Quote from Sacrieur

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.

I do. Ignoring that 5y(2x+5) factors into 10xy+25y, and nowhere does it say multiply anything by y, I can't imagine what you were reading that gave you that equation, especially one apparently showing that when x=2, and n=5, the y value must be a fraction, when it's generally assumed that only integer values apply. It's impossible for the N (which I assume is the result that he's supposed to be finding the X and Y for) to be 5 and the X and Y to be anything but 0 and 5, respectively. Regardless:

Quote

Multiply X by 2.
Add 5 to X.
Multiply X by 5.
Add Y to X

Let's break this into 4 steps.
1: 2x
2: 2x+5
3: 5(2x+5)
4.0 beta: 5(2x+5)+y=n
4: 10x+25+y=n

Finally, unless we assume that people will always use the same X or Y, we can't make a classic Cartesian graph like that. We'd need a 3D visual reference, as X and Y are independent values (both from each other and N) with a N dependent value.

Quote from Vrael

X = 4641, Y = 12

4641 * 10 = 46410 + 25 + 12 = 46435 + 12 = 46447.
46,447 = 60,345?

EDIT: Sac, I just realized what you might've done. Did you work the initial problem from the bottom up, like so:
x+y
5(x+y)
5(x+y)+5
2(5(x+y)+5)?

Actually, scratch that, that factors into 10x+10y+10. Still not sure how you got the 10xy + 25y.

Post has been edited 1 time(s), last time on Oct 3 2011, 5:34 am by poison_us.

Memory failed me, I thought I was multiplying y, not adding. This turns it into a linear equation with all of the same pitfalls as the first.

Oh, I missed that we had a new sequence of operations lololol. My 60335 was for the old one with 13's and stuff.

I still fail to see the pitfalls, other than how you intend to represent the problem. As X and/or Y become(s) large, N also becomes large. Since X and Y are related through entirely independent and arbitrary rates, you can't make a simple Cartesian representation of their relationship that is valid, as I maintained before. That is, without assuming that X or Y is a constant, which neither are. As N gets large, there will always be exactly floor((N-25)/10)+1 valid terms for both X and Y. Even at N=100, predicting the X and Y values will be a 1/64 chance (1/8 for both, 1/8² = 1/64).

I realized that...

Well... This trick was supposed to be done with a dice (1 - 6). Then, I decided to try it with other numbers and it seemed to work... Until I realized there were more than 1 possible answer that give the same result. I guess it would be better if I listed all the possible answer... The trick wouldn't be as impressive...

For 60345
I made a simple program to calculate it all.

You made a program based off of my functions above, didn't you?

Yep but it doesn't really make any difference. Yours was just shorter and better. (No misleading information.)



This is a very bad code. It was done in a few seconds. It's still pretty fast though so you won't notice any slow down until you start trying very high numbers.

Post has been edited 1 time(s), last time on Oct 3 2011, 3:18 pm by Apos. Reason: Noticed there was a mistake in the code.

Any particular reason you iterate over the y values instead of just solving for y? Do you only want natural numbers or something? I just don't really see a reason for the inner loop.

As I said, I coded it in a few seconds only. You are right though. It's weird that I didn't think about doing that in the first place. It would definitively save a lot of calculation time. (I did want to exclude decimal numbers but I guess it doesn't matter at this point.)

I was just did a brute force search.

... I think this time it should be better...

Think of a 3 or a 4 digit number, name it X. (It has to contain more than 1 different digits. So 1111 would not work.) (When I say 3 or 4 digit numbers, it can be any digit numbers. It just requires a bit more work. Even 10 digits would work.) (When I say it takes more work, it means that it adds 1 - 2 extra seconds.)
Now, reorder the numbers in X to create Y.

Subtract the bigger number by the smaller number.

Then, remove any 0 in your answer. And cross out one of the digit. Randomize that new answer again and tell me what it is.

I'll be able to tell you which number you crossed out.



Post has been edited 3 time(s), last time on Oct 4 2011, 11:26 pm by Apos.

If you do the math for it, it's pretty obvious this one will always work. Good job.

That's easy .. Transposed numbers are always divisible by 9.

I guess that math games are hard to do properly. It's only good if people can't calculate it afterwards... (As long as they are possible )

I guess this thread should be called post your math games instead.

Alright, 10 digits it is. Call them a-j.

abcdefghij → ABCDEFGHIJ (because you said rearrange, not reverse). If(ABCDEFGHIJ > abcdefghij, ABCDEFGHIJ-abcdefghij, else abcdefghij-ABCDEFGHIJ) = klmnopqrst. Remove any zero, so let's just say m and r = 0. Cross out a random digit, so klnopqs. Rerandomize: klnopqs → KLNOPQS. Please note that there is no relation other than position between upper- and lowercase variables.

Alright, since I can't mathematically prove it is a vanishingly small chance with exhaustively large numbers, I'll bite. Let's start with a decently large random.org number, randomize with random.org (both times), and randomize the number to cross out.

SPOILER.87446
SPOILER84674
SPOILER2772
SPOILER272

722.

Hmm, that just won't do, the number is relatively easy to see. Let's try this one, with a much larger number. Oh, and no spoiler text.

6653749

You struck out a 5.

How'd you come about getting that? And can you tell me which slot it was in (or between which numbers)? It's right, I'm just curious as to how the math is applied. I get that it's got to do with transposed numbers and divisible by 9, but I always thought you'd need to know the original number.

It seems more complicated if you think of the digits as digits of a singular number, but the way to look at it is like, instead of 24561, it's

20000 + 4000 + 500 + 60 + 1

Which you could take as 10000X + 1000Y + 100Z + 10A + B.

Then you rearrange the variables, which is the same as rearranging the digits originally, and might get

10000Y + 1000C + 100X + 10 A + Z.

Assuming the first number is larger, it's (10000X + 1000Y + 100Z + 10A + C) - (10000Y + 1000C + 100X + 10A + Z).

No matter how you rearrange them, every variable's coefficient ends up either canceling out or being a multiple of 10 subtracted by another multiple of 10. Each one ends up being 10ⁿ - 10^m, which can only result in every digit of the result being composed entirely of 9's possibly followed by 0's, which can always be factored by 9.

If you take the previous example, the result is 9900X - 9000Y - 999C + 99Z.

This can be factored into 9(1100X - 1000Y - 111C + 11Z).

The only thing that matters is that no matter what, the final answer will always be divisible by 9.

Since adding 9 to itself any number of times results in the digits of the number always adding up to 9 (assuming that you continually add the digits of a multiple digit sum as well), the digits of the final answer must add up to 9. 0's are irrelevant since he's just summing all your digits.

So basically he just takes your number, adds the digits of your answer, then the digits of the sum, and continues until he gets one number. Then he subtracts that number from 9, and that has to be the digit you struck out.

Well duh, should've seen that coming.

Here is an other one (not really )

Choose any number you like.

Multiply it by 3.

Square it. (X * X)

Cross out one of the digits and tell me your result. (You can randomize it just like before.)

Nothing.