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A Portland Community College mathematics telecourse.

A Course in Arithmetic Review.

Produced at Portland Community College.

Let's say that for some reason we wanted to know

if 108 or any other number for that matter

were divisible evenly by these particular numbers.

Now one obvious way to check that

is so try to see if they divide in evenly

either by hand or by calculator.

But there are some nice tricks which if learned

simply allow you to look at this number

and answer yes or no with each ones of these.

The one for 2 you already know.

You simply look at the one's digit and ask, is it even?

In this case, [ 2 ] it is. So the answer is yes.

2 will divide evenly.

That we call a 'divisibility test'

and the divisibility test for 2 was

if the one's digit is even, it's divisible by two.

That's what this lesson is about to give you a series

of simple divisibility tests for various easy numbers,

and we've just had one already.

But there is also an equally easy test

to check divisibility by three.

A number is divisible by 3,

if the sum of its digits is divisible by 3, our number was 108.

If you were to add its digits, you would get 9. [ 1 + 0 + 8 = 9 ]

9 is divisible by 3, hence this test says 108 is divisible by three.

Check it to see. [ 108 ũ 3 = 36 ]

And it is. It goes in 36 times.

So the question "Is 108 divisible by three?"

The answer is 'yes.'

Now how about 5?

Most people know that one as well.

If the one's digit is 0 or 5 it's divisible by 5.

It isn't, so the answer to this is no.

So let's formally add that to our list of divisibility test.

The one's digit is either 0 or 5.

If the answer is yes.

Then yes, the number is divisible by 5.

To check divisibility by 4, you ask, Is the number formed

by just its tens and ones digit is divisible by 4?

Let's check just that portion alone first.

Here's our number. [ 108 ]

Let's look at that portion

formed by the ones and the tens digits [ 08 ]

which is still 8, of course.

Is that divisible by four? [ 8 ũ 4 ]

The answer's yes.

Therefore, the whole number, 108 is divisible by 4.

As a matter of fact, 4 divides into 108 27 times.

Sometimes that gets to be rather messy,

so there is an appended comment here that says this:

or if another number formed by it minus any multiple of 4.

Now remember what we mean by multiples of 4.

4, 8, 32, 40, 400.

Any number that 4 will go into evenly is a multiple of 4.

Now let's look at what this statement's really saying.

If another number formed by it minus any multiple of 4.

Now if we wanted to check to see if this is divisible by four,

our first part of our rule says

to just look at those two right there

and ask is that [ 92 ] divisible by 4?

But that's not so easy in its own right.

So let's take a number that we know is a multiple of four, 80.

That's 4 times 20. [ 4 x 20 = 80 ]

Subtract that and we get 12. [ 92 - 80 = 12 ]

Now that's a smaller number.

Now is that [ 12 ] divisible by 4? [ 12 ũ 4 ]

And the answer's yes.

That tells us the whole number [ 1392 ] is divisible by 4.

This fact in particular will help you

in many of your business courses

because by definition a leap year,

that's the year that has 366 instead of 365 days,

by definition that's any year divisible by four.

So if we needed to know ahead of time in a business ledger

whether 1996 will be a leap year,

this test tells us to look just at the end two digits [ 96 ]

Will 4 divide that? [ 96 ũ 4 ]

And if you can't see that very readily,

take a multiple of four and in this case 80 again.

Subtract from that and we get 16. [ 96 - 80 = 16 ]

Now, is that divisible by 4? [ 16 ũ 4 ]

And the answer's yes. So, yes, that's [ 1996 ] a leap year.

A very handy trick for those of you going on into business.

The similar divisibility test for 8,

if a number formed

by the right end three digits is divisible by 8,

then the whole number is divisible by 8.

This example. [ 82,134 ]

If the number formed by the end 3 digits [134] is divisible by 8,

then the entire number [ 82,134 ] is divisible by 8.

But that's not so obvious [ 134 ũ 8 ] in its own right.

So our second phase or that number

which is any multiple of 8 less than it.

So if this [ 134 ] was still too large for us to mentally handle,

take any multiple of 8.

Well 80, that's 10 times 8. [ 10 · 8 = 80 ]

Subtract that [ 134 - 80 ]

and we get 4, 8 from 13 is 5. [ 134 - 80 = 54 ]

Is that [ 54 ] divisible by 8?

Now if it's still not obvious, take another multiple of 8.

Let's take 40.

That's 5 times 8. [ 5 · 8 = 40 ]

Is that divisible by 8?

The answer is no.

Therefore, the original number [ 82,134 ] is not divisible by 8

even though it is divisible by two.

Checking just that [34] we find that this is not divisible by 4.

So the whole number will not divide by 4 or 8, but it will by 2.

So in asking about our original number 108,

we have to ask about the whole thing

because that is the last three digits.

Well probably just as fast to go ahead

try mentally to divide it by 8

but we could subtract this multiple 80 from it very quickly.

8 from 10 is 2. [ 108 - 80 = 28 ]

28 is not divisible by 8.

Therefore, 108 is not divisible by 8.

We had skipped 6.

But let's skip that just for a moment and jump down to 9.

A number is divisible by 9

if the sum of its digits is divisible by 9.

That's essentially the same test one had for 3.

So the sum of our digits here is 1 and 8 is 9. [ 1 + 0 + 8 = 9 ]

9 is divisible by 9 so, yes, 108 is divisible by 9.

Now, we still left out some.

