Question

In mod n we can make an algorithm that gives us a new number. What value does x need to be to make this algorithm give unique results?

Answer 1

f is the algorithm acting on a*b*c and it takes each of the numbers in any order. \[f(a*b*c)=(a+x)*b*c+(c+x)*b+(b+x)\]

Answer 2

Other examples using the algorithm: \[f(a*b*c)=(a+x)*b*c+(b+x)*c+(c+x)\] \[f(a*b)=(a+x)*b+(b+x)\]

Answer 3

what does this mean ?
"we can make an algorithm that gives us a new number. "

Answer 4

Just that we are applying these rules. So suppose we have 12, we can factor it into 4*3 and do the algorithm on it as (4+x)*3+(3+x) or we can factor it as 2*2*3 and get (2+x)*2*3+(3+x)*2+(2+x) and either way will give us the same answer. =)

Answer 5

I don't know if that sounds too complicated or confusing, but the answer is simple haha.

Answer 6

well not complicated i only i don't understand the question xD

Answer 7

like what is the algorithm ?
what is the unique number (or what does it represent )
finally what are a,b,c,... in mod n represent

Answer 8

wait i mean new instead of unique

Answer 9

a, b, and c are all integers, but we could have any number of integers.

What the algorithm does is it takes any amount of integers, takes one at random adds x to it, and then multiplies that sum by the rest of the numbers. Then it takes an unused number and repeats this until it runs out of numbers.

So another way to write the algorithm is start with these integers \[\Large n_1, n_2, n_3,... n_i\] then picking one at random (like 3rd one why not ) we get the first term as this: \[\Large (n_3+x)*(n_1n_2n_4 \cdots n_i)+\] The next term repeats this with another random number, so the next term could be the first one: \[\large (n_3+x)*(n_1n_2n_4 \cdots n_i)+ (n_1+x)*(n_2n_4 \cdots n_i)+ \cdots\] until eventually you run out.

Answer 10

\[f(a_1 \cdot a_2 \cdot a_3 \cdot a_4 \cdots a_n) \\ =(a_1+x) \cdot a_2 \cdot a_3 \cdot a_4 \cdots a_n+(a_2+x) \cdot a_3 \cdot a_4 \cdots a_n +(a_3+x) \cdot a_4 \cdots a_n + \\ +(a_n+x)\]
So is what we are looking at?

Answer 11

Yep, that's one way to look at it. But we can also do the terms out of order or even several terms at the same time! =D

\[\Large f(a_1*a_2*a_3) = (a_1*a_3+x)*a_2+(a_2+x)\]

Answer 12

Seems ridiculous right? Hahaha

Answer 13

so you are saying for this algorithm x=12 works

and you are looking for a set of values that do not work?

Answer 14

No, all values work all the time. What's the value of x that makes this algorithm work correctly for all values?

Answer 15

doesn't your first sentence before the question say it works for all values?

Answer 16

x=1 ? xD
randomly

Answer 17

When I said the cases for 12 @freckles notice that there's an x in there. That's what we have to figure out, what are all the possible values of x. 12 was not equal to x, what I was saying there was we had a_1 and a_2 equal to 4 and 3 or a_1 to a_3 equal to 2,2, and 3.

Answer 18

@ikram002p Hahaha there are multiple right answers, but you have to prove it's true. =P

Answer 19

was only wanna see black swan view :3
like checking out values that works and thats not
but now i understand it =)

Answer 20

I still don't get it so we have f(12)=15+4x
Like I don't get what x is suppose to be here

Answer 21

x is a constant that makes it so no matter how you decide to do the algorithm, it will always give the same result.

Answer 22

f(12)=15+4x is only one way to do the algorithm

Answer 23

f(4)=f(2*2)=(2+x)*2+(2+x)=6+3x
So do we want to find for example when f(4)=f(12)?

Answer 24

No, look back at the very beginning where I do several examples again of alternate ways of doing the algorithm. I think if I say much more I'll give it away.

