Here’s a cute little problem from the MATHCOUNTS 2002-2003 School Handbook, Counting and Combinatorics Stretch:

How many perfect square factors does the number 46,656 have?

Related Post: Some Interesting Factoring Problems

Advertisements

14 Comments
leave one →

Is the answer 16?

Here’s what I did.

First I factored 46656 into 2^6 x 3^6.

Then I noted that for a factor to be a square it needs to have an even number of powers of 2 AND an even number of powers of 3.

So, the even exponents of 2 are 0, 2, 4, and 6. And the even exponents of 3 are 0, 2, 4, and 6.

So, all of the square factors occur when we take combinations of exponents from the two sets. Each set has 4 values so there are 4×4 or 16 possible square factors.

I assume the problem considers 1 and 46,656 to be factors of 46,656.

Yup, that’s the right answer, and that’s how I solved it as well.

How would you explain to middle schoolers the part about needing even powers of each prime factor?

Hmmm….

I’d try something like this:

For a number to be a square you must be able to create two identical sets of factors that when multiplied together are that square.

36 is a square because you can factor it into 6×6.

100 is a square because you can factor it into 10×10.

576 is a square because you can factor it into 24×24.

So, the square root factors appear twice.

Now, in the cases where the square root is not itself prime you can factor the square root into the product of powers of primes.

In the case of 576, the square root is 24 = 2^3 x 3.

Notice that when you square 2^3 x 3 that you *double* each factor so that (2^3 x 3)^2 = (2^6 x 3^2).

Both powers of primes in the square are always even because you’ve *doubled* them in the square root to get the square.

So, my explanation would center around the fact that squaring a number is the same as doubling the power of each of its factors.

Another approach might be in several steps:

1. p^2 = p*p for a prime number. Note that the power of p is even, 2, in p^2.

2. If a number, q, is the product of two distinct primes, p1 and p2, then q^2 = (p1 x p2)^2 = p1^2 x p2^2. Note that both powers are even.

3. If a number r, is the product of two primes raised to any powers such that r = (p^a x q^b) then r^2 = p^(2a) x q^(2b). 2a and 2b are always even since what it means for a number to be even is for it to be 2 times something.

4. Trust that any integer can be factored into the product of powers of primes in only one way.

Then, if N = p1^a1 x p2^a2 x p3^a3 x … x pn^an

where p1, p2, …, pn are the prime factors of N and a1, a2, …, an are the corresponding powers of the primes then N^2 = p1^2a1 x p2^2a2 c p3^2a3 x … x pn^2an. Note that all powers in N^2 are even.

I teach it a similar way, using parentheses. I say that the basic form of any square number is:

(_____) x (_____)

How can we fill in the factors?

Put a 2 in each parenthesis:

(2) x (2) = 2^2

Add another 2 to each:

(2 x 2) x (2 x 2) = 2^4

or we might try:

(2 x 2 x 3) x (2 x 2 x 3) = 2^4 x 3^2

How many options do we have in all? I teach this after they have learned how to count prime factors in general. It is a small step from there to recognize that in this case we are looking only for the factors with even exponents. Not all the students make that step on their own, but they all recognize it as logical as soon as one of them says it.

Sol, the algebraic method you posted is beyond the middle schoolers I work with. The method I have tried is similar to your first method and Denise’s. What I find is that they have some confusion about is that you can take all the prime factors, re-arrange them however you want, multiply two subsets of them together, and get two numbers that will multiply to the original number. That’s just not obvious, but it is important for understanding the method of counting factors in general.

What I have tried is create worksheets with “clouds” (similar to Denise’s parentheses) on them where they put some of the prime factors into each cloud (but each factor must go into exactly one cloud), then see what each cloud multiplies to, and see that those two factors always multiply to the original number. Since exponent notation is relatively novel to them as well, this also helps them practice the notion that 2^3 means there are three 2’s to be multiplied.

I’ve never actually done the extension to square factors with my kids, but if they got the general counting factors thing, I think they would get this, too. But I have to say that I don’t think this always makes sense to all the kids. Some of them still seem rather mystified by the whole rearranging prime factors to make other factors thing.

Some of my high school students are still mystified by this too – but now it’s confusion about factors of polynomials. Some are still amazed that the product of two linear factors of a polynomial is a factor of the original polynomial.

I think it stems from a lack of understanding about factors in general. Hopefully having a solid understanding will help them in the future!

That’s an interesting point, Jackie. And it gives me good reason to keep “harping on” playing with and understanding factors, which at this level seem mainly to have applications in “contest math”.

It’s interesting because when I started teaching problem solving, our goals were mainly to reinforce their regular pre-algebra skills, and to hone their problem-solving skills, independent of topic area. But more and more I see that playing with these sometimes seemingly-random “contest math” topics not only improves their overall problem solving and thinking skills for all future topics, it also does directly “prime the pump” for a number of future topics they will study in high school.

A deeper understanding of factors and how they “work together” does seem critical to a basic sense of numeracy. Perhaps that’s why those contest writers seem to like to include lots of problems that involve factorization at the middle school level. 😉

I just added a link to an old post that points to a good resource on this topic.

i need the name of the factors of a perfect square!!!! plz answer this..

I just stumbled upon this thread when I looked up something about prime factors of a perfect square. I just had to comment, I am looking up stuff for my college algebra class! I cannot believe you guys are teaching this stuff in middle school! I never even got this far in high school that I can remember. That was a ways back, I am 33. Still, all of these concepts are totally new and not easy to master for me. Middle school, wow!?!!???! I feel like an idiot.

Hey Bethany, don’t feel bad. Most middle schools still don’t teach this, but I and some of the posters on this thread like to push “problem solving” for elementary through high school, and this is a type of problem that a middle school student knows enough math to do, and “just” needs to hone the intuition, problem solving experience, etc. to be able to solve.

What about numbers whose factors cannot be completely expresses as a power of a square – say 1000 – 2^3 * 5^3? How do we calculate the no. of perfect square factors for such a number?

Neo, a square factor could have either zero or two 2’s and zero or two 5’s. So there are two choices of how many 2’s there could be times two choices of how many 5’s there could be for a total of 4 perfect square factors. Those would be 2^0 * 5^0 = 1, 2^0 * 5^2 = 25, 2^2 * 5^0 = 4 and 2^2 x 5^2 = 100. So 1000 has only the same square factors as 100 does — adding the other “odd” factors doesn’t increase the number at all.

Suppose if there is odd power of a co-prime factor as

Find out the numbers of perfect square factors as in (2^5) x (3^6) x (5^2) ?