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Number Puzzle Solved with and without Algebra


Date: 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/   
    
Associated Topics:
Middle School Algebra
Middle School Puzzles
Middle School Word Problems

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