Math 20 - Lesson 11


Uploaded by PCCvideos on 09.09.2009

Transcript:
A Portland Community College mathematics telecourse.
A Course in Arithmetic Review.
Produced at The Portland Community College.
Here we have a rather simple problem.
The technique we will use to solve this problem systematically
in fact is going to be very useful
for other, tougher problems down the road.
So let's attack this problem then back off
and look at how we did it.
Okay, first we want to find all the divisors or factors of 64.
Well, one obvious one is 1.
And 1 goes into 64, of course, 64 times. [ 64 ÷ 1 = 64 ]
2 is another one.
2 divides into 64, 32 times. [ 64 ÷ 2 = 32 ]
3, should we try to divide that into 64? [ 64 ÷ 3 = ? ]
It won't go.
4 goes into 64, 16 times. [ 64 ÷ 4 = 16 ]
Now note, we're already beginning to see a pattern.
In trying every single whole number one at a time
to see if it divides in, if it does,
by dividing it, we get as a gift a second divisor.
So when you find one divisor,
another divisor is given to you free.
Now take that fact into your mind.
When you get one factor of a number,
it gives you a second one free, in this case, that's larger.
But notice as the first factor gets larger,
the second factor gets smaller.
That's a second important fact.
In trying 5, that won't work. [ 64 ÷ 5 = ? ]
6 won't divide. [ 64 ÷ 6 = ? ]
7 won't divide. [ 64 ÷ 7 = ? ]
8 will divide into 64, 8 times. [ 64 ÷ 8 = 8 ]
We try 9 won't. [ 64 ÷ 9 = ? ]
10 won't. [ 64 ÷ 10 = ? ]
11 won't. [ 64 ÷ 11 = ? ]
Nor 12, [ 64 ÷ 12 = ? ]
13, [ 64 ÷ 13 = ? ]
14, [ 64 ÷ 14 = ? ]
15. [ 64 ÷ 15 = ? ]
But 16 will. [ 64 ÷ 16 = ? ]
16 will go in 4 times. [ 64 ÷ 16 = 4 ]
But note this, that that [ 4 · 16 ]
is essentially the same as this [ 16 · 4 ] simply commuted.
That's not surprising
as this one got bigger, this one got smaller.
And as this one gets bigger and this one gets smaller,
there reaches a point where the two
either are equal as in this case or are nearly equal.
Then from that point on
as this one gets bigger, this one has to get smaller,
so all of these others that I had previously will be retraced.
Okay, we do have all of our divisors now.
We have the 1, 2, 4, 8.
Then going backwards 8, 16, 32, 64.
Now at first thought to find all the divisors of 64,
it might appear we'd have to try 1, 2, 3 all the way up to 64.
But it turns out we didn't have to go very far,
only up to 8 to find all of them
because the smaller ones were giving me the larger ones free.
So now let's see if we can isolate how far it is we have to go
before we can quit and say we've gone far enough.
Let's recall loosely what we meant by square root [√]
We can verbalize "the square root of 64"
or you might recall the symbol for it was this: [ √64 ]
The 'square root' of a number is that number which times itself,
so these must be the same, [ (same) · (same) ]
will equal the insides.
And of course in this case that's 8,
because 8 times 8 is 64. [ 8 · 8 = 64 ]
Square rooting of course is so important
that most calculators will have a square root key.
Mine's right there. [√]
So if I were to punch in 64, then hit the square root key [√]
sure enough there's 8. [6][4][√] = 8
Most square roots don't come out even like that.
If I were to punch in 26. [2][6]
Hit square root, I get a decimal. [2][6][√] =
Just have that intuitive loose feel for what a square root is.
And notice this: the point at which I could quit here
happens to be the point where these two are about the same
or that's another way of saying
that this is about the square root of 64.
That leads us into a convenient rule:
when trying to find all the factors of a given number,
one need only try as far as
the approximate whole number square root of that number.
The larger factors will be automatically extracted
by the smaller ones.
Now let's summarize that through another example.
The nearest square root is 72.
8 times 8 is 64. [ 8 · 8 = 64 ]
That's too small.
However, 9 times 9 is 81. [ 9 · 9 = 81 ]
That's too big.
So the square root of 72 is approximately equal to 8. [ √72 ≈ 8 ]
So that means as I try to find all the divisors of 72,
I need only go as far as 8.
If there are any higher ones,
these lower ones will pull them out.
