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 plus 1, x plus
2, x plus 3, and x plus 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.