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This was the question:
9/14/06
Dear Math Mojo,
Hi, I am the mother of a 6th grader with these math problems listed below,
I am not a math expert, I am good at everything else but math, can you help
me find the prime factorization of the questions listed below.
1. 36
2. 293
3. 180
4. 760
5. 525
6. 216
7. 165
8. 231
Thanks in advance,
Signed,
The factors, ma'am, just the factors
Professor Homunculus' answer:
Warning - This answer contains one or more LEFASIs (Long Explanation For A Simple Idea) .
Hi, Factors ma'am,
You are about to get A LOT more than you asked for. I just finished 3 hours of typing this answer. I hope you can use it. It has a lot of mojo tricks in it.
It will be very long-winded to read, but the concepts are simple once you catch on.
Let me know how you did.
Have you checked out the mathmojo page about why
learn prime factors?
?
Here's some help with the first number on your list:
First, you must know what a factor is. In general terms, a factor is a whole number, not including 1, which goes into a number evenly (that is, with no remainder).
According to Isaac Asimov, on page 52 of his lamentably out-of-print book “The Realm of Numbers”:
...Because any number can be divided by itself and by one, the Greeks (who loved to play games with factors) usually disregarded those two numbers as factors. There is no interesting patern, after all, in something that hits all numbers without distinction. (incedentally, for every factor in number has, there is another that is the negative of the first. For instance, 60 can be divided by -2, - 3, - 4, and so on. The Greeks, however, didn't recognize negative numbers, and they don't really introduce anything new, so we usually disregard them as factors, also. )
Second, you must know that a prime number is a number
that has two and only two factors, which are one, and itself. Take the number
3. It is only divisible by 1 and 3, so it is prime.
The number four, on the other hand, can be divided by 1, 4, but also 2, (it
has three factors) so it is not prime.
So far so good ?
Here is an important point: The number 1 is generally not considered a prime number. Why? Because the definition of a prime is that has two and only two factors. But the number 1 has only 1 factor, which is 1. Therefore it doesn't fit the definition of a prime number.
One good thing about that is that it keeps us from always having to use 1 in the solution for prime factors of number. It is simply assumed that 1 is a factor of all whole numbers (which it is) but it would be redundant to include it in every solution for primes, so we don't.
Prime factors are the factors of a number that also happen to be prime.
Take the number 30. What numbers go into 30 evenly (besides one and itself)?
2 goes in evenly, 3 goes in evenly, 5 goes in evenly, 10 goes in evenly, and
15 goes in evenly. But 10 and 15 are not prime numbers, so they are not prime
factors of 30 (or anything else!)
So 2, 3 and 5 are prime factors of 30.
Prime factorization also requires us to tell how
many times each of those prime factors go into a number, along with the other
prime factors.
Take the prime factors, and multiply them by themselves
2 x 3 x 5 = 30. So you only need one of each to have the prime factorization
of 30.
But if I had the number, say, 40, we'd have the prime factors:
2, and 5. (all other factors of 40 are not prime).
But 2 x 5 is not 40.
So how can we get 40 with just twos and fives?
Well, we'd need 2 x 2 x 2 x 5 to make 40. (That's the same as 8 x 5).
A simpler way to write 2 x 2 x 2 x 5 is to write 2 ^3 x 5 (which is read, "Two
to the third power times 5.")
So the prime factors of 40 are 2 and 5, but the prime factorization of 30 is
2^3 x 5.
Are we good so far?
To lock it in, let's take your first number:
36.
The factors are 2, 3, 4, 6, 9, 12, and 18.
Not all of those numbers are prime. Which are?
2 and 3 are the only prime factors of 36.
How many of each do you need to make 36?
2 x 2 x 3 x 3 will give you 36.
That is 2^2 x 3^2, and that's your answer.
An easy recipe to prime factorize a number is this:
List the prime factors. To do this, start with 2, see if that will go into
the number. If it does go in, divide the number by 2.
Now you have a new number. Repeat the above step until 2 will no longer go
into the number evenly.
For the answer so far, write the number 2 (if it actually does go into the
original number at all) and after the caret (the ^) write the amount of times
it went in.
Then use the next highest prime number, which is 3, and repeat the above steps with the number 3.
After that use the next highest prime number and repeat the above. (The next
highest prime number is not 4!)
Let's try the whole thing with the number 293:
Does 2 go into 293 evenly? No. You can tell this immediately because 293 is
not an even number. So 2 is not a prime factor of 293.
Try 3. Does 3 go into 293?
The best way to tell is if you learned the ways to test for divisibility. If
you haven't, or you don't know what that means, you will just have to divide
293 by 3 and see if it comes up with no remainder. (If you can't divide, give
up on factorization for now, and go get good at division -
Check out:
http://mathmojo.com/basic_operations/division_mojo/long_division/long_division.html
).
You'll see that 3 is not a factor of 293.