And of course one goes into 72, 72 times. [ 72 ÷ 1 = 72 ]
2 goes into 72, 36. [ 72 ÷ 2 = 36 ]
And so far trying each one of these,
either by hand or with a calculator, we get this:
3 works and it gives us 24. [ 72 ÷ 3 = 24 ]
4 works and it also gives us 18. [ 72 ÷ 4 = 18 ]
5 does not work, but 6 does, and it gives us 12. [ 72 ÷ 6 = 12 ]
7 doesn't, but 8 does. [ 72 ÷ 8 = 9 ]
And since this [ 8 ] is the nearest whole number
square root of that, I can quit there.
All the larger ones have been given to me already:
9, 12, 18, 24, 36, and 72.
So if our problem was to find
all the divisors or factors of 72,
we have them very systematically and very quickly,
and writing them in order
we get these 12 different
divisors of 72. [ 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 ]
Now if we were to phrase the problem this way:
"Write 72 as the product of two factors in all possible ways,"
can you see we're just asking the same question
from a different angle that, in fact,
this procedure has already answered that question as well.
Let's summarize that through one last problem.
If we wish to write 105 as a product of two factors
in all possible ways,
that's the same thing as asking "Find all the divisors of 105."
Starting with 1, 2, 3, and so on.
And the question is how far do we have to go
before we can say hey, we have gone far enough?
Ask yourself, what's the approximate square root of 105?
Well 10 times 10 gives me 100, [ 10 · 10 = 100 ]
and 11 times 11 will give me [ 11 · 11 ]
a number more than 105, [ 11 · 11 = 121, 121 > 105 ]
so 10 is as far as I have to go.
So all I need to try are these 10 numbers and ask in each case:
do they divide into this? [ 105 ÷ ]
Well 1 goes in for sure. [ 105 ÷ 1 = 105 ]
1 is the one number that divides into every number.
Them trying these one at a time [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
either by calculator or by actually trying to divide.
Now in another lesson or so
we're going to develop some mental shortcuts
to check the divisibility by smaller numbers.
But if we try it,
we'll find that one [ 2 ] doesn't go in. [ 105 ÷ 2 = ? ]
3 does, 35 times. [ 105 ÷ 3 = 35 ]
4 does not go in, [ 105 ÷ 4 = ? ]
but 5 does, 21 times. [ 105 ÷ 5 = 21 ]
6 we'll find won't, [ 105 ÷ 6 = ? ]
but 7 does, 15 times. [ 105 ÷ 7 = 15 ]
8 won't. [ 105 ÷ 8 = ? ]
9 won't. [ 105 ÷ 9 = ? ]
And 10 won't. [ 105 ÷ 10 = ? ]
And I found that out
by simply trying to divide all of these into 105.
Actually I didn't.
Many of these I could just look at 105
and tell whether one of these would or would not divide,
and those shortcuts will be the gist of our next few lessons.
So we have four pairs of factors of 105.
It is well to get used to the phrase: 'to factor,' as a verb.
This will be used constantly in all math courses
from Algebra I higher.
Consider the phrase: "I wish to factor 12."
Just exactly what are we stating here?
Loosely at first, and we'll tighten this grip later,
what we're asking for here
is to write 12 as an indicated product.
3 times 4 is one possibility, [ 3 · 4 ]
and when we can write an expression
or in this case a single number as a indicated product,
we are said to have factored that number.
Of course, I could have done it several ways.
2 times 6. [ 2 · 6 ]
And of course the most obvious way
is 1 times 12. [ 1 · 12 ]
Please note this: the word to 'factor'
is placing our emphasis on the fact that I can write this [ 12 ]
as an indicated product. [ 1 · 12, 2 · 6, 3 · 4 ]
So we say that 3, one of our multipliers [ 4 · 3 ]
that would get me to the 12, is a factor of 12.
4 is a factor of 12 [ 4 · 3 ]
and the same with 2, 6, [ 2 · 6 ] 1 and 12. [ 1 · 12 ]
But you can see in this sense
if 3 or 4 or 2 or 6 is a factor of 12, it's also a divisor.
When we're using the word 'factor,'
we're trying to emphasize
that ultimately we want a product situation.
If we use the word 'divisor,'
we're trying to focus our attention on the fact
that a given number divides into our given number or expression.
So essentially the real difference
between these two terms is where our focus is.
Here our focus is on, ultimately, an indicated product.
And here our focus is on a process
which a number will do relative to a given number,
that is, divide evenly into it.
This term will be used so frequently in the future
that we can safely say a student who's not comfortable with this
will have some difficulties at the beginning of algebra.
So if we were faced with a request to find and list
all the divisors of, say 30,
that would be semantically, at this point,
the same as asking for all the factors of 30.
So in this particular case
we always start with the most obvious one: 1 times anything.
This [1] is always a factor of any other number.
And then we began to count from 2 to 3 and so on
until we get to the nearest square root of 30.
So we can see in this case by inspection alone
that the nearest approximate square root of 30
rounded down is 5, because 5 times 5 is 25, [ 5 · 5 = 25 ]
which makes it too low.