This one is almost universally known.

If the units digit is zero, it's divisible by 10

so the answer to divisibility here is no.

Now these three we don't need a special test.

All we need to remember is that they are made up, product wise,

of numbers we've already used.

6 is nothing more than 2 times 3. [ 6 = 2 · 3 ]

And 2 goes in, [ 108 ũ 2 ] 3 goes in, [ 108 ũ 3 ]

and 2 times 3 is 6 [ 2 · 3 = 6] So 6 goes in as well.

12 is 4 times 3. [ 12 = 4 · 3 ]

4 goes in, [ 108 ũ 4 ] 3 goes in [ 108 ũ 3 ]

So 12 goes in. [ 108 ũ 12 ]

And the divisibility for the three

was a sum of the digits is divisible by 3.

The test for 4 was the end two digits is divisible by 4.

Now can you see what we're going to do with 15?

15 is 3 times 5. [ 15 = 3 · 5 ]

3 goes in, 5 does not go in, so 15 is not a divisor.

Frequently when we try to divide a number by lower numbers,

we try first of course 2, 3, 4, 5, 6 and on up to about 10 or 12.

And these tests says that those checks can generally be made

by observation and just a mental act.

One doesn't need to divide or to use a calculator.

If you learn those tests, it will save you much time.

Let's go through those tests once more

with six rather large numbers.

Let's do the test one at a time now to really nail it down.

To check the divisibility by two

we don't have to divide each of these by 2 by hand or by calculator.

All we need to do is look at each end or units digit,

and ask: is it even?

If the answer is yes and 8 is even, then that's divisible by 2.

So is that. [ 156 ]

So is that. [ 3012 ]

So is that. [ 5174 ]

That isn't. [ 9117 ]

This isn't. [ 105,215 ]

So these first four [ 148, 156, 3012, 5174 ] are divisible by two

so on these last two we wouldn't waste our time even trying.

Not even with a calculator.

And see this is much faster to determine

than to even use a calculator.

So once again to check the divisibility by 2,

you merely look at the units digit and ask: is it even?

Yes or no.

How about the same six numbers?

Are they divisible now by 3?

To check the divisibility of 3 now we must add the digits

so 1 and 4 is 5 [ 1 + 4 = 5 ]

and 8 is 13. [ 5 + 8 = 13 ]

That is not divisible by 3.

Neither is the number made up of them 148.

Checking here, one and 5 is 6 [ 1 + 5 = 6 ]

and 6 is 12. [ 6 + 6 = 12 ]

Yes, 3 will go into 12, [ 12 ũ 3 ]

so yes, 3 will divide that [ 156 ]

Same thing here. [ 3012 ]

3 and 1 is 4 [ 3 + 1 = 4 ]

and 2 is 6. [ 4 + 2 = 6 ]

3 will divide into 6, [ 6 ũ 3 ]

So yes, 3 divides into that [ 3012 ũ 3 ]

Now a somewhat faster approach to this,

as soon as we have a sum that 3 will divide into,

in a sense cast it away.

So 3 goes into 3, [ 3 ũ 3 ] so throw it away.

1 and 2, 3 goes into that, so I have my yes.

So here 1 and 5 is 6. [ 1 + 5 = 6 ]

3 goes into that.

Cast it away.

3 goes into 6 twice [ 6 ũ 3 = 2 ]

So yes, it divides into the whole thing.

So trying that trick here 5 and 1 is 6, 3 divides that.

7 and 4 is 11.

3 does not divided that, so we will not divide into 5,174.

3 goes into 9.

1 and 1 is 2 and 7 is 9, so three will divide into that.

1 and 5, 3 goes into that. [ 1 + 5 = 6 ]

2 and 1, 3 divides into that. [ 2 + 1 = 3 ]

5, 3 will not go into that. [ 5 ũ 3 ]

So 3 will not divide the whole thing [ 105215 ũ 3 ]

Did you see our approach?

I could add 1 and 5 is six, [ 1 + 5 = 6 ]

and two is eight, [ 6 + 2 = 8 ]

1 is 9 and 5 is 14. [ 1 + 5 + 2 + 1 + 5 = 14 ]

But rather than doing that,

if I get a sub sum, divisible by three,

cast it away and start over.

2 and 1 is 3. [ 2 + 1 = 3 ]

Sub sum.

3 divides it cast it away. [ 3 ũ 3 ]

5, 3 won't go into that. [ 5 ũ 3 ]

It won't go into the whole expression.

So 3 will go into these three: [ 156, 3012, 9117 ]

but not these three [ 148, 5174, 105,215 ]

Same numbers again.

Now are each of them divisible by four?

Our test for that was to look at the end two digits alone.

4 will go into 48. [ 48 ũ 4 ]

So will go into 148. [ 148 ũ 4 ]

Will 4 go into 56? [ 56 ũ 4 ]

If that's not quite so obvious,

then subtract a multiple of 4

which in this case is 40, [ 56 - 40 = 16 ] which gives me 16.

4 will go into 16. [ 16 ũ 4 ]

Therefore, it will go into 156. [ 156 ũ 4 ]

Again look at the end two digits [ 12 ]

4 will go into 12. [ 12 ũ 4 ]

It will go into here. [ 3012 ũ 4 ]

Again, the end 2 digits [ 74 ]

If it's not obvious subtract a multiple of 4.

