| Author |
Message |
    
Dave Ross
Supporter
Username: Dave
Post Number: 4158
Registered: 5-2001
| | Posted on Thursday, February 13, 2003 - 11:25 am: |     |
http://www.cyberglass.co.uk/assets/Flash/psychic.swf |
jgberkeley
Supporter
Username: Jgberkeley
Post Number: 2750
Registered: 5-2001
| | Posted on Thursday, February 13, 2003 - 12:00 pm: | |
Beyond me, Clever! |
peteglider
Citizen
Username: Peteglider
Post Number: 35
Registered: 8-2002
| | Posted on Thursday, February 13, 2003 - 12:14 pm: | |
I think there are a limited number of possibilities from this number game -- for example, start with any number from 13 to 19 and the answer is 9, 20 to 29 = 18, etc.
So the trick is in just displaying the same symbol for any of the number combos (most of those numbers listed will never happen as a result of the calculation!)
Fun! |
jamie
Moderator
Username: Jamie
Post Number: 242
Registered: 6-2001
| | Posted on Thursday, February 13, 2003 - 12:16 pm: | |
Unreal - it got me 6 times in a row so far. It's bizarre! |
njjoseph
Supporter
Username: Njjoseph
Post Number: 1876
Registered: 5-2001
| | Posted on Thursday, February 13, 2003 - 12:17 pm: | |
Not so hard. Take a look at the repeats in the items. Take your numbers, do the math, then look at the chart. Don't click, but do it 5 times with different numbers, and you'll see that you get the same symbol. When you click, you get that symbol.
Now comes the fun: notice that the chart changes everytime you click, so that it seems like it's predicting what appears to be your new symbol. For example, do 34. 34 - (3+4) = 27. Look at the symbol. Click. Get the symbol in the ball. Now look at 27 in the chart. Different. Click again. Get the symbol.
Enjoy! |
jamie
Moderator
Username: Jamie
Post Number: 243
Registered: 6-2001
| | Posted on Thursday, February 13, 2003 - 12:24 pm: | |
Pete,
Even if there were only say 5 symbols, the chances of it guessing the right one would be 1 in 5. It hasn't missed yet, this is truly bizarre.
|
tom
Citizen
Username: Tom
Post Number: 615
Registered: 5-2001
| | Posted on Thursday, February 13, 2003 - 12:25 pm: | |
I get it. No matter what number you choose, the result is going to be a multiple of 9! |
jgberkeley
Supporter
Username: Jgberkeley
Post Number: 2751
Registered: 5-2001
| | Posted on Thursday, February 13, 2003 - 12:26 pm: | |
Ahhh, Njjoseph, peteglider,
I just wanted so much to just believe!
Now, back to duct tape and plastic. |
aquaman
Citizen
Username: Aquaman
Post Number: 54
Registered: 8-2001
| | Posted on Thursday, February 13, 2003 - 12:30 pm: | |
Oh it's clever, but Dave wants an explanation, so here goes.
There are only 10 possible answers: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81. All multiples of 9. Each has the same symbol, so no matter what your result is, the symbol matches. When you try again, the common symbols change.
Clever
|
aquaman
Citizen
Username: Aquaman
Post Number: 55
Registered: 8-2001
| | Posted on Thursday, February 13, 2003 - 12:32 pm: | |
oops i see someone beat me to it |
jamie
Moderator
Username: Jamie
Post Number: 244
Registered: 6-2001
| | Posted on Thursday, February 13, 2003 - 12:39 pm: | |
ok, I get it now, didn't realize the chart changed. duh |
Lambda Lover
Citizen
Username: Lambdalover
Post Number: 1
Registered: 4-2003
| | Posted on Sunday, May 18, 2003 - 12:11 am: | |
I guess this is a dead thread by now, but I thought I'd chime in with a complete explanation in case it's not clear from the posts already here. You're exactly right about the multiples of 9 (incidentally, the author of the flash page doesn't predict 0 correctly) but I thought I'd share a deeper look into it (my own immediate reactions).
First let's deal with the author's immediate problem ... which says we have to take a 2 digit number and subtract from it the sum of its digits. We can say that for integers x, y in [0..9] we have:
((10 * x) + y) - (x + y)
This is just the expression for the problem; let's clean it up:
10 * x + y - x - y
10 * x - x
x * (10 - 1)
x * 9
And since x can be any integer in [0..9], then the result is any multiple of 9 from 0 to 81. QED.
In fact, we can predict exactly which multiple of 9 it'll be from the leading digit (the x). It will always be the product of 9 with that leading digit. So you'll find that picking 00 to 09 gives you 0, 10 to 19 gives you 9, 20 to 29 gives you 18, etc.
So that was kind of neat, but now (if we're mathematically inclined) we might be asking ourselves if this is just an interesting coincidental property of 2 digit numbers or if it holds for larger numbers also.
Well one way we can approach this problem is to use digit sequences. Let there be a digit sequence A whose ith digit is represented by A[i] and which has a total of N digits. We could calculate the actual value of A like this:
Sum( 10^(N - i) * A[i], 1 <= i <= N )
So, for example, let's say we want to represent the number 1234. Then A = [1, 2, 3, 4] and the whole number itself would be computed as:
10^3 * 1 + 10^2 * 2 + 10^1 * 3 + 10^0 * 4 = 1234
OK that's easy enough, so now with this tool in hand we can go about figuring out this trick for digit sequences of any length. Using this notation, the problem can be expanded to numbers with any number of digits like this:
Sum(10^(N - i) * A[i], 1 <= i <= N) - Sum(A[i], 1 <= i <= N)
For those of you who remember your sums, that can be simplified to:
Sum(10^(N - i) * A[i] - A[i], 1 <= i <= N)
Sum(A[i] * (10^(N - i) - 1), 1 <= i <= N)
So what this is telling us is that even as the numbers get bigger (N >> 2) we'll get back ever increasing multiples of 9 (because 10^x - 1 is a multiple of 9 for any integer x). Unfortunately we don't have anything as simple as that the multiple of 9 is given by the leading digit, but we can see why that's true (only the last digit is ignored, the rest are actually relevant).
There are some other cool places you can go with this. For example, rather than using base 10 digit sequences we might solve the problem for any base B; you'll find that the result is always a multiple of B - 1. Another interesting problem dealing with multiplying by sequences of 1s is hiding in this one. Unfortunately I don't have the time to go into all of them ... but this is a good introduction to problems with digit sequences.  |
ReallyTrying
Citizen
Username: Reallytrying
Post Number: 67
Registered: 1-2003
| | Posted on Thursday, May 22, 2003 - 1:47 pm: | |
I am feeling so unbelievably ignorant right now.. What does ^ mean? And <=? (Does the latter mean "less than or equal to"? |
jimmurphy
Citizen
Username: Jimmurphy
Post Number: 127
Registered: 5-2001
| | Posted on Thursday, May 22, 2003 - 1:58 pm: | |
RT:
^ is used for exponents (i.e. 9^2 would be 9 squared or nine to the second power).
You're correct on the other one.
Jim |
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