6 times 6 is 36, [ 6 · 6 = 36 ] which is too high.
So it's quite sufficient
if we find the nearest square root rounded down,
which in this case is 5,
so all we need to do is to try all the whole numbers
up to the nearest square root of this number rounded down.
Then we simply try.
Will 2 divide into 30 [ 30 ÷ 2 = ? ]
The answer's yes. [ 30 ÷ 2 = 15 ]
So we say that 2 is not only a divisor of 30,
2 is a factor of 30, as is 15. [ (2)(15) ]
3 will also divide evenly into 30, 10 times. [ 30 ÷ 3 = 10 ]
So we would say 3 and 10 [ (3)(10) ] are each factors of 30,
and this product [ (3)(10) ] we call a factorization of 30.
So to say that we have a factorization of a number
is a way of saying I have now written that number
as am indicated product so far in three different ways.
Okay. Continuing our quest for all of the factors of 30.
4 will not go into 30 evenly, [ 30 ÷ 4 = ? ]
but 5 will 6 times. [ (5)(6) ]
We've come to the end of our list so our question:
Find and list all the factors.
Well, we have found 1, 2, 3, 5.
Now going back up our list 6, 10, 15, and 30.
So in proper
ascending sequential order [ 1, 2, 3, 5, 6, 10, 15, 30 ]
we have all of the factors or divisors of our given number 30.
This lesson is as much about terminology as it is processes.
So if one can answer the question
one not only knows that a student can do something,
but that he or she understands the terminology.
So note, this question could have been phrased
"Write 30 as a product of two factors in all possible ways."
Now what, how are we asking
for our data to be presented here differently than here?
Here we wanted a list of all of the divisors,
which we also call factors,
so we listed every single one of them.
But here we don't want a list of singletons;
we want a list of products.
Hence, we're really getting down
to the fundamental meaning of the word 'factor,'
and we want the products, two at a time, all possible ways.
Well in this case we have it already.
There's 1 times 30. [ 1 · 30 ]
2 times 15. [ 2 · 15 ]
3 times 10. [ 3 · 10 ]
And finally 5 times 6. [ 5 · 6 ]
So here they're asking for 'products'
and there are only four of them [ 1 · 30, 2 · 15, 3 · 10, 5 · 6 ]
Here they're asking for a list of individual divisors,
and there are 1,
2,
3,
4,
5,
6,
7,
8 of them. [ 1, 2, 3, 5, 6, 10, 15, 30 ]
Do you see the difference
between the phrasing of these two questions?
It's very important that you see very subtle difference
in shades of communications
because this is what will make math easy or hard in your future.
Let's try to find all the factors
of a more involved larger number.
No new technique, just a larger number.
But let's see if we can systemize our search
and at the same time see how we can use a calculator
to help speed our process of finding all the factors,
and remember to find all the factors
is the same thing as asking for all of its divisors.
Now first we know we're going to start from 1, then try 2,
then try 3, all the way up to the nearest square root of 840.
So we want to ask first
What is the nearest square root of 840 rounded down?
So we want the approximate square root of that number.
So on your calculator somewhere there is a key
that looks like this [√] A square root key.
So if you can find it, and on my calculator
it's that key right there, [√]
all you need do is enter the number that you wish to square root.
840, hit the [√], and I get 28.98 [8][4][0][√] 28.98
so I round to the nearest whole number downwards,
so that would be 28. [ 28.98 ≈ 28 ]
So this tells me all I need to do
is try all the numbers up to 28
and see if each of them, one by one, will go into 840.
Well, that's still 28 divisions I must try.
We can see that's a lot of division to do.
One by one we want to try each of these
to see if they will divide evenly into this,
so basically we're going to use that number there [ 840 ] 28 times
each time with one of these.
So if have you a calculator with a memory,
on this particular calculator it's a key that says
x arrow memory.
Let me write that down a little bit larger.
There, the key looks like this: [ X→M]
Now on some calculators it will be M sub i [Mᵢ]
which means 'memory in.'
Other calculators will simply have an M plus, [M+]
which means to 'add to memory.'
And of course if memory has nothing in it,
then you are simply putting in your current number.
On my particular calculator if I take my given number
I'm going to use over and over and over,
840, [8][4][0] then push my [x to memory]
and x stands for the number you currently have, which is 840,
so I push my x to memory key. [X→M]
Now that tells me that not only is this number on my screen,
it's inside the calculator in memory.
To prove that to myself, let me clear the calculator
and push my RM key. [RM]
It's a key that looks like this. [RM]
RM, which means 'recall memory,'
on some calculators it will be MR for 'memory recall.'
So check your calculator.
