#1
Ok, this is really confusing the **** out of me and my girlfriend (who I'm trying to help with this so she doesnt get a zero on it)

Ok this is one of the calqulater tricks.

1: Enter 8 digit or less number into calc
2:double
3: add 15
4: triple
5: add 33
6: devide by 6
7: subtract origional number

Always ends up 13.

Well my girlfriend needs a diffrent one of those thats 5 steps and ends up 13. I cant figure it out and she cant, so I turn to the pit. HELP US!
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#2
this can be solved with google and wiki surely. i don't know any off the top of my head.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
#3
Sorry man, Im terrible at math. I agree with looking it up on google. Or like me, make something completly up.
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#4
Instead of add 33, make it add 66, and the next step, subtract 33.

Smartassery ftw!!!
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#5
I'm sure it can be solved algebraically, but I'm really not in the mood to try and figure it out, sorry. Here's a headstart:

For x<100,000,000:

[((2x+15)3 + 33)/6] - x = 13

So you've just got to try some other numbers there that will still make it equal to 13. I can see right now that you can factor a 3 out of the top of the first fraction part and get:

[((2x+15)+11)/2] -x = 13

Maybe that's the answer?

edit: Wait! I got it!
That simplifies to:
[(2x + 26)/2] - x = 13

So the answer is:
1: Enter 8 digit or less number into calculator
2: Double
3: add 26
4: divide by 2
5: subtract original number

edit2: So yeah, it works because the x's eventually subtract out, like so:
[(2x + 26)/2] - x = 13
Factor out the 2 on top: [2(x+13)/2] - x = 13
Cancel the 2's: x + 13 - x = 13
Subtract x's: 13=13
True.

In case your teacher asks.
Last edited by rockon1824 at Nov 19, 2007,
#6
i think he/she is just asking for other functions that, when applied consecutively, always give the same number. not why that particular one works.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
#7
Wait...I tried 8675309 (Jenny's phone #) and I got 513...so...no?
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#8
Quote by rockon1824
I'm sure it can be solved algebraically, but I'm really not in the mood to try and figure it out, sorry. Here's a headstart:

For x<100,000,000:

[((2x+15)3 + 33)/6] - x = 13

So you've just got to try some other numbers there that will still make it equal to 13. I can see right now that you can factor a 3 out of the top of the first fraction part and get:

[((2x+15)+11)/2] -x = 13

Maybe that's the answer?

edit: Wait! I got it!
That simplifies to:
[(2x + 26)/2] - x = 13

So the answer is:
1: Enter 8 digit or less number into calculator
2: Double
3: add 26
4: divide by 2
5: subtract original number


I LOVE YOU!! And so does my girlfriend! Threesom!!!
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#10
Quote by rockon1824
It's like math prostitution. I feel dirty all over.

You're welcome. You have no idea how smart solving that made me feel.


I dont blame you. It confused the **** out of me, and I'm past that class.

And math prostitution FTW. Thanks.
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#11
Quote by rockon1824
It's like math prostitution. I feel dirty all over.

You're welcome. You have no idea how smart solving that made me feel.


you factored out things, but where did those factors go?
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
#13
I'm sure it can be solved algebraically, but I'm really not in the mood to try and figure it out, sorry. Here's a headstart:

For x<100,000,000:

[((2x+15)3 + 33)/6] - x = 13

So you've just got to try some other numbers there that will still make it equal to 13. I can see right now that you can factor a 3 out of the top of the first fraction part and get:

[((2x+15)+11)/2] -x = 13

Maybe that's the answer?

edit: Wait! I got it!
That simplifies to:
[(2x + 26)/2] - x = 13

So the answer is:
1: Enter 8 digit or less number into calculator
2: Double
3: add 26
4: divide by 2
5: subtract original number


Wouldn't it be:
(2x+15)/2 + (11/6) - x = 13?
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#14
sorry its like 12 40 am here... not good for maths. anyway, i still maintain that's not what he was looking for rockon...
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
#15
Quote by americablanco
Wouldn't it be:
(2x+15)/2 + (11/6) - x = 13?

..no.

Look:

(((2x+15)3 + 33)/6) - x = 13

Factor out the 3 in the bolded stuff:
3[(2x+15) + 11] all over 6

Then the 3 cancels out with the six to leave a 2 on the bottom, under the stuff in brackets.

Then you have (2x + 15) + 11 on top. Since nothing else can be done in the parentheses, you can just add the 15 and the 11, leaving you with:

[(2x + 26)/2] - x =13

Quote by Sol9989
sorry its like 12 40 am here... not good for maths. anyway, i still maintain that's not what he was looking for rockon...

He was just looking for the steps...I just felt like explaining all the algebra behind it because I have no life and don't feel like doing my AP Calc homework at the moment...Math for math I guess.
#16
Rockon, you are a genius. And I mean it



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#17
..no.


yes, b/c after (2x+15)/2 + (11/2) - x

you can split the (2x+15)/2 into ---> [(2x)/2] + (15/2) + (11/2) - x

which would simplify to 13 after cancelling the x's and adding 15+11 then dividing by two

P.S.
Yes, I spelled out '2'


**Correction**
Split the (2x+15)/2 into ---> [(2x)/x] + (15/2) ONLY, the (11/2) - x is the rest of the problem.
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Last edited by americablanco at Nov 19, 2007,
#18
Quote by americablanco
yes, b/c after (2x+15)/2 + (11/2) - x

you can split the (2x+15)/2 into ---> [(2x)/2] + (15/2) + (11/2) - x

which would simplify to 13 after cancelling the x's and adding 15+11 then dividing by two

P.S.
Yes, I spelled out '2'


**Correction**
Split the (2x+15)/2 into ---> [(2x)/x] + (15/2) ONLY, the (11/2) - x is the rest of the problem.

What? Where are you getting that x from, the denominator's a two. As for the rest of your post, it's the exact same thing I did except that you split the fraction into two terms, which doesn't change the value.
#20
Quote by rockon1824
What? Where are you getting that x from, the denominator's a two. As for the rest of your post, it's the exact same thing I did except that you split the fraction into two terms, which doesn't change the value.


Sorry, That x is a 2, and after cancelling the two's and subtracting the "last step x" from the "original x" you end up with 13.

Again, sorry for typing an 'x' where the two was supposed to be. I don't know why, but my middle finger reacts to 'x' A LOT when i try to type '2'
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Team Pale Yellow?
------m-------m------
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#21
Quote by americablanco
Sorry, That x is a 2, and after cancelling the two's and subtracting the "last step x" from the "original x" you end up with 13.

Again, sorry for typing an 'x' where the two was supposed to be. I don't know why, but my middle finger reacts to 'x' A LOT when i try to type '2'

Okay, so like I said, they both come out to 13, so it's two different ways of getting the same answer.
#22
lol for any number - 1. subtract the original number (u get 0 )
2. add 1
3. subtract 1 (get 0 again)
4. multiply by 13 (0 x 13 still 0 )
5. add 13.


lol
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