40 again in this case. [ 74 - 40 = 34 ]

4 will not divide into 34. [ 34 ũ 4 ]

Hence it will not divide into the entire number. [ 5174 ũ 4 ]

Now you can see 4 won't go into these end two digits [ 17 ]

because it's not even even so for sure if 2 won't go into it,

4 can't because 4 is made up of two 2s factor wise. [ 2 · 2 = 4 ]

And in the same sense,

I would eliminate that because this isn't even an even number.

So the first three are divisible by 4. [ 148, 156, 3012 ]

Again the divisibility test is to check the end two digits

ask: Will 4 go into them?

If it's a little bit difficult to see that

then subtract some convenient multiple of 4

and take the smaller difference and check it by 4.

Again, same six numbers.

Are they divisible by 5?

Well, this is one of our easier tests.

We simply look at the end digit and check to see if it's 5 or 0.

And if it is not, then it's not divisible by 5.

Don't waste your time.

But this [ 105215 ] does end in 5 or 0.

So this is the only number of the six that's divisible by 5.

Hence if I needed to, I would go ahead and divide it,

but I wouldn't even waste my time dividing these

because they're not even possible.

How about divisibility by 6?

We didn't have a special test for that, although one does exist.

But because 6 factor wise

is really made up of the factors 2 and 3, [ 2 · 3 = 6 ]

we can check each of these tests singularly,

and if it's divisible by 2 and 3,

2 times 3 is 6 [ 2 · 3 = 6 ] hence is divisible by 6.

So first our divisibility test of 2 passes here [ 148 ũ 2 ]

Then our divisibility of 3,

1 and 4 is 5 and 8 is 13, [ 1 + 4 + 8 = 13 ]

so 2 divides this [ 148 ] but 3 won't.

Hence, this is not divisible by 6 because I must have both of these.

Two divides this.

1 and 5 and 6 is 12. [ 1 + 5 + 6 = 12 ]

3 goes into that. [ 12 ũ 3 ]

So 2 goes into this. [ 156 ũ 2 ]

3 goes into this. [ 156 ũ 3 ]

So this is divisible by 2 and 3 which is 6. [ 156 ũ 6 ]

Again the 2 is obvious. [ 3012 ũ 2 ]

3 and 1 is 4 and 2 is 6 [ 3 + 1 + 2 = 6 ]

so 2 goes in, 3 goes in, 6 goes in.

Again we see that the two goes in [ 5174 ]

so 5 and 1 is 6 and 7 is 13 and 4 is 17, [ 5 + 1 + 7 + 4 = 17 ]

which is not divisible by three 3, [ 17 ũ 3 ] so this is out.

I need both 2 and 3.

2 won't go into that. [ 9117 ]

6 won't.

2 won't go into that. [ 105215 ]

6 won't.

So, for 6 we check 2 and 3 divisibility test each separately.

How about by 7?

You will note that we didn't have a divisibility test for 7.

One does exist by the way.

But it is so involved

that most courses of this type don't give it to you.

It's faster to go ahead and divide it

than it is to try to remember the rule.

However, your teacher might want to give it to you

on her or his own.

It's rather an interesting test.

But in this case we simply take on a calculator or by hand.

Calculator is obviously faster.

148 divide by 7, equals. [1][4][8][ũ][7][=]

If I get decimals,

then that tells me this is not evenly divisible by 7.

So the next one 156 divide by 7 equals [1][5][6][ũ][7][=]

Decimals.

That's out.

The next one.

3012 divide by 7, equals. [3][0][1][2][ũ][7][=]

Decimals.

That's out.

5174 divide by 7 equals. [5][1][7][4][ũ][7][=]

Decimals.

That's out, so we're not having much luck with 7s are we?

9117 divide by 7 equals. [9][1][1][7][ũ][7][=]

Decimals.

That's out.

Just one more to try.

105,215 divide by 7 equals. [1][0][5][2][1][5][ũ][7][=]

Decimals.

That's out.

So none of these were divisible by 7.

So, with a 7 just try it.

It still takes a bit of effort,

but that's as fast as any method at this point.

How about by 8?

Our test for that was to test the end three digits,

which in this case [148] and this case [156] is the entire thing.

And ask is that the divisible by 8?

If it's messy then take a multiple of 8.

Maybe 40, maybe 80, maybe 120.

These are all multiples of 8.

Subtract one of those from that and take the difference

and see if 8 will go into that.

So let's get one as close to this as possible.

That's 120, so if we

subtract 120 from this, [ 148 - 120 = 28 ] we'll get 28.

And 8 won't go into 28. [ 28 ũ 8 ]

So it won't go into this one.

Again let's take a multiple close to this.

Now don't be tempted that because 8 goes into 56

it should go into this.

It must go into all three of these.

156.

So let's subtract 120, [ 156 - 120 = 36 ] which gives me 36.

8 will not divide 36, [ 36 ũ 8 ]

so it will not divide the whole number. [ 156 ũ 8 ]

Now here the last 3 digits only make up the number 12.

And we can see that 8 won't go into 12.

Hence, it won't go into this one, either.

On this one [ 5174 ] the end three digits. [ 174 ]

Let's take a multiple that's close to this.

In that case, that would be 160 if you see it.

See, that's 20 times 8, [ 20 · 8 = 160 ]

and that gives me 14. [ 174 - 160 = 14 ]

8 will not divide into 14. [ 14 ũ 8 ]

8 won't go into this. [ 5174 ]

Can't go into here [ 9117 ] because it's even, not even,

and 8 is an even divisor.

