Factoring vs. an Equation
Date: 10/01/2002 at 09:08:29 From: Sarah Subject: Mathematics Hello, Dr. Math! The product of 2 consecutive odd numbers is 255. Find the numbers.
Date: 10/01/2002 at 10:23:24 From: Doctor Ian Subject: Re: Mathematics Hi Sarah, Forget about consecutive odd numbers for a moment. Can you find _any_ two numbers whose product is 255? Let me know what you come up with. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Date: 10/01/2002 at 13:21:47 From: Sarah Subject: Mathematics Just tell me the equation for that - any possible EQUATION.
Date: 10/01/2002 at 14:42:43 From: Doctor Ian Subject: Re: Mathematics Hi Sarah, The reason you're given problems to solve isn't so you can just find the answers and give them back to the teacher. The teacher already _knows_ the answers. The reason you're given problems to solve is so you can learn how to approach problems that you've never seen before, and don't know how to solve. I'm trying to help you learn to do that. You're supposed to find two consecutive odd numbers whose product is 255. One way to do that is to find _all_ the factors of 255, and see which ones are consecutive odd integers. Now, 255 is divisible by 5: 255 = 5 * 51 and 51 is divisible by 3: 255 = 5 * 3 * 17 and all of these are prime factors. So there are only a few ways to multiply numbers together to get 255. 1 * (3 * 5 * 17) 3 * (5 * 17) 5 * (3 * 17) 17 * (3 * 5) Does one of these fit the problem description? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Date: 10/02/2002 at 10:02:24 From: Sarah Subject: Mathematics In an examination I simply don't have the time to do factorisation!
Date: 10/02/2002 at 10:56:43 From: Doctor Ian Subject: Re: Mathematics Hi Sarah, Actually, the way I showed you is _faster_ than setting up and solving an equation. Here's how that would look: The product of two consecutive odd numbers is 255. What are the numbers? If the first number is n, the second is n+2. Their product is n(n+2) = 255 n^2 + 2n - 255 = 0 (n + a)(n - b) = 0 which means that a - b = 2 a * b = 255 So now you have two equations to solve, instead of one. And you have to solve the second equation by finding the factors of 255, so you can see which ones differ by 2. (If you look carefully, you'll see that you're really right back where you started.) All I'm saying is, since you're going to have to find the factors of 255 _anyway_, why not just _start_ there, and then pick the factors that fit the problem description? It's often the case for a problem on a test that there are at least two ways to do the problem. The first way is to set up an equation and solve it, which works (if you set up the right equation), but which takes a lot of time. The second way is to try to look at the problem from the point of view of the person who made it up. If you wanted to _create_ a problem like this, you wouldn't do it by setting up the equation. You'd do it by picking a pair of consecutive odd numbers, like 7 and 9, or 11 and 13, or 15 and 17. So all you have to do to _solve_ the problem is find those numbers. Prime factoring is a quick way to do that, which is why your teachers have tried to get you to learn it over the years. One of the quickest ways to factor a number is to make use of divisibility rules. The first one I used was very simple: Every number that is divisible by 5 must end in either 0 or 5: 5, 10, 15, 20, 25, 30, 35, ... So I know that 255 must be divisible by 5, and I factor a 5 out of it: 51 _____ 5 ) 255 Note: This isn't actually the way I do the division; I do it this way: 5 * 40 is 200, so the answer is 40, plus however many times 5 goes into 55, which is 11, so the answer is 40 + 11 = 51. In other words, I use the distributive property, (1/5)*255 = (1/5)*200 + (1/5)*55 to do the division. This way, I don't need to use paper to do it. So now I have 255 = 5 * 51 Another very common divisibility rule is that if the digits of a number add up to something divisible by 3, then the number is divisible by 3: Number Sum of digits ------ ------------- 3 3 6 6 9 9 12 1 + 2 = 3 15 1 + 5 = 6 18 1 + 8 = 9 21 2 + 1 = 3 : 2196 2 + 1 + 9 + 6 = 18 => 1 + 8 = 9 The reasons for this are a little complicated, but the trick itself is pretty easy to apply. Anyway, I can use this trick to see that 51 is divisible by 3, so I can factor a 3 out of it: 17 ____ 3 ) 51 (As before, I don't actually use long division; I pick something I know is close, e.g., 3 * 15 = 45, so then I just have to divide 3 into 6, and add the two quotients: 15 + 2 = 17.) So now I have 255 = 3 * 5 * 17 And I know that 17 is prime. So now I know that _any_ factor of 255 must be the product of the prime factors, which means I just have to try to construct two consecutive odd numbers from these. One of the numbers will clearly be 17, which leaves 3*5, or 15, as the other number. A third very useful rule is that any number that is divisible by 2 must end in 0, 2, 4, 6, or 8. Our FAQ has a whole section on divisibility rules. A lot of them are pretty specialized, but the rules for divisibility by 2, 3, 5, and 9 are worth learning by heart. I know it must seem at times that math class is just about turning you into a kind of robot that sets up and solves equations. But that's not how it's supposed to be! Math is a set of tools that are designed to let you get right to the heart of a problem, and solve it with a minimum of effort. And so there are two parts to learning math: the first is learning to use the tools; the second is learning to recognize which tools to use in a given situation. Unfortunately, in many math classes, almost all the effort is directed at the first part, when in fact, it's the second part that's most important. In this case, it's as if you're trying to open a door that's been locked. You can get a big hammer and try to knock down the door (e.g., set up and solve an equation); or you can check above the door frame to see if someone left the key there (e.g., check the prime factors to see if there is an obvious answer). And sometimes you can just look for an open window, bypassing the door completely. In this case, you could do something like this: 10^2 is 100, so 9 * 11 would be too small. 20^2 is 400, so 19 * 21 would be too big. 255 ends in 5, so one of the numbers has to be 15. So it has to be 13 * 15, or 15 * 17. When you approach it this way, solving a math problem is less like what a robot does, and more like what a detective does, and it's when you start to see it that way that it becomes fun, instead of boring. Anyway, I hope this helps. Write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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