I found an old magic box for kids. I decided to read the manual. All the tricks were big fails accept for one that I thought was pretty clever.

Here are the instructions:

• Choose a random number (X)
  
• Multiply by 3
  
• Add 5
  
• Multiply by 4
  
• Add X
  
• Add a random number (Y)
  
• Subtract 20
  

Note: It doesn't work with decimal numbers.

If you post your final number, I can tell you what is your X and Y.

(I guess this topic ends when someone figures out the explanation.)

Post has been edited 1 time(s), last time on Aug 24 2011, 2:25 pm by Apos.

My final number is 170.

X = 10
Y = 40

I may have made a mistake, but I think I got it right.

Let's see the mathematical logic.
x = first number, y = second.
x = x (first step)
3x
3x+5
4(3x+5) = 12x+20
13x+20
13x+20+y
13x+20+y-20 = 13x+y

The number we end up telling you is 13(first number) + (second number). Simple enough, as the factors of 13 are pretty unique, but you have no way of knowing anything more. We could say x=1 and y= multiples of 13 to fuck with you.

I'll bite, how about 260? To be fair, I thought I'd pick a decently low number. My x could range from 0 to 20, and my Y could range from 0 to 260. Random guessing would yield a probability of 1 in 5,200. Have a blast

X = 20
Y = 0

Quote from Apos

Quote from poison_us

I'll bite, how about 260?

X = 20
Y = 0

No. I sent TiKels the numbers, just so I can't lie to you.

EDIT: haha, just realized I lied when I said the probability was 1/5,200. It's probably more like 21², or 441. I forgot there can only be so many y values that come out to 260, since the x values must be a multiple of 13.

Post has been edited 1 time(s), last time on Aug 24 2011, 2:49 pm by poison_us.

My numbers were x=1 and y= 157
OR
x=13 y=1

I guess this trick is busted...

Let's say X = 5 and Y = 5
5 * 3 = 15
15 + 5 = 20
20 * 4 = 80
80 + 5 = 85
85 + 5 = 90
90 - 20 = 70

Here is the way it was supposed to work:

As I sent to TiKels, my numbers were 5 and 195.
The trick is inherently flawed, as you simply cannot provide enough information to the person presenting the trick.

Choose a random number (X)
Multiply by 3
Add 5
Multiply by 4
Add X
Add a random number (Y)
Subtract 20

[4(3x + 5) + x] + y - 20 = n

y = n + 20 - [4(3x + 5) + x]

y = n + 20 - (12x + 20 + x)

y = n + 20 - (13x + 20)

y = n + 20 - 20 - 13x

y = n - 13x

---

This is a planar cartesian linear equation. With known n, it becomes a linear equation.





Therefore, where n is a whole number greater than or equal to two, it is impossible to determine the x and y coordinates.

Which is what I was saying, minus all the fancy graphics and explanations.

This, my friend, is why you don't just copypaste tricks from the internet without analyzing them first. Did you think SEN would be impressed by it?

Did you even bother to read the OP?

I got an other trick. I'm not sure if it's much better than the first one... But it's pretty cool (Until the brains come in ).

Choose two numbers. X and Y
Multiply X by 2.
Add 5 to X.
Multiply X by 5.
Add Y to X (Which is now the result of all the previous operations.)

Tell me your results and I should be able to tell you what your X and Y is. (With a bit of work.) (The higher your two numbers, the harder it seems to be but it's still doable.)

Post has been edited 1 time(s), last time on Oct 2 2011, 8:54 pm by Apos. Reason: Nothing to say.

Quote from Apos

Double X by 2.

Assuming you mean 2x:
10x+25 + y = 135.

X = 10, Y = 10?

Post has been edited 1 time(s), last time on Oct 2 2011, 9:01 pm by Apos. Reason: Forgot to answer

60345

Quote from Apos

Quote from poison_us

Quote from Apos

Double X by 2.

Assuming you mean 2x:
10x+25 + y = 135.

X = 10, Y = 10?

No, 9/20.

Quote from Vrael

60345

x=1, y=60,335

X = 4641, Y = 12

5y(2x + 5) = n
5y = n/(2x + 5)
y = n/5(2x + 5)

With n = 5

y = 1/(2x + 5)



I hope you see the problem with this.