<p>Tell me how you can derive this then from foiling? I want to know. I have been trying to figure out WHY it works for a while. I am going to keep going at it till I get it. It is not derived from foil though. THere are no two numbers that are like (a-b)^2.
Look people
Foil would be for 116 (120-4)^2
Then 14400 - 120 (4) (2) + 16
Now look at mine
116^2
12136 + 1320 = the same thing as foil with less work. It cannot be gotten from foil because you do not ADD A^2 and B^2. And its easier because there is one less calculation. The only thing I am wondering is why it works.. its still confusing me.
And momofchris is stil not listening to me
No point in even responding to someone who is completely oblivious to me.</p>
<p>NM momofchris removed her post.</p>
<p>116^2 = (10<em>11 + 6)^2 = 100</em>11^2 + 6^2 + 10<em>2</em>11<em>6 = 100</em>121 + 36 + 10*132 = 12136 + 1320
that is why it works, gyros.</p>
<p>you're still wondering? post #4</p>
<p>and post #23, and post #24</p>
<p>that should keep you busy for a while.</p>
<p>River could you explain that in letters? I understood it that way in numbers but trying to just find it in letters though.. and post 4 does not answer my question.</p>
<p><a href="http://en.wikipedia.org/wiki/Foil_rule%5B/url%5D">http://en.wikipedia.org/wiki/Foil_rule</a></p>
<p>Sigh I guess its useless trying to explain it to you people. And why did you delete the wikipedia post before mom and repost it now?</p>
<p>Oh ya anyhow I think I liked it because I tied it into my application on how it helped certain EC's and I linked it with my essay (talked about clubs that require very fast mental math so thats how i linked it).</p>
<p>gyros, post 4 does explain it</p>
<p>(AB)^2 = (10A + B)^2 = 100A^2 + B^2 + 2<em>10</em>A*B
what you've noticed is that you can just place A^2 in front of B^2 in the sum if the latter has two digits</p>
<p>But I dont get it...that is just foil..... mine is not foil. there is no (A + B)^2 thing anywhere in my problem.. i am not trying to sound bad here but re-read my post
(AB)^2 = A^2B^2 (just place them next to each other. this is not foil) + 2ab 0 (add 0 after multiply those 3).
It looks like foil but it is not. There is no 3rd term
For instance
89^2, A = 8 B= 9
6481 + 1440. This is much easier than foil and its not foil at all. I think you guys are not getting what i am saying. And my thing does work for three digit and four digit numbers too.
SORRY i just read your post now.</p>
<p>Oh well i am getting it by what you are saying. However its still a bit cloudy. I just want this to end now before I create too much animosity between me and the MIT board (is there even any?). ANyhow now time to study for SAT's ugh... oh well. 30 days till the test !!</p>
<p>I think it's cool that you made up a formula. Just yesterday i thought i have found a new formula for vapor pressure, but then i read it in the textbook....</p>
<p>Well its not new. Everyone in this topic knows about it (most do i mean...). Man and I used it as a hook..</p>
<p>hey, i used to think that if one travels at close to light speed and shoot a light ray, then the light ray will exceed light speed. But turns out it's a counter example used in text book.......</p>
<p>
[quote]
there is no (A + B)^2 thing anywhere in my problem
[/quote]
Yes, there is. Think about it!
Let's take your 128^2 thing as an example. What you are doing basically is just (120 + 8)^2. You really should be able to see that. 120^2 = (12*10)^2 = 144 * 100 = 14400, and that's why you can just append the 64 to 144, because 0+64 = 64.
Anyway, you <em>are</em> using (a + b)^2. And please stop being that aggressive; nobody here wants to make you down. Maybe you just seemed a bit too self-sure in your first posts...</p>
<p><a href="AB">quote</a>^2 = A^2B^2 (just place them next to each other. this is not foil) + 2ab 0 (add 0 after multiply those 3).
[/quote]
</p>
<p>Not to criticize you, but you're obviously not one of those "genius" math students.</p>
<p>Your statement "just place them next to each other" is equivalent to saying 100times the first number + the second number in mathematical terms.
(this is precisely why you have to add a 0 before the second term if it is a single digit square, because 100 times the first will give you a zero in the middle)</p>
<p>And the second part of your "formula," 2 times the product of the number and then place a 0 at the end is equivalent to saying 20 times the product.</p>
<p>It is a method that is simply derived from FOIL. I understand that you are not using FOIL exactly as taught in school when you use the method, but it is still the same principle.</p>
<p>I'm just trying to help you out here; people at MIT would definitely not consider this anything special.</p>
<p>There are kids who prove hypothesis that people have been trying for years, and kids who have certain methods named after them at IMO.</p>
<p>You used this as a "hook." I visited MIT for the last 3 days and talked with many of the professors and students (I went for a conference). My impression is that MIT is looking for students take big risks, try new things, and can bounce back when they are shot down. The first thing I realized when I visited is that almost nothing in my life that I have done / learned can compare to what I will in college. This "hook" concept is the result of misinterpreting different patterns of people who have gotten in.</p>
<p>"My impression is that MIT is looking for students take big risks, try new things, and can bounce back when they are shot down."</p>
<p>This in my opinion is an excellent perspective. Bottom line is almost all applicants are intelligent, hardworking and qualified. The concept that a single hook no matter how good will get you into MIT is using flawed logic. In regards to the "new solution" issue, a superior applicant will investigate such possibilities but when confronted with overwhelming data would admit their shortcomings.</p>
<p>Gyros, this was just embarrassing..</p>
<p>From what I've seen of you (and in terms of posts on CC, I've seen a lot),
you get yourself stuck into corners and into denial ALL the time.
Please, work on it?</p>