Same thing here [105215 ]

Not even.

8 cannot go into it, so 8 will not divide into any of these.

So the test for 8 is: will the number

made up of the last three digits divide by 8?

And specifically, if it's not even you throw it out to begin with.

And how about divisibility by 9?

Recall that test was add up the sum of all digits.

So 1 and 4 is 5 and 8 is 13 [ 1 + 4 + 8 = 13 ]

and ask will 9 divide into this sum, 13? [ 13 ũ 9 ]

If the answer's no then it won't divide into the number 148.

Here 1 and 5 is 6 and 6 is 12. [ 1 + 5 + 6 = 12 ]

9 won't go into 12. [ 12 ũ 9 ]

It won't go into that. [ 156 ũ 9 ]

Here 3 and 1 is 4 and 2 is 6. [ 3 + 1 + 2 = 6 ]

9 won't go into 6. [ 6 ũ 9 ]

It won't go into this. [ 3012 ũ 9 ]

Here 5 and 1 is 6 and 7 is 13 and 4 is 17. [ 5 + 1 + 7 + 4 = 17 ]

9 won't go into 17. [ 17 ũ 9 ]

This is out.

9 and 1 is, now note, 9 goes into 9, so I'll cast that out.

1 and 1 is 2 and 7 is 9. [ 1 + 1 + 7 = 9 ]

9 will go into 9. [ 9 ũ 9 ]

Hence, it will go into 9,117. [ 9117 ũ 9 ]

So see if I wanted to see if 9 will go into these,

I would have been wasting my time trying them.

I can tell that without trying them.

But now I know it will go into this, so I can try.

If I want the other factor of this divided by that,

all I'd have to do is take 9,117 divide by, [9][1][1][7][ũ][9][=]

and we're claiming that this will go in evenly.

And sure enough it does.

1,013.

How about this one?

1 in 5 is 6 and 2 is 8, [ 1 + 5 + 2 = 8 ]

1 is 9 and 5 is 14. [ 1 + 5 + 2 + 1 + 5 = 14 ]

9 will not go into 14. [ 14 ũ 9 ]

It won't go into this. [ 105215 ũ 9 ]

Let's prove that.

Not proving it, illustrate it on our calculator perhaps

so 1,0, 5, 2, 1, 5 divided by 9 [1][0][5][2][1][5][ũ][9]

equals [=]

I get my decimals.

Hence, it won't.

It would have been faster to check that by hand

than it would be to even use the calculator.

So the test for divisibility of 9, add together the digits.

If their sum is divisible by 9, so is the original number.

How about by 10?

That's the fast one. Isn't it?

We just look at the units digit and ask are any of them zero?

The answer's no.

Then none of them are divisible by 10.

That's the fast one.

Probably you knew that before you even got into this course.

How about divisibility by 11?

Well 11 ranks with 7.

There is a divisibility test,

but it's complicated enough that we usually don't teach it.

In that case we simply say look, just try it and see.

So 148 divide by 11 equals, [1][4][8][ũ][1][1][=]

you get decimals. No it won't.

So one by one you simply have to try to divide these by 11

either by hand or faster by calculator,

and had you tried it you would find this would be the only one.

Let's try it.

105,215 divide by 11 equals, [1][0][5][2][1][5][ũ][1][1][=]

and there it is: 9,565.

To double-check times 11 equals [ 9,565 [x][1][1][=]

and sure enough we're right back to where we were.

But the test, divisibility test for 11 again,

is messy enough we don't teach it.

We simply say try it and see.

It takes time, but less time

than trying to remember a rather involved divisibility test.

Now you see this process can go on and on.

In fact, entire books can be written about it,

so let's finish this lesson by our last divisibility test of 12.

There are many, many others but we'll conclude here.

Remembering that 12 is made up of the prime number,

we'll talk more about that later.

3 and the number 4. [ 3 · 4 = 12 ]

We simply ask will 3 and 4 each divide?

If they will, then 12 will because 12 is made up of 3 times 4.

So 4, yes because 4 will divide into 48.

3, 1 and 4 is 5, and 8 is 13. [ 1 + 4 + 8 = 13 ]

So 4 will go in but 3 won't, so that's out.

Okay 4 will go into 56. [ 56 ũ 4 ]

3 will go into: 1 and 6 plus 6 is 12. [ 1 + 5 + 6 = 12 ]

So this [ 156 ] is divisible by 12.

Let's check it on our calculator to get a comfortable feel for that.

156 divide by 12. [1][5][6][ũ][1][2][=]

And sure enough 13 times.

Okay. 4 will go into the end two digits [ 12 ũ 4 ]

so this is divisible by 4.

3 and 1 and 2 is divisible by 3. [ 3 + 1 + 2 = 6 ]

So this [ 3012 ] is divisible by 12.

This case [ 5174 ] let's check for 3 first.

5 and 1 is 6 and 7 is 13 and 4 is 17. [ 5 + 1 + 7 + 4 = 17 ]

3 won't go into 17.

So we won't waste our time with that.

They both have to go in.

So in this case this isn't even even

so I can't get a 4, so I don't care if 3 goes in.

12 will not.

Same thing here.

If this is odd and 12 is even then for sure, 12 can't go in.

So we have two of these six [ 156, 3012 ] divisible by 12.

Practice these

and you'll see why this is important in the very next lesson.

