Factoring StrategiesDate: 11/01/2002 at 09:02:37 From: Verna Snedacar Subject: Factoring strategies Hi Dr. Math, I am returning to school and I am having a very hard time in a math for elememtary teachers class. The objective is to teach for understanding. My assignment is to explore strategies that help you find a number having 5 factors and a number having 13 factors and to explain my thinking and strategies. So far I think I have found that the number has to be a square number because square numbers have an odd number of factors. For example, 16 = 1, 2, 4, 8, and 16, but I need to know why. Is there a way I can know a number has 13 factors without wrecking my brain factoring all of the square numbers? Please don't use x = y z and so on; I do not understand what all that means. I also don't know what this symbol means (^) as I explore your site. I really need some help. Thanks. Date: 11/01/2002 at 13:19:49 From: Doctor Peterson Subject: Re: Factoring strategies Hi, Verna. It would be a great idea to become at least somewhat comfortable with the idea of variables (x, y, z, and so on), because they are an extremely useful part of the language of mathematics. As for a^b, you can find it explained in our FAQ: Typing math http://mathforum.org/dr.math/faq/faq.typing.math.html We use it to represent exponents without having to take up two lines; 3^2 means 3 squared. It's hard to discuss this problem without using variables; that's sort of like telling certain people to talk without moving their hands! But I'll try. That means I'll mostly have to use examples. Let's look at the factors of a square number, first; say, 6^2 = 36: 1*36 2*18 3*12 4*9 6*6 9*4 12*3 18*2 36*1 Here I listed not just the factors, but the factor pairs. That is, for each factor, there is another number you multiply it by to get 36, and I included it in my list. We have 9 factors, and 9 factor pairs. Notice that every factor appears twice in the list, once as the first member of a pair, and once as the second. That means that the rows can be paired off - except for the 6*6 row: 1*36 --------+ 2*18 ------+ | 3*12 ----+ | | 4*9 --+ | | | 6*6 o | | | | 9*4 --+ | | | 12*3 ----+ | | 18*2 ------+ | 36*1 --------+ The only exception is the 6*6 row, because there both occurrences of 6 are in the same row. Now, when you can pair everything up, you have an even number; when you can't you have an odd number. In this case, we have an odd number of rows, namely 9 factors. If the number were not a square, we would have an even number of rows, because there would be no special row like 6*6. For example, here are the factor pairs for 35: 1*35 ---+ 5*7 -+ | 7*5 -+ | 35*1 ---+ Here every factor pairs up with another in a different row, and there is an even number of them. Does that demonstrate why you need a square in order to have an odd number of factors? Now things will get a little more complicated, because I'm going to show how you can find the number of factors without listing them. Let's work with 36 again. First we'll find the prime factorization: 36 = 2*2*3*3 Now, when we make a factor pair, what we are really doing is just splitting these prime factors up between two numbers. For example, the factor pair 3*12 is 36 = 3 * 2*2*3 where I took 3 for the first factor, and all the rest for the second. Do you see why this has to be true? When you multiply two numbers together, the prime factors of the product are the prime factors of the factors, combined; we're just pulling them apart. Now, how many ways can you split up the prime factors of a number? Let's make a table, listing how many of each prime factor we use for the first factor: number number product of of 2's of 3's 2's and 3's --------+--------+------------ 0 | 0 | 2^0*3^0 = 1 0 | 1 | 2^0*3^1 = 3 0 | 2 | 2^0*3^2 = 9 1 | 0 | 2^1*3^0 = 2 1 | 1 | 2^1*3^1 = 6 1 | 2 | 2^1*3^2 = 18 2 | 0 | 2^2*3^0 = 4 2 | 1 | 2^2*3^1 = 12 2 | 2 | 2^2*3^2 = 36 I can choose anything up to the maximum number of 2's in 36, namely 0, 1, or 2 of them; and the same with the 3's. Altogether, this gives me 3 times 3, or 9, ways to choose the prime factors I want to use; and that gives me 9 factors of 36. Just in case you're not fully comfortable with powers (especially when the exponent is 0), here's another way to look at the same thing: product of 2's 3's 2's and 3's --------+--------+------------------- | | 1 = 1 | 3 | 1 * 3 = 3 | 3*3 | 1 * 3*3 = 9 2 | | 1 * 2 = 2 2 | 3 | 1 * 2 * 3 = 6 2 | 3*3 | 1 * 2 * 3*3 = 18 2*2 | | 1 * 2*2 = 4 2*2 | 3 | 1 * 2*2 * 3 = 12 2*2 | 3*3 | 1 * 2*2 * 3*3 = 36 I'll leave it to you to generalize this idea, and try it on some other numbers. The idea is that you count the number of each prime factor in your number; the number of factors will be one more than the number of each prime factor, all multiplied together. In this case, we had two 2's and two 3's, so (2+1)*(2+1) = 3*3 gives 9 factors. For 35, we have one 5 and one 7, so the number of factors is (1+1)*(1+1) = 2*2 = 4, as we saw. Now you need to apply this to your problem. If a number has 13 factors, what does that tell you about its prime factors? Can you find a number like this? Can you find more of them? If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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