# the sum of the positive odd integers less than 100 is subtracted from the sum of positive even integers less or equal to 100 what is the resulting difference ? Sorry... (U__U) I'm very bad at...

the sum of the positive odd integers less than 100 is subtracted from the sum of positive even integers less or equal to 100 what is the resulting difference ?

Sorry... (U__U) I'm very bad at word equations, can I also have some tips to make solving these equations easier ? Thank you

*print*Print*list*Cite

The typical way to do these type of problems is what kspcr did. A similar method:

Consider the odd numbers: 1, 3, 5, 7, 9, 11, . . ., 97, 99

Start pairing them off, as in:

1+99

3+97

5+95

7+93

and so on

You would have 25 pairs of numbers that all add to 100. Add all the pairs up, you get 2500.

Consider the same exact steps with the even numbers:

2+100

4+98

6+96

8+94

and so on.

You would get 25 pairs of numbers totaling 102. Adding those up, we get 2550. Subtracting the evens from the odds:

2550 - 2500 = 50

As a matter of fact, it's been said that Gauss, a famous mathematician from years ago, was punished by his math teacher in elementary school for misbehavior. For his punishment, his teacher gave his the work, "Add all the numbers up from 1 to 10", thinking Gauss would add them all up individually, taking about 30 minutes to an hour. What Gauss did was something similar to what we did above and got done in about 3 minutes, leaving for the day.

The reason to do word problems is because they are applications of the number problems. There's a number problem in there; you just need to get it out.

Why ones like these just thrown out there for students to solve? It would get you using all your math knowledge. For instance, if you are covering quadratics in school, doing number problems with quadratics, then you get to the word problems, odds are the word problems are going to involve quadratics.

But, in real life, you don't always know what problems you are going to come up against. So, you need to be able to call on all of your knowledge. And, what did we use here? Finding a pattern, addition, and subtraction, all basic math skills.

What does it have to do with everyday life? First, I will say I agree on the job odds are you will never get a problem like this. However, let me ask this? Football players lift weights. Let's consider the bench press, where the football players lie on the back and lift weights upward with their arms. Why do this? Nowhere in a football game does their job call on lying on their back and lift metal weights upward. So why do bench press? Because it makes them stronger. And if they get stronger, they become better football players.

That's essentially the same thing going on in math classes. Yes, you may not get the exact same type of problems in real life that you get in math classes, exactly like the bench press isn't the exact same movement they do in the games. However, you will get problems in real life that you have to solve, just like the football players will get in situations they need to use their strength. And just like how football players become better when they bench press, all people become better problem solvers "in real life" when they can become better problems solvers "in math class". Math classes work the brain for real life problems just like lifting weights work football players for their games.

We are asked to find the difference between the sum of the positive even integers less than or equal to 100 and the sum of the positive odd integers less than 100.

(1) Can you solve a simpler problem? Is there a pattern?

2-1=1 n=2

(4+2)-(3+1)=2 n=4

(6+4+2)-(5+3+1)=3 n=6

(8+6+4+2)-(7+ ... +1)=4 n=8

It appears that the answer will be n/2. You might try to see a reason for this:

Pairing terms in the last example yields (8-7)+(6-5)+(4-3)+(2-1)=4.

For the stated problem, n=100 so the difference should be 50.

(2) Can you solve a more general problem? For an even n:

2+4+6+ ... +n is an arithmetic series as is 1+3+5+ ... +n-1. The sum of an arithmetic series is `S=n(a_1+a_n)/2`

So the difference between the series is:

`1/2n((2+n)/2)-1/2n((1+n-1)/2)=1/4n^2+1/2n-1/4n^2=1/2n`

Note that there are n/2 terms in each series.

So in the specific case of the problem, n=100 so the answer is 50.

Another way to solve this is to realize that these are arithmetic series. (A series that increases by the same amount every time.) The formula for the sum of an arithmetic series is `(n/2)*(a_1 + a_n)`

In English, this means we take the average of the first and last term of the series, and multiply it by how many things we have in the series.

For the odds: `(50/2)*(1 + 99) = 2500`

For the evens: `(50/2)*(2 + 100) = 2550`

The difference between the sum of the evens and the odds is 50!

I really dislike word problems, as well. I don't get the point of them! Application of learned techniques? Okay... but I don't really see how it is application.

This problem was written to mess with you. What the question **wants** you to think and do is to waste time. Taking a first glance at this problem, you know it'll take you a LONG time to finish it unless you find some sort of pattern.

In order to find a pattern, we can take the 100 and simplify it to 10. If we do this, *maybe* we might be able to see some sort of trend.

So, using 10 instead of 100:

Lets take consecutive odd and even pairs.

So, we can say the numbers 1 and 2, for instance. 1 is odd, and 2 is even. Since the question says odd numbers SUBTRACTED from even numbers, that means with these two numbers, the equation would be even number - odd number, or 2-1. We know that the difference of those two numbers is 1.

Next, we can take 3 and 4. Why not? So, 4-3 ALSO equals 1!

Hmm. Both of the differences of these equal 1. Lets take another two numbers; one even and one odd. Let's make it a little different; lets take 9 and 10 as those consecutive integers. 10-9. That equals 1, as well!

We can see that each of this odd-even number pair has a difference of 1. Let's convert the problem back to where it was in the beginning. LESS THAN 100.

All that this question is really asking is to find the sum of all the even-odd number pairs there are under 100. I really dislike the wording they use in this problem. It makes it a ton more confusing.

So, under 100, there are 50 odd numbers and 50 even numbers. Since each even-odd pair includes both even AND odd, you would count 1 for every 1 odd and 1 even. So, since there are 50 odd numbers and 50 even numbers, there are 50 pairs. Since there are 50 pairs, and each pair equals 1 (remember the even-odd pattern we did earlier), then the answer is 50 * 1 = **50!**

Hope I helped!

This is simple

the sum of the positive odd integers less than 100

= **(1 + 3 + 5 + 7 + … + 99)**

the sum of positive even integers less or equal to 100

= **(2 + 4 + 6 + 8 + … + 100)**

**To find the difference between **

** = (sum of even integers) - (sum of odd integers)**

= (2 + 4 + 6 + 8 + … + 100) - (1 + 3 + 5 + 7 + … + 99)

= **(2 - 1) + (4 - 3) + (6 - 5) + (8 - 7) + … + (100 - 99) **

**= 1 + 1 + 1 + 1 + … + 1**

**There are 50 (even - odd) combinations so ,**

**= 1 + 1 + 1 + 1 + … + 1 = 50**

**so the resulting difference = 50 **

**:)**