So until then, this is your host Bob Finnell.

A Course in Arithmetic Review.

Produced at Portland Community College.

Let's say that for some reason we wanted to know

if 108 or any other number for that matter

were divisible evenly by these particular numbers.

Now one obvious way to check that

is so try to see if they divide in evenly

either by hand or by calculator.

But there are some nice tricks which if learned

simply allow you to look at this number

and answer yes or no with each ones of these.

The one for 2 you already know.

You simply look at the one's digit and ask, is it even?

In this case, [ 2 ] it is. So the answer is yes.

2 will divide evenly.

That we call a 'divisibility test'

and the divisibility test for 2 was

if the one's digit is even, it's divisible by two.

That's what this lesson is about to give you a series

of simple divisibility tests for various easy numbers,

and we've just had one already.

But there is also an equally easy test

to check divisibility by three.

A number is divisible by 3,

if the sum of its digits is divisible by 3, our number was 108.

If you were to add its digits, you would get 9. [ 1 + 0 + 8 = 9 ]

9 is divisible by 3, hence this test says 108 is divisible by three.

Check it to see. [ 108 ũ 3 = 36 ]

And it is. It goes in 36 times.

So the question "Is 108 divisible by three?"

The answer is 'yes.'

Now how about 5?

Most people know that one as well.

If the one's digit is 0 or 5 it's divisible by 5.

It isn't, so the answer to this is no.

So let's formally add that to our list of divisibility test.

The one's digit is either 0 or 5.

If the answer is yes.

Then yes, the number is divisible by 5.

To check divisibility by 4, you ask, Is the number formed

by just its tens and ones digit is divisible by 4?

Let's check just that portion alone first.

Here's our number. [ 108 ]

Let's look at that portion

formed by the ones and the tens digits [ 08 ]

which is still 8, of course.

Is that divisible by four? [ 8 ũ 4 ]

The answer's yes.

Therefore, the whole number, 108 is divisible by 4.

As a matter of fact, 4 divides into 108 27 times.

Sometimes that gets to be rather messy,

so there is an appended comment here that says this:

or if another number formed by it minus any multiple of 4.

Now remember what we mean by multiples of 4.

4, 8, 32, 40, 400.

Any number that 4 will go into evenly is a multiple of 4.

Now let's look at what this statement's really saying.

If another number formed by it minus any multiple of 4.

Now if we wanted to check to see if this is divisible by four,

our first part of our rule says

to just look at those two right there

and ask is that [ 92 ] divisible by 4?

But that's not so easy in its own right.

So let's take a number that we know is a multiple of four, 80.

That's 4 times 20. [ 4 x 20 = 80 ]

Subtract that and we get 12. [ 92 - 80 = 12 ]

Now that's a smaller number.

Now is that [ 12 ] divisible by 4? [ 12 ũ 4 ]

And the answer's yes.

That tells us the whole number [ 1392 ] is divisible by 4.

This fact in particular will help you

in many of your business courses

because by definition a leap year,

that's the year that has 366 instead of 365 days,

by definition that's any year divisible by four.

So if we needed to know ahead of time in a business ledger

whether 1996 will be a leap year,

this test tells us to look just at the end two digits [ 96 ]

Will 4 divide that? [ 96 ũ 4 ]

And if you can't see that very readily,

take a multiple of four and in this case 80 again.

Subtract from that and we get 16. [ 96 - 80 = 16 ]

Now, is that divisible by 4? [ 16 ũ 4 ]

And the answer's yes. So, yes, that's [ 1996 ] a leap year.

A very handy trick for those of you going on into business.

The similar divisibility test for 8,

if a number formed

by the right end three digits is divisible by 8,

then the whole number is divisible by 8.

This example. [ 82,134 ]

If the number formed by the end 3 digits [134] is divisible by 8,

then the entire number [ 82,134 ] is divisible by 8.

But that's not so obvious [ 134 ũ 8 ] in its own right.

So our second phase or that number

which is any multiple of 8 less than it.

So if this [ 134 ] was still too large for us to mentally handle,

take any multiple of 8.

Well 80, that's 10 times 8. [ 10 · 8 = 80 ]

Subtract that [ 134 - 80 ]

and we get 4, 8 from 13 is 5. [ 134 - 80 = 54 ]

Is that [ 54 ] divisible by 8?

Now if it's still not obvious, take another multiple of 8.

Let's take 40.

That's 5 times 8. [ 5 · 8 = 40 ]

Is that divisible by 8?

The answer is no.

Therefore, the original number [ 82,134 ] is not divisible by 8

even though it is divisible by two.

Checking just that [34] we find that this is not divisible by 4.

So the whole number will not divide by 4 or 8, but it will by 2.

So in asking about our original number 108,

we have to ask about the whole thing

because that is the last three digits.

Well probably just as fast to go ahead

try mentally to divide it by 8

but we could subtract this multiple 80 from it very quickly.

8 from 10 is 2. [ 108 - 80 = 28 ]

28 is not divisible by 8.

Therefore, 108 is not divisible by 8.

We had skipped 6.

But let's skip that just for a moment and jump down to 9.

A number is divisible by 9

if the sum of its digits is divisible by 9.

That's essentially the same test one had for 3.

So the sum of our digits here is 1 and 8 is 9. [ 1 + 0 + 8 = 9 ]

9 is divisible by 9 so, yes, 108 is divisible by 9.

Now, we still left out some.

This one is almost universally known.

