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Solve this in front of me. And give me another example

okay umm lemme show u why i think this relation is important one

okay ill show u all my work so far, its not much lol

:)

im looking at only the increasing case, and the decreasing case will be the same numbers

123 increasing
321 decreasing
456 inc
654 dec ( non bouncy) all )

yes that is correct

okay now lets move to the 9 digit case

here is where we can see the pattern

123456789
987654321

leets only look at the increasing for now

Ok.

dont think about the decreasing its just a mirror and repeated

Yeah, you just turn left to right and so on.

okay umm wait this is important first

if u look at the 10 digit case

|dw:1435690693214:dw|

there is 1 difference between all the digits,

the sum of the difference have to add upto 9, and all of them have to be more than 0

What do you mean by all of them have to be more than 0 ? why not 0 ?

which is 1, not 0 anymore, but anything more than 1 means we will fail

because if the difference is 0 then we have 1 1 1 1 1 which is bouncy

12345 all have diff 1 for their digits

following so far?

You mean the difference between the numbers can't be zero. Fine I understood all.

okay now look at this

The sum of difference 12345 is 4

|dw:1435690935221:dw|

so for the 9 digit case we know that 1 is the biggest number

when you pick 1, the difference have to add up to 8 and we have 8 places so everything is 1

but when you pick 0, the differences can add up to 9 and we have 8 places

But if we used 2, won't the number above 23456789---
will be bouncy if we added any other digit ?

yes that is why 2 is not possible

are u good upto here?

this is where the parition question starts

Yeah. But you must give me 1 or 2 easy questions as not to lose the info I gained.

okay i think we should just work on this simpler problem

a+b+c+d = k a**
**

lets say K = 7

a+b+c+d=7
how many solutions are there to this?
where
a**
**

Bouncy or non bouncy numbers ?

no this is just a separate question

you have to find hte number of integer solutions to that equation

a**
**

Oh

a = 4 ,b = 2, c = 1 ,d = 0

you have to find the number* of integer solutions

a+ b = 2 , a < b < solve this.

there is only 1 solution to that

oh and a b c d > 0 ofcourse

Ok .. because only when a = 2
But what about this a + b + c = 50 a < b

or equal to 0

yeah thats the question im asking u lol

50 is a big one try just 7 for now

Ok a = 4, b = 2, c = 1, d = 0 (1)
a = 5, b= 2, c =1 , d= =0
Maybe 2 or so

i think a logical way to break this question down is to look at the differences

a+b+c=7
okay
say a=0
this means we have a difference of 7 to distribute

Ok , so failed because it doesn't satisfy the condition.

0 , 1, 6
0,2,5
0,3,4
no other solutions with a = 0

hmm okay u think about it for a bit, i am gonna look at something

Damn -_- I saw the condition a > b > c >d

thats okay too same thing

xD dont get caught on details lol

we need to work faster rn

Lol, details is important. But what about a+b+c+d = 10^9
won't I die before completing it ?

are*

i found 11 increasing non bouncy numbers for digit 9 , im gonna work on 8 now

try this problem instead ditch the older one
a + b+ c = 7
and a,b,c >0 and

so we are allowing the equal case now

yes thats where i am right now

now instead of this, people have found a formula to do that counting for the number of solutions

which is our goal

AND YOU LEFT ME CALCULATING !!!!!!!!!!!!!!!!! >:(

But at least I learnt the basics :P

this stuff is an intro to something called paritioning theorems in mathematics

people have been working on approximations formulas for this count

approximations ? So answers are not accurate ?

the degree of error is quite low, so for the number we are dealing with the error will not effect it

What's that formula ?

i dont know, there are a lot apparently

okay but theres still work to be done here before the formula

there are more simpler ways to count

Like ?

okay like take this case for 9 digit

the difference if you pick 0 as the start

are
1,1,1,1,1....,2
there is one of 2
that means there are 9 ways with a dfference of 2 and all 1s

the other 2 cases are when all the differences are 1
which gives us 2 more cases so a total of 9

now for the 8 digit case, we can use some combinatorics to count faster

and so on
for 7

explain " there is one 2"
we can have a 3 somewhere or a 2 somewhere

if u look at the differences

when u have 0 as a start

|dw:1435693429832:dw|

and we can have 2 and 2

how many ways to distribute 2, 2s in 8 positions

it would be 8*7/2

I am dumb lol. |dw:1435693617081:dw|

oops lol it should be 1 1, 2

|dw:1435693673194:dw|

And you didn't tell me where the 3 gone ?

what do u mean?

1 2 4 56789 ?

no look at the differences

we are trying to find the unique differences and see the number of permutatiosn each have

and that completes the case for a=0 for the 8 digit case

for a=1 there are
8+1 way
for
a=2 there is 1 way
so a total of
1+8+8+8*7/2+9+1= eight digit CASE

Give me simple exercise

a simple*

a+b+c=5
a,b,c>0

a+b+c=5
a,b,c>0
Total number of solutions = ?

hahahahahahahahahaaaaaaaaaaaa, Nice solution ? I am using dumbs way.

here is something that might be useful im not sure yet

if you take the 2nd differences we can allow 0 so this is great

then we can use a nice combinatorics trick

like look at this other question okay where we allow 0

a+b+c = 10 lets say

you can think of this problem like this think of the 10 as being 10 ones

|dw:1435694528337:dw|

1 1 1 1 1 1 1 1 1 1
and the number of places we can put 2 plus signs to get unique solution

will tell us total number of solutions

1 1 1 1+ 1 1 1 1+ 1 1
this is
saying
4 + 4 + 2

Wait wait, here is my problem "put 2 signs to get unique solution"

I understood !

HAHAHAHA , I UNDERSTOOD Xd

Give me example fast I understood xD

You were explaining to me combinations and permutations ?

the solution to his, if its loonies must be

billion + 9999 choose 9999

1,000,000,000 P 10,000

not Permute but choose

but if u want to allow cents too then
its ((billion*100 )+ 9999) choose 9999

But Permutations are right here ?

when we say chosoe and permute we are actually not thinking about permute and choosing

in your mind you know the formula you get and u just say was it a choose or permute

But how about the a+b+c+d = k ? I still can't think about it

okay so here is what im thinking about a way to solve this

When It was 7 you told me write them as 1111111
now 11+111+11 and calculate 2+3+2

wait i wanna work something out on paper first

Take your time.

I feeel like this will eventually lead to our own counting method for this whole thing

What did you find ? :>

okay okay look at this!! there is only 1 case we even need to lookat!

okay for example this one
a+b+c=7
lets say
and a,b,c>= 1

now we consider the simplst case

1+1+5

okay now here is what we do

|dw:1435695479325:dw|

now that 2nd difference can be solved with this method

5 choose 1 ways

so there msut be 5 solutions to this

let see if its right!

|dw:1435695530619:dw|

this gives us all the solutions for fixing first number

lets try a bigger example

|dw:1435695642003:dw|

i just need to deal with that first digit part lemme see about that one thing hmm

I must give you 100 medal for that work , awesome.

hahaha itss nothinngg

this stuff is very interestin :D i feel like we found something very neat!

:D

Oh, before I forget. Dan are you good at partial integration ?

okay wait here is the bad part

we solved a slightly different problem

lol because 112233 is still classified as a increasing lmao

Oo

but this is kind of good because we found this really cool relationship!

ill change that question and post it ON OS, let other ppl try it!

As you like.