Answer 25

ok for
f(a,b)=(a-x)b+(b-x)
=(b-x)a+(a-x)
ab-xb+b-x=ab-ax+a-x

b(x-1)=a(x-1)
hmmm lol
i have condition on a,b which sounds not right

Answer 26

Here are three different ways of doing the algorithm, notice I changed the order of picking, for the first I picked a, c, b and the second one I picked a,b,c. We could also have picked more than one at a time, so a*c, b which I have also added below \[f(a*b*c)=(a+x)*b*c+(c+x)*b+(b+x)\]\[f(a*b*c)=(a+x)*b*c+(b+x)*c+(c+x)\]\[f(a*b*c)=(a*c+x)*b+(b+x)\]

Answer 27

Hehehe. =) There is one trick I'll give you that might help you to figure it out.

f(a)=f(a*1)

That must be true for the algorithm to work, otherwise you will be able to repeat the algorithm infinitely and get different results. =P

Answer 28

:P

Answer 29

im not getting idea lol @Kainui
rust already holds in xD

Answer 30

ok i should think of it out of algebra

Answer 31

You're actually both on the right track haha, it's sort of weird to explain but once you see the answer it should make sense.

Answer 32

"Just that we are applying these rules. So suppose we have 12, we can factor it into 4*3 and do the algorithm on it as (4+x)*3+(3+x) or we can factor it as 2*2*3 and get (2+x)*2*3+(3+x)*2+(2+x) and either way will give us the same answer. =)
" so f(4*3)=15+4x and f(2*2*3)=20+9x
so f(12) doesn't have a unique representation for all x
but 15+4x=20+9x
when x=-1

Answer 33

it has unique representation when x=-1 is what I meant to say

Answer 34

"b(x-1)=a(x-1)
hmmm lol
i have condition on a,b which sounds not right "

@ikram002p you were right about this, notice you were just using -x instead of +x like @freckles so really you both are at the same answer. =)

Answer 35

:O i used ur first comment hmm

Answer 36

So you're both right, you've found one possible value for x. Haha yeah I changed it to make it harder. =P

Answer 37

1 mod n lol

Answer 38

-1 mod n

Answer 39

So what is n when x=6?

Answer 40

is it 5
since 6-5=1

Answer 41

n=a*b*c..... ?

Answer 42

x mod n
means x+kn where k is an integer right

Answer 43

@freckles is right, that is one possible answer, but there's one more possibility it could be.

Answer 44

6 mod 1 ?

Answer 45

Ok here's my solution: \[\Large f(a)=f(a*1) \\ \Large f(a) = (a+x) \\ \Large f(a*1)=(a+x)*1+(1+x) \\ \Large 1+x \equiv 0 \mod n \\ \Large \therefore x = n+1 \text{ or } x =n-1\]

Answer 46

so x= 1 or -1 :P

Answer 47

ok tgat was nice

Answer 48

Yeah, for the integers the +1 doesn't work however, and if you look at it, -1 version is the telescoping series!

Answer 49

In other words, the algorithm is simply this: \[\Large f(n)=n-1\]

Answer 50

hmmm i wonder how u keep say such smart stuff :P

Answer 51

lol I have too much free time XD

Answer 52

This all came from back when we were looking at factorial bases, remember how each "digits" place held only a factorial number of digits? I noticed it had the same property as 1000-1=999 where each previous place was filled except each one had a factorial in it. So to show mathematically what I mean: \[1000-1 = (10-1)*10*10+(10-1)*10+(10-1)=999\] or with factorials: \[\small 5!-1 = (4-1)*3*2*1+(3-1)*2*1+(2-1)*1 = 3*3!+2*2!+1!\]

Answer 53

oh yes i remember now !!!

Answer 54

do you have any cool stuff with ur factorial base ?
i would like to see

Answer 55

Nope, nothing really past that I think other than the repeating decimal representation of e which you already know about. =P

Answer 56

oh :O
ok

Answer 57

But if you have any ideas, I think there's some thing with factorials that this might relate to in finding primes, I forget though what it's called.

Answer 58

Like n!-1 has a lot of primes for it, I don't know much past that hmm \[\Large n!-1 = \sum_{i=1}^{n-1}i*i!\]

Answer 59

humming

Answer 60

i think i'll sleep now lol
11:49 pm :P

Answer 61

gn

Answer 62

Awww I just found it, Wilson's Theorem. Alright goodnight.

Answer 63

welson is like this
(n-1)!=-1 mod n

Answer 64

oh wait i remember ur proof :P
ok

Answer 65

My proof? I don't know what you mean

Answer 66

there was a question when u showed factorial base for first time (caaling it ur proof since its original work)
however i remember when u typed these stuff

Answer 67

Oh ok I see. I don't remember what all I said exactly, but I remember this much at least lol. Anyways I think this is a fun way to play around with numbers, I don't know if it's really useful or not but it is interesting and unexpected.

Answer 68

@freckles you still there?

Answer 69

Sorry I abandoned you guys.