If the units digit is zero, it's divisible by 10

so the answer to divisibility here is no.

Now these three we don't need a special test.

All we need to remember is that they are made up, product wise,

of numbers we've already used.

6 is nothing more than 2 times 3. [ 6 = 2 · 3 ]

And 2 goes in, [ 108 ũ 2 ] 3 goes in, [ 108 ũ 3 ]

and 2 times 3 is 6 [ 2 · 3 = 6] So 6 goes in as well.

12 is 4 times 3. [ 12 = 4 · 3 ]

4 goes in, [ 108 ũ 4 ] 3 goes in [ 108 ũ 3 ]

So 12 goes in. [ 108 ũ 12 ]

And the divisibility for the three

was a sum of the digits is divisible by 3.

The test for 4 was the end two digits is divisible by 4.

Now can you see what we're going to do with 15?

15 is 3 times 5. [ 15 = 3 · 5 ]

3 goes in, 5 does not go in, so 15 is not a divisor.

Frequently when we try to divide a number by lower numbers,

we try first of course 2, 3, 4, 5, 6 and on up to about 10 or 12.

And these tests says that those checks can generally be made

by observation and just a mental act.

One doesn't need to divide or to use a calculator.

If you learn those tests, it will save you much time.

Let's go through those tests once more

with six rather large numbers.

Let's do the test one at a time now to really nail it down.

To check the divisibility by two

we don't have to divide each of these by 2 by hand or by calculator.

All we need to do is look at each end or units digit,

and ask: is it even?

If the answer is yes and 8 is even, then that's divisible by 2.

So is that. [ 156 ]

So is that. [ 3012 ]

So is that. [ 5174 ]

That isn't. [ 9117 ]

This isn't. [ 105,215 ]

So these first four [ 148, 156, 3012, 5174 ] are divisible by two

so on these last two we wouldn't waste our time even trying.

Not even with a calculator.

And see this is much faster to determine

than to even use a calculator.

So once again to check the divisibility by 2,

you merely look at the units digit and ask: is it even?

Yes or no.

How about the same six numbers?

Are they divisible now by 3?

To check the divisibility of 3 now we must add the digits

so 1 and 4 is 5 [ 1 + 4 = 5 ]

and 8 is 13. [ 5 + 8 = 13 ]

That is not divisible by 3.

Neither is the number made up of them 148.

Checking here, one and 5 is 6 [ 1 + 5 = 6 ]

and 6 is 12. [ 6 + 6 = 12 ]

Yes, 3 will go into 12, [ 12 ũ 3 ]

so yes, 3 will divide that [ 156 ]

Same thing here. [ 3012 ]

3 and 1 is 4 [ 3 + 1 = 4 ]

and 2 is 6. [ 4 + 2 = 6 ]

3 will divide into 6, [ 6 ũ 3 ]

So yes, 3 divides into that [ 3012 ũ 3 ]

Now a somewhat faster approach to this,

as soon as we have a sum that 3 will divide into,

in a sense cast it away.

So 3 goes into 3, [ 3 ũ 3 ] so throw it away.

1 and 2, 3 goes into that, so I have my yes.

So here 1 and 5 is 6. [ 1 + 5 = 6 ]

3 goes into that.

Cast it away.

3 goes into 6 twice [ 6 ũ 3 = 2 ]

So yes, it divides into the whole thing.

So trying that trick here 5 and 1 is 6, 3 divides that.

7 and 4 is 11.

3 does not divided that, so we will not divide into 5,174.

3 goes into 9.

1 and 1 is 2 and 7 is 9, so three will divide into that.

1 and 5, 3 goes into that. [ 1 + 5 = 6 ]

2 and 1, 3 divides into that. [ 2 + 1 = 3 ]

5, 3 will not go into that. [ 5 ũ 3 ]

So 3 will not divide the whole thing [ 105215 ũ 3 ]

Did you see our approach?

I could add 1 and 5 is six, [ 1 + 5 = 6 ]

and two is eight, [ 6 + 2 = 8 ]

1 is 9 and 5 is 14. [ 1 + 5 + 2 + 1 + 5 = 14 ]

But rather than doing that,

if I get a sub sum, divisible by three,

cast it away and start over.

2 and 1 is 3. [ 2 + 1 = 3 ]

Sub sum.

3 divides it cast it away. [ 3 ũ 3 ]

5, 3 won't go into that. [ 5 ũ 3 ]

It won't go into the whole expression.

So 3 will go into these three: [ 156, 3012, 9117 ]

but not these three [ 148, 5174, 105,215 ]

Same numbers again.

Now are each of them divisible by four?

Our test for that was to look at the end two digits alone.

4 will go into 48. [ 48 ũ 4 ]

So will go into 148. [ 148 ũ 4 ]

Will 4 go into 56? [ 56 ũ 4 ]

If that's not quite so obvious,

then subtract a multiple of 4

which in this case is 40, [ 56 - 40 = 16 ] which gives me 16.

4 will go into 16. [ 16 ũ 4 ]

Therefore, it will go into 156. [ 156 ũ 4 ]

Again look at the end two digits [ 12 ]

4 will go into 12. [ 12 ũ 4 ]

It will go into here. [ 3012 ũ 4 ]

Again, the end 2 digits [ 74 ]

If it's not obvious subtract a multiple of 4.

