Number Puzzle Solved with and without AlgebraDate: 12/10/2001 at 22:37:35 From: Angela Subject: Pre-algebra My class is doing pre-algebra in school and she gave us this homework that I don't really understand. I'll give you a question or two. Can you answer them and tell me the strategy you used, so I can use it on all the problems I have in school? I am a number. Add 25 to me. Divide the sum by 2. You end with 75. What number am I ? I am a number. Add four times me to 25. The sum is equal to 35 minus 2. What number am I ? Date: 12/11/2001 at 08:47:55 From: Doctor Ian Subject: Re: Pre-algebra Hi Angela, One way to start attacking a problem is to guess, and see how your guess turns out. Let's guess that the number is 1. Then 1 is a number. Add 25 to 1, to get 26. Divide 26 by 2, to get 13. We don't end with 75, so 1 can't be the number. 13 is a lot lower than 75, so we probably need a higher starting number. What about 100? 100 is a number. Add 25 to 100, to get 125. Divide 125 by 2, to get 62 1/2. Now, this still isn't large enough. But we've got a clue. We're going to end up dividing the number by 2, so we want to end up with an even number after adding 25. Which means we have to start with an odd number. How about 151? 151 is a number. Add 25 to 151, to get 176. Divide 176 by 2, to get 88. Now we're too high, so we're probably looking for a number between 100 and 151. So that's one way to go about solving the problem - figure out how to _check_ an answer, and then try to guess what the answer is, being smart about limiting our guesses to realistic ones. It's not usually the _fastest_ strategy, but it's perfectly acceptable when you don't know what else to do, and playing around this way can help you spot patterns that you can use to develop a more efficient strategy. Can we find a pattern like that? In fact, we can. We know that we have to end up with 75, right? And we're going to get there by dividing some other number by 2. That is, we're looking for a number such that ? / 2 = 75 Now, we can figure that out: The missing number must be 150. So now we can forget about the last step of the problem. The problem now looks like this: I am a number. Add 25 to me. You end with 150. What number am I? Now we're looking for s number such that ? + 25 = 150 Can you figure out what number that would be? Let's look at the second problem, and see if we can use the same trick: I am a number. Add four times me to 25. The sum is equal to 35 minus 2. What number am I? There's no reason not to change '35 minus 2' to 33 in the next-to-last line: I am a number. Add four times me to 25. The sum is equal to 33. What number am I? Again, let's start with a guess: 1 is a number. Add 4*1 to 25, to get 29. The sum isn't equal to 33, so 1 isn't the number. In this case, another guess gets us the answer quickly: 2 is a number. Add 4*2 to 25, to get 33. The sum is equal to 33. So the missing number is 2. But let's pretend for a minute that it wasn't that easy. We know that the final sum is going to be 33. And we're going to get that by adding 25 to something, so we want to find a number such that ? + 25 = 33 A little thought convinces us that the number we need is 8. So now we have a shorter problem: I am a number. Multiply me by 4 to get 8. Even if we didn't already know that the answer is 2, it wouldn't be hard to figure out from this simplified version of the problem. So, that's how you can solve problems like this without using algebra. Just for fun, I'll show you how you'll learn to do it later on. The key idea is that there is some number, and I don't know its value, but I have some information about it. So I need a name for it. We can talk about people without knowing exactly who they are, right? For example, I can talk about 'the guy who drives the school bus past my house each morning', even though I don't know who he is or anything else about him. Or I can talk about 'the junior Senator from North Dakota' even though I couldn't tell you who that is. But I know that he exists. When we do this with numbers, we use letters like 'x' or 'n' to represent the numbers whose values we don't know. In this case, we might say: This number we're looking for, let's call it 'N'. That allows me to write the problem this way: I am a number, called N. Add 25 to me, to get (25+N). Divide the sum by 2, to get (25+N)/2. You end with 75, so it must be true that (25+N)/2 = 75. So now I have an equation, and there is exactly one value of N that will make it true. As before, I could try to guess the values; or I could use algebra to find it. That consists mostly of looking for operations that I can apply to both sides of the equation, without changing the value of the equation. For example, the equation 5 = 5 is obviously true, right? And it remains true if I multiply both sides of the equation by any number: 3*5 = 3*5 15 = 15 4*15 = 4*15 60 = 60 It also remains true if I subtract any number from both sides: 60 - 4 = 60 - 4 56 = 56 So let's look at (25+N) / 2 = 75 I can multiply both sides of the equation by 2: 2 * (25+N) / 2 = 2 * 75 (25+N) = 150 That looks a little simpler. Now I can subtract 25 from each side: 25 + N - 25 = 150 - 25 N = 125 So I found the number without any guessing. Which is pretty much the point of algebra. Of course, I'm not done until I _check_ my answer, by substituting it for N in the original equation: (25+125) / 2 = 150 / 2 = 75 It's important to get into the habit of doing this, because it's easy to slip up and make a mistake. For example, it would be easy to do something like this: (25 + N) / 2 = 75 2 * (25 + N) / 2 = 2 * 75 (2*25 + N) / 2 = 150 Applied the distributive property incorrectly! (2*25)/2 + N/2 = 150 25 + N/2 = 150 25 + N/2 - 25 = 150 - 25 N/2 = 125 2 * N/2 = 2 * 125 N = 250 Now, if I just assume that I did all the steps correctly, I'll believe that this is correct. If it's just a test problem, I'll get a lower grade, which is unfortunate, but not too big a deal. But if I'm computing something like how many millions of dollars we should lend a corporation, or how many cars can safely be on a bridge at any one time, a lot of people are going to be hurt by my mistake! A simple check will tell me whether I'm correct: (25+250)/2 = 75 275/2 = 75 137.5 = 75 No! This tells me that I made a mistake somewhere, which means I should go back and try again. So that's where you're ultimately going with these kinds of problems. The problems are designed to give you some practice thinking about numbers whose values you don't know yet. Everyone finds a different way to deal with that concept. Some people imagine little bags of marbles, where you know that there is some number of marbles in there, but you don't know what that number is. Other people imagine yardsticks where you can't read the markings. Some people imagine a chalkboard with the number written on it, but in a blurry way, so you can't see what it is. Once you find something that works for you, the next step will be to get used to attaching names to those uknown numbers; and then you start learning rules for how to solve particular kinds of equations that result from translating 'word problems' like these into equations. 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/