Integer sums

Adding sums of consecutive integers

Welcome to the presentation
on finding sums of integers.

You're probably wondering why
are we doing this within

the context of averages.

Well, if you think about it,
all an average is is you take

a sum of a bunch of numbers
and then you divide by the

number of numbers you have.

What we're going to do here is
do a couple of algebra problems

that involve just the sum parts
first, and actually they can

carry over into average
problems as well.

Let's get started
with a problem.

Let's say I told you that I had
the sum of five consecutive

integers is equal to 200.

What is the smallest -- I
apologize for my handwriting

-- what is the smallest
of the five integers?

Well there's a couple of ways
to do this, but I guess the

most straightforward way is
just to do it algebraically,

I would say.

So let's say that x is the
smallest of the integers,

right, so x is actually
what we're going to want to figure out.

Well if x is the smallest, what
are the other four going to be?

We have a total of five.

Well, they're consecutive.

Consecutive just means that
they follow each other,

like 5, 6, 7, 8, 9, 10.

All of those are consecutive
integers, right?

And if you remember, integers
are just whole numbers, so it

can't be a fraction
or a decimal.

So if x is the smallest, so
then the next integer is

going to be x plus 1.

And the one after that's
going to be x plus 2.

And the one after that's
going to be x plus 3.

And the one after that's
going to be x plus 4, right?

It might seem confusing I'm
writing all of these x's.

But if you think about it, if x
was 5, then this would be 6,

this would be 7, this would
be 8, and this would be 9.

And that's all I'm
writing here, right?

So these would be, assuming
that x is the smallest of the

integers, the five integers
would be x, x+1, x+2, x+3, and x+4.

And we know that the sum
of these five consecutive integers is 200.

What is the sum of these
five, I guess we could say, numbers or expressions?

Well let's see, we have
five x's -- 1, 2, 3, 4, 5.

So x plus x plus x plus x
plus x is equal to just 5x.

Or you could just
say 5 times x.

And then that's plus 1
plus 2 is 3, 3 plus 3

is 6, 6 plus 4 is 10.

So the sum of these five
integers is going to be 5x plus

10, and all I did is add up the
x's and added up the constants.

And we know that that
is going to equal 200.

Now this is just a level
two linear equation.

We can just solve for x.

So we get 5x is equal to 190
-- I just subtracted 10 from both sides, right?

And then x is equal to --
let me divide 5 into 190.

5 goes into 19 three
times, 3 times 5 is 15.

9 minus 5 is 4,
bring down the 0.

5 goes into 40, eight times.

So x is equal to 38.

Pretty straightforward
problem, don't you think?

Now what if I were to ask you
what is the average of the five consecutive numbers?

Well now, there's two
ways of doing this.

Now that we already know that x
is 38, we know that the other

numbers are going to be -- well
this is 38, 39, 40, 41, 42.

Well we could just average
these four numbers.

You could just say 38 plus
39 plus 40 plus 41 plus 42.

And well we already know
what those -- I don't

even have to do the math.

You already know that they
average up, they sum up to 200

and then we divide the sum by
5, because there are 5 numbers.

So the average is 40.

There are a couple ways you
could think about that.

One, you see 40's just a middle
number so that makes sense.

And the only time we can really
say it's the middle number

is when the numbers are
distributed evenly around 40.

If we had a number that was
much smaller than 40 or

something, you couldn't
just necessarily pick the middle number.

But in this case these are
consecutive and makes sense.

Another way we could have done
this problem, if you were, say,

taking the SAT and they were to
ask you the sum of five

numbers is 200, what's the
average of the numbers?

Well you say, well, all I have
to do is divide that 200

by 5 and I'll get 40.

Let's do another problem
and I'll make it a

little bit harder.

Let's say the sum of seven odd
numbers, and let me make up a

good -- I hope this one works,
I'm going to try to do it in

my head -- is 217, what
is the largest number?

I shouldn't say number
-- seven odd integers.

Actually it becomes a much
harder problem if it was just

seven odd -- well actually, the
only thing that could be odd

are integers anyway, so you
could almost assume it.

But the sum of seven
odd integers is 217.

What is the largest
of the integers?

As you can tell I'm
doing this on the fly.

Actually my wife just diagnosed
me with, she thinks I

have benign vertigo.

I got very dizzy this morning
when I went to work, so you

have to forgive me
for that as well.

That's impairing me even more.

So let's do this problem.

Let's say that x
is the largest.

Then what would the
number right below x be?

Would it be x minus 1?

Well, if x is an odd
number, x minus 1 would

be an even number.

So in order to get the number
right below it, we have to

do x minus 2 to get
another odd number.

My apologies -- it should
say the sum of seven

consecutive odd.

I don't know if
you assumed that.

I'm trying my best
today to confuse you.

The sum of seven consecutive
odd integers is 217.

What is the largest
of the integers?

So if x is the largest, then to
next smallest one would be x

minus 2, right, because it's
consecutive odd numbers,

not just consecutive.

So consecutive odd numbers are
like 1, 3, 5, 7 -- you're

skipping the evens, right?

So that's why you're going up
or down by two, depending

how you view it.

So the next one down would be
x minus 2, then we'll have x

minus 4, x minus 6, x minus
8, x minus 10, x minus 12.

I think that's it.

One, two, three, four,
five, six, seven, right.

Those are seven numbers.

They're separated by two.

X is the largest
of them, right?

We can assume that they're
odd because apparently

the problem will work
out so that they're odd.

So what is the sum of
these seven numbers?

Well the seven x's
just add up to 7x.

And then let's see, 2 and 4 is
6, 6 and 6 is 12, 12 and 8 is

20, 20 and 10 is 30,
30 and 12 is 32.

So 7x minus 32 is equal to 217.

We just solved for x.

7x is equal to -- let's see,
if we add 32 to both sides

of this equation we get 249.

Let's see, 7 goes into
249 -- is that right?

Right.

So 7 goes into 249 -- did I
do this addition properly?

I want to make sure.

2 plus 4 is 6, 6 plus 6 is 12,
12 plus 8 is 20, 20 plus 10

is 30, 30 plus 12 is 42.

Oh, here you go.

See, my mathematical spider
sense could tell that something

was fishy about this.

So that's 7x minus 42.

So if we add 42 to both sides
it's 7x is equal to 259.

See how brave I am, I do
this thing in real time.

259.

So 7 goes into 259 -- let's
see, 7 goes into 25 three

times, 3 times 7 is 21, 49 --
it goes into it 37 times.

So we get x is equal
to 37 and we're done.

So just to review because I
think had a lot of errors in

this problem when
I presented it.

The question was the sum
of seven consecutive odd integers is 217.

What is the largest
of the integers?

I said x is the largest, and
then if x is the largest, the

next smaller one
will x minus 2.

Because we're not saying
just consecutive integers,

we're saying consecutive
odd integers, right?

So if x is 37, which is what we
solved for, then x minus 2 is

35, this is 33, this is 31,
this is 29, this is

27, this is 25.

And then we just added up all
the x's and I'll add up all

the constants and said,
well they add up to 217.

And then we just solved for x.

I think you're now ready to
try some of these problems.

Have fun.