40 again in this case. [ 74 - 40 = 34 ]

4 will not divide into 34. [ 34 ũ 4 ]

Hence it will not divide into the entire number. [ 5174 ũ 4 ]

Now you can see 4 won't go into these end two digits [ 17 ]

because it's not even even so for sure if 2 won't go into it,

4 can't because 4 is made up of two 2s factor wise. [ 2 · 2 = 4 ]

And in the same sense,

I would eliminate that because this isn't even an even number.

So the first three are divisible by 4. [ 148, 156, 3012 ]

Again the divisibility test is to check the end two digits

ask: Will 4 go into them?

If it's a little bit difficult to see that

then subtract some convenient multiple of 4

and take the smaller difference and check it by 4.

Again, same six numbers.

Are they divisible by 5?

Well, this is one of our easier tests.

We simply look at the end digit and check to see if it's 5 or 0.

And if it is not, then it's not divisible by 5.

Don't waste your time.

But this [ 105215 ] does end in 5 or 0.

So this is the only number of the six that's divisible by 5.

Hence if I needed to, I would go ahead and divide it,

but I wouldn't even waste my time dividing these

because they're not even possible.

How about divisibility by 6?

We didn't have a special test for that, although one does exist.

But because 6 factor wise

is really made up of the factors 2 and 3, [ 2 · 3 = 6 ]

we can check each of these tests singularly,

and if it's divisible by 2 and 3,

2 times 3 is 6 [ 2 · 3 = 6 ] hence is divisible by 6.

So first our divisibility test of 2 passes here [ 148 ũ 2 ]

Then our divisibility of 3,

1 and 4 is 5 and 8 is 13, [ 1 + 4 + 8 = 13 ]

so 2 divides this [ 148 ] but 3 won't.

Hence, this is not divisible by 6 because I must have both of these.

Two divides this.

1 and 5 and 6 is 12. [ 1 + 5 + 6 = 12 ]

3 goes into that. [ 12 ũ 3 ]

So 2 goes into this. [ 156 ũ 2 ]

3 goes into this. [ 156 ũ 3 ]

So this is divisible by 2 and 3 which is 6. [ 156 ũ 6 ]

Again the 2 is obvious. [ 3012 ũ 2 ]

3 and 1 is 4 and 2 is 6 [ 3 + 1 + 2 = 6 ]

so 2 goes in, 3 goes in, 6 goes in.

Again we see that the two goes in [ 5174 ]

so 5 and 1 is 6 and 7 is 13 and 4 is 17, [ 5 + 1 + 7 + 4 = 17 ]

which is not divisible by three 3, [ 17 ũ 3 ] so this is out.

I need both 2 and 3.

2 won't go into that. [ 9117 ]

6 won't.

2 won't go into that. [ 105215 ]

6 won't.

So, for 6 we check 2 and 3 divisibility test each separately.

How about by 7?

You will note that we didn't have a divisibility test for 7.

One does exist by the way.

But it is so involved

that most courses of this type don't give it to you.

It's faster to go ahead and divide it

than it is to try to remember the rule.

However, your teacher might want to give it to you

on her or his own.

It's rather an interesting test.

But in this case we simply take on a calculator or by hand.

Calculator is obviously faster.

148 divide by 7, equals. [1][4][8][ũ][7][=]

If I get decimals,

then that tells me this is not evenly divisible by 7.

So the next one 156 divide by 7 equals [1][5][6][ũ][7][=]

Decimals.

That's out.

The next one.

3012 divide by 7, equals. [3][0][1][2][ũ][7][=]

Decimals.

That's out.

5174 divide by 7 equals. [5][1][7][4][ũ][7][=]

Decimals.

That's out, so we're not having much luck with 7s are we?

9117 divide by 7 equals. [9][1][1][7][ũ][7][=]

Decimals.

That's out.

Just one more to try.

105,215 divide by 7 equals. [1][0][5][2][1][5][ũ][7][=]

Decimals.

That's out.

So none of these were divisible by 7.

So, with a 7 just try it.

It still takes a bit of effort,

but that's as fast as any method at this point.

How about by 8?

Our test for that was to test the end three digits,

which in this case [148] and this case [156] is the entire thing.

And ask is that the divisible by 8?

If it's messy then take a multiple of 8.

Maybe 40, maybe 80, maybe 120.

These are all multiples of 8.

Subtract one of those from that and take the difference

and see if 8 will go into that.

So let's get one as close to this as possible.

That's 120, so if we

subtract 120 from this, [ 148 - 120 = 28 ] we'll get 28.

And 8 won't go into 28. [ 28 ũ 8 ]

So it won't go into this one.

Again let's take a multiple close to this.

Now don't be tempted that because 8 goes into 56

it should go into this.

It must go into all three of these.

156.

So let's subtract 120, [ 156 - 120 = 36 ] which gives me 36.

8 will not divide 36, [ 36 ũ 8 ]

so it will not divide the whole number. [ 156 ũ 8 ]

Now here the last 3 digits only make up the number 12.

And we can see that 8 won't go into 12.

Hence, it won't go into this one, either.

On this one [ 5174 ] the end three digits. [ 174 ]

Let's take a multiple that's close to this.

In that case, that would be 160 if you see it.

See, that's 20 times 8, [ 20 · 8 = 160 ]

and that gives me 14. [ 174 - 160 = 14 ]

8 will not divide into 14. [ 14 ũ 8 ]

8 won't go into this. [ 5174 ]

Can't go into here [ 9117 ] because it's even, not even,

and 8 is an even divisor.

