<p>Just went through a math section and got my first perfect score on a math section! Anyways, this problem took a while to do. I ended up adding up all of the odd positive integers and all of the even positive integers in my calculator. Isn't there a faster way to do this?</p>
<p>
[quote]
The sum of the positive odd integers less than 100 is
subtracted from the sum of the positive even integers
less than or equal to 100. What is the resulting
difference?
[/quote]
</p>
<p>hmmm, first I did what they asked but just went up to 10.</p>
<p>1 + 3 + 5 + 7 + 9 = 25
2 + 4 + 6 + 8 + 10 = 30</p>
<p>then I noticed that each even number will always be one more than the odd one corresponding to it. (like the first term of evens (2) is one greater than the first term of odds (1) and so on…)</p>
<p>so it’d look like</p>
<p>2 + 4 + 6 + 8 …
<p>for each pair its 1 and there’s 50 pairs in 100 so the answer is 50?</p>
<p>Yeah, I think it’s 50. Each group of ten results in an additional difference (pun!) of 5. There are ten groups of ten, so 5*10 = 50.</p>
<p>The answer is 50. </p>
<p>Look at what happens to
1,3,5 = total 9
2,4,6 = total 12
All the even numbers are 1 more than the odd numbers. Since there are 3 numbers, the total is simply 3. </p>
<p>Now, how many odd numbers are there between 1-99? The answer is 50. That is the same number of even between 2-100. </p>
<p>So the answer is 50 x 1 = 50. </p>
<p>Another way is to calculate the sums via the following operations:
[(1+99)/2]x50=2500
[(2+100)/2]x50=2550 </p>
<p>2550-2500 = 50</p>
<p>^ I bow down. :)</p>
<p>Thanks guys :)</p>
<p>2-1=1
4-3=1
6-5=1
…
100-99=1</p>
<p>50 equations. 50 x 1 = 50.</p>