Let's go to the next prime number, which is 5. You can tell if a number is
divisible by 5 if it ends in a five or a zero. This number doesn't, so 5 is
not a factor.
So far, you have nothing to write in the answer. (You must have a tough teacher
- this is a hard one! - good for your teacher, s/he obviously wants to challenge
your mind. Hotcha!)
What's the next prime? Is it 6? No, because 6 has, besides itself and one, 2 and 3 as factors. (Remember, a prime has only 1 and itself as factors).
So the next prime number is 7. Divide 293 by 7, and you'll get 21, with a remainder of 6, so it doesn't go in evenly, and is therefore not a factor of 293.
8, 9 and 10 all have factors other than 1 and themselves, so they are not prime.
The next highest prime number is 11. Divide 293 by 11 and get 26, with a remainder of 7, so that won't work.
12 is not prime. (By the way, an number that isn't prime is known as a "composite" number.)
13 is prime. Divide 293 by 13 and get 22 with a remainder of 7, so 13 is NOT a factor. (Boy, this teacher must be REALLY tough!)
14, 15, and 16 are composite numbers.
The next highest prime is 17. Divide 293 by 17 and get 17 with a remainder of 4. Wow, this is getting to be a pain. Who is this teacher?
18 is composite.
19 is prime. Divide 293 by 19 and still get a remainder. (Now I'm getting mad!)
20, 21 and 22 are composite.
Divide 293 by 23 and get 12 with a remainder. (GRRRRR!)
24, 25, 26 are composite.
27 (Yeah, this one must work, right? Wrong! It's composite. It's 3 x 9.)
28 is composite.
By inspection, you can see that 29 won't work. Multiply 29 by 10, get 290, which won't work (it's too low). Adding another 29 will make it too high. So 29 doesn't go into 293 evenly.
30 is composite.
31 is prime. 293 divided by 31 will be 9 with a remainder. Still no go.
32, 33, 34, 35, 36 are composite.
293 divided by 37 is 7 with a remainder. Darn!
38,39,40 are composite.
293 divided by 41 is 7 with a remainder.
42 is composite.
293 divided by 43 is 6 with a remainder.
44,44,46 are composite.
293 divided by 47 is 6 with a remainder.
48, 49 and 50 are composite.
293 divided by 51 is 5 with a remainder.
(Whoops! I didn't even have to do this. 51 is not prime. It is composite. It
has factors of 3 and 19.)
52 is composite.
293 divided by 53 is 5 with a remainder.
54, 55, 56 are composite
57 is also composite. It doesn't LOOK composite, does it? But do the same thing with that number as we are doing for 293 and you will find out that it is 3 x 19).
58 is composite.
293 divided by 59 is 4 with a remainder (but this one was close, too!)
60 is composite.
293 divided by 61 is 4 with a remainder.
62, 63, 64, 65, 66 are composite.
293 divided by 67 is 4 with a remainder.
68, 69, 70 are composite.
293 divided by 71 is 4 with a remainder.
72 is composite.
293 divided by 73 is 4 with a remainder. Sheesh!
74, 75, 76, 77, 78 are composite.
293 divided by 79 is 3 with a remainder.
80, 81, and 82 are composite.
293 divided by 83 is 3 with a remainder.
84, 85 86, 87, and 88 are composite.
293 divided by 89 is 3 with a remainder.
90, 91, 92, 93, 94, 95, and 96 are composite.
293 divided by 97 is 3 with a remainder.
98, 99, and 100 are composite.
293 divided by 101 is 2 with a remainder.
102 is composite.
293 divided by 103 is 2 with a remainder.
104, 105, 106 are composite.
293 divided by 107 is 2 with a remainder.
108 is composite.
293 divided by 109 is 2 with a remainder.
110, 111, 112 are composite.
293 divided by 113 is 2 with a remainder.
114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126 are composite.
293 divided by 127 is 2 with a remainder.
128, 129, 130 are composite.
293 divided by 131 is 2 with a remainder. (What is this teacher's PROBLEM? Does he hate kids?) (Just kidding! It's a great lesson!)
131, 132, 133, 135, 136 are composite.
293 divided by 137 is 2 with a remainder.
138 is composite.
293 divided by 139 is 2 with a remainder
140, 141, 142, 143, 144, 145, 146, are composite.
Any number higher than 146 is not going to be a factor of 193. Can you see
why?
It is because Any number higher than half of the original number cannot possibly
be a factor of the number because it cannot possibly divide into it evenly.
So we can stop here. *
That means that 293 is the only prime factor of 293.
Which means that 293 is a prime number.
Which means that your teacher is a really sneaky guy, (or girl).
What did we learn from this.
Another thing - 293 is by far the hardest number on your list. You won't have ay problems like that with the other numbers. I promise!
Here's a tip : whenever you come to a large prime factor of a number, make
a list of all the multiples of that number up to half of the original number.