Same thing here [105215 ]

Not even.

8 cannot go into it, so 8 will not divide into any of these.

So the test for 8 is: will the number

made up of the last three digits divide by 8?

And specifically, if it's not even you throw it out to begin with.

And how about divisibility by 9?

Recall that test was add up the sum of all digits.

So 1 and 4 is 5 and 8 is 13 [ 1 + 4 + 8 = 13 ]

and ask will 9 divide into this sum, 13? [ 13 ũ 9 ]

If the answer's no then it won't divide into the number 148.

Here 1 and 5 is 6 and 6 is 12. [ 1 + 5 + 6 = 12 ]

9 won't go into 12. [ 12 ũ 9 ]

It won't go into that. [ 156 ũ 9 ]

Here 3 and 1 is 4 and 2 is 6. [ 3 + 1 + 2 = 6 ]

9 won't go into 6. [ 6 ũ 9 ]

It won't go into this. [ 3012 ũ 9 ]

Here 5 and 1 is 6 and 7 is 13 and 4 is 17. [ 5 + 1 + 7 + 4 = 17 ]

9 won't go into 17. [ 17 ũ 9 ]

This is out.

9 and 1 is, now note, 9 goes into 9, so I'll cast that out.

1 and 1 is 2 and 7 is 9. [ 1 + 1 + 7 = 9 ]

9 will go into 9. [ 9 ũ 9 ]

Hence, it will go into 9,117. [ 9117 ũ 9 ]

So see if I wanted to see if 9 will go into these,

I would have been wasting my time trying them.

I can tell that without trying them.

But now I know it will go into this, so I can try.

If I want the other factor of this divided by that,

all I'd have to do is take 9,117 divide by, [9][1][1][7][ũ][9][=]

and we're claiming that this will go in evenly.

And sure enough it does.

1,013.

How about this one?

1 in 5 is 6 and 2 is 8, [ 1 + 5 + 2 = 8 ]

1 is 9 and 5 is 14. [ 1 + 5 + 2 + 1 + 5 = 14 ]

9 will not go into 14. [ 14 ũ 9 ]

It won't go into this. [ 105215 ũ 9 ]

Let's prove that.

Not proving it, illustrate it on our calculator perhaps

so 1,0, 5, 2, 1, 5 divided by 9 [1][0][5][2][1][5][ũ][9]

equals [=]

I get my decimals.

Hence, it won't.

It would have been faster to check that by hand

than it would be to even use the calculator.

So the test for divisibility of 9, add together the digits.

If their sum is divisible by 9, so is the original number.

How about by 10?

That's the fast one. Isn't it?

We just look at the units digit and ask are any of them zero?

The answer's no.

Then none of them are divisible by 10.

That's the fast one.

Probably you knew that before you even got into this course.

How about divisibility by 11?

Well 11 ranks with 7.

There is a divisibility test,

but it's complicated enough that we usually don't teach it.

In that case we simply say look, just try it and see.

So 148 divide by 11 equals, [1][4][8][ũ][1][1][=]

you get decimals. No it won't.

So one by one you simply have to try to divide these by 11

either by hand or faster by calculator,

and had you tried it you would find this would be the only one.

Let's try it.

105,215 divide by 11 equals, [1][0][5][2][1][5][ũ][1][1][=]

and there it is: 9,565.

To double-check times 11 equals [ 9,565 [x][1][1][=]

and sure enough we're right back to where we were.

But the test, divisibility test for 11 again,

is messy enough we don't teach it.

We simply say try it and see.

It takes time, but less time

than trying to remember a rather involved divisibility test.

Now you see this process can go on and on.

In fact, entire books can be written about it,

so let's finish this lesson by our last divisibility test of 12.

There are many, many others but we'll conclude here.

Remembering that 12 is made up of the prime number,

we'll talk more about that later.

3 and the number 4. [ 3 · 4 = 12 ]

We simply ask will 3 and 4 each divide?

If they will, then 12 will because 12 is made up of 3 times 4.

So 4, yes because 4 will divide into 48.

3, 1 and 4 is 5, and 8 is 13. [ 1 + 4 + 8 = 13 ]

So 4 will go in but 3 won't, so that's out.

Okay 4 will go into 56. [ 56 ũ 4 ]

3 will go into: 1 and 6 plus 6 is 12. [ 1 + 5 + 6 = 12 ]

So this [ 156 ] is divisible by 12.

Let's check it on our calculator to get a comfortable feel for that.

156 divide by 12. [1][5][6][ũ][1][2][=]

And sure enough 13 times.

Okay. 4 will go into the end two digits [ 12 ũ 4 ]

so this is divisible by 4.

3 and 1 and 2 is divisible by 3. [ 3 + 1 + 2 = 6 ]

So this [ 3012 ] is divisible by 12.

This case [ 5174 ] let's check for 3 first.

5 and 1 is 6 and 7 is 13 and 4 is 17. [ 5 + 1 + 7 + 4 = 17 ]

3 won't go into 17.

So we won't waste our time with that.

They both have to go in.

So in this case this isn't even even

so I can't get a 4, so I don't care if 3 goes in.

12 will not.

Same thing here.

If this is odd and 12 is even then for sure, 12 can't go in.

So we have two of these six [ 156, 3012 ] divisible by 12.

Practice these

and you'll see why this is important in the very next lesson.

So until then, this is your host Bob Finnell.