It will be different on each calculator somewhat,
but if I were to push 'recall memory' see the 840 is still there.
Now if I clear it again, push the recall memory again;
it's still there.
And it stays there until I choose to erase it
or put something else in its place.
So that allows me to put a large number in
and not have to re-enter it over and over and over.
It will come out any time I want it now
by pushing the memory recall.
So let me begin a sort of a coded set of instructions
to keep track of what we're about to do.
This says I have taken 840 [ 840 [X→M] ]
and put it into memory on my calculator.
And it will be there no matter how much I clear it
until I change it.
Now I wish to see if 1 will divide into 840,
and of course it will 840 times.
2 will go into 840.
I can see that mentally, so 2 goes into 840 4, 2, 0 times.
Now it might not be quite as obvious whether 3 will go into 840,
so what I will do is to recall memory [RM]
Okay. I pushed [RM] the recall memory key.
See, my 840 comes back at me.
Then I will push the division key [÷] which is right here.
It might be difficult for you to see on the screen,
so I'm going to divide.
And I want to divide by 3, so I push the [3], then equal [=]
if there is no decimal,
that tells me that 3 divided into there evenly 280 times.
So I have 2 more factors of 840.
Now I don't need this anymore, so I can erase it.
Now I recall, [RM] first we recall memory,
divided by 3, [RM][÷][3][=] and we got the 280.
So now I push the recall memory key [RM] once again.
There's the 840 again.
Then the division key. [÷]
Then the 4. [4]
Of course we can see that 4 will go into 840. Can't we?
4 goes into 80 two times [ 80 ÷ 4 = 20 ]
4 goes into 4 once [ 4 ÷ 4 = 1 ]
so we should have 210.
So I hit recall memory which gave me 840,
the division key, now 4 equals, [ 840 [÷][4][=] ]
and sure enough there is my 210. [ 840 [÷][4][=] 210 ]
Then I can clear that.
I can see that 5 will go into this [ 840 ]
because this ends in a zero or a 5.
But rather than doing it mentally or on paper,
I can simply push the recall memory again,
then the divide again, and then 5, [ [RM] 840 [÷][5] ]
then the equal, and I get 168. [ 840 [÷][5][=] 168 ]
So I have two more factors of 840.
So recall memory, divide by, 6 equal, 140. [ [RM][÷][6][=] 140 ]
So I have 2 more factors.
Then recall memory, divide, 7, equal 120. [ [RM][÷][7][=] 120 ]
So I have two more factors. [ 7 · 120 ]
Then recall memory, divide by 8, =, 105. [RM][÷][8][=] 105
I have two more. [ 8 · 105 ]
Recall memory, divide by 9, equals. [ [RM][÷][9][=] ]
It won't go in evenly.
Now we can see that 10 goes into 840, 84 times. [ 840 ÷ 10 = 84 ]
And you can see after a while
this becomes automatic and you proceed just this fast.
Recall, memory, divide, 11, equal. [ [RM][÷][11][=] ]
It won't go in.
Clear memory.
Okay. Recall memory, divide, 12, equals. [ [RM][÷][12][=] 70 ]
It goes in 70 times. [ 12 · 70 ]
Then recall memory, divide, 13, equals. [ [RM][÷][13][=] ]
It won't go in.
Recall memory, divide by 14, equals. [ [RM][÷][14][=] ]
It will go in. [ 14 · 60 ]
See, we're pulling these out rather quickly aren't we?
So our repeating sequence is recall memory, [RM]
divide by the number [÷] equals [=]
So recall memory, divide by the number, = [RM][÷]number[=]
15 goes in 56 times. [ [RM][÷][15][=] 56 ]
Okay. Recall memory, divide, 16, equals. [ [RM][÷][16][=] ]
It doesn't go in evenly because of the decimal point.
Recall memory, divide, 17, equals. [ [RM][÷][17][=] ]
Won't go in.
Recall, divide, 18, equals. [ [RM][÷][18][=] ]
Won't go in.
Recall, divide, 19, equals. [ [RM][÷][19][=] ]
Won't go in.
Recall, divide, 20, equals, 42. [ [RM][÷][20][=] 42 ]
So I have 2 more factors of my 840. [ 20 · 42 ]
And so you continue this process
until you've exhausted your list,
and actually after a while it becomes a very fast procedure.
So you see the number 840 has quite a few divisors or factors.
And we found all of them very, very systematically.
So this system is what we're really after in this lesson
as much as the factors and divisors themselves.
In fact, possibly more so.
The vocabulary, which tells you what we want,
that too, is a major part of this lesson.
In the next lesson we'll also show you some mental shortcuts
to assist you in this process as well.
So, this is your host, Bob Finnell.
We'll see you at that next lesson.