For example, when we get to 13, make a list of its multiples:
13
26
39
52
65
78
91
104
117
130
143
156
(that's enough, because we have reached half the original number (293). No numbers higher than this can possibly be a factor of 293.So now you know, whenever you come to any of these numbers as you do the above method, you'll know that they are composite numbers, and you don't have to divide to test them. It will save you some work. Especially with numbers like 91, 117 and 143 which are not otherwise easily recognized as composites.
Now do the same thing for 17, 19, 23, etc. It really will save you work.
17
34
51
68
85
102
119
136
153
(that's enough, because we have reached half the original number (293). No numbers higher than this can possibly be a factor of 293.
19
38
57
76
95
114
133
152
(that's enough, because we have reached half the original number.)
23
24
69
92
115
128
151
That's enough.
29
58
87
116
145
174
Enough.
31
62
93
124
155
Enough.37
74
111
148
185
Enough.41
82
123
164
Enough.43
86
129
172
Enough.47
94
141
188
Enough.53
106
159
Enough.59
118
177
Enough.61
122
183
Enough.67
134
201
Enough.71
142
213
Enough.73
146
219
Enough.79
158
Enough.83
166
Enough.87
174
Enough.91
182
Enough.
Once you have made the lists like the ones above, you'll know that when you come to those numbers, you don't have to divide them into 293, because none of them are prime.
There is one more hint for today: It is a cool thing to memorize all the primes up to 100. It's a nifty little thing to have in your head. It'll come in handy for all sorts of things (but I can't think of a single one!) It still is kind of cool, though.
They are:
1 (which doesn't count in prime factorization), 2, 3, 5, 7, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.
I suspect that one of the things your teacher was trying to get you to do in this assignment is become familiar with all the prime numbers up to around 100, as well as get familiar, by your own effort, with the things I've tried to explain here.
Or else it was dumb luck that the assignment made you cover all of them.
Or else it was out of pure SADISM! Here's a super tip (it's kind of advanced, and a sixth grader doesn't need to know it - ask the teacher if s/he even knows it - but it is cool to know):
Let's say you want to know if 71 is a factor of 293, without dividing. Try
this:
Multiply just the LAST DIGIT of 71 by 1, 2, 3, 4, 5, 6, 7, 8, and 9.
When you do that, you can see that only doing it by 3 will give you a product
that ends in 3, so you can safely say that 71 will not possibly divide evenly
into 293 (which ends in "3") by any of the other digits. So you only
have to see if 71 x 3 is 293. If it isn't, then it isn't a factor of 293.
71 x 3 = 213, so it is not possibly a factor of 293.
(This only works with numbers that are ten times or less larger than the number you are trying to check as a factor. What I mean is that it won't work if you want to see if 71 is a factor of 2,314, because 2,314 is more than ten times 71). There is a way to check and see if larger and smaller numbers work, but it is a tiny bit too complicated to go into here. See if you can work it out yourself, if you like.
Let's try it with another number.
We want to know if 17 is a factor of 128. 128 is less than ten times 17, so this method will work. We multiply the last digit of 17 (which is 7) times 1, 2, 3, 4, 5, 6, 7, 8, and 9. The only one of those that will give us an 8 (which is the last digit of 128) as the last digit of the product, is 4 (because 7 x 4 = 28).
So now we multiply 17 times 4 and get 68, which is not 128. So 17 is CANNOT possibly be a factor of 128.
This method may seem complicated, but it can save you a lot of dividing.
Factors ma'am , I hope you got something out of this lesson. I enjoyed making it. I'll be putting it up in the "interesting lessons" section of mathmojo.com soon.
I appreciate your question. It is input from readers like you that makes this site fun to work on.
HOTCHA!
Yours truly,
Brian Foley (a.k.a. Professor Homunculus)
Feel free to send me your child's answers to the other examples, and I'll be happy to check them.
All the best,
Brian
* You are going to hate me for this -
Remember way back when we were factoring 293? Remember where I wrote:
Any number higher than 146 is not going to be a factor of 193. Can you see
why?
"It is because Any number higher than half of the original number cannot
possibly be a factor of the number because it cannot possibly divide into it
evenly.It turns out that we could have stopped at 17.
Once you reach a factor that is the square root (or larger) of the original number, you need to look no further. This is why:
The square root of 293 is around 17. (17 squared is 289.) We have already tested 17 and found that it is not a factor.
Consider this now - If a number higher than 17 was a factor of a 293, the other number would have to be lower than 17. We know this because if we multiply 17 by 17 we already get 289, so if we multiply 17 (or any number larger than 17) by anything higher than 17 we'd get something larger than 293.
Therefore, we'd have to multilpy 17 (or any number larger than 17) by some number that is smaller than 17. But we know that no prime numbers that are smaller than 17 are factors of 293, therefore there is no chance that a larger number could be multiplied by one of them to give you 293.
So you can stop at 17.
As usual - Long explanation for a simple Idea (LEFASI).
You can find another example, with some other hints on the page about What's the prime factorization of 1 trillion using exponents?
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