dan815
  • dan815
a+b+c+d = k a
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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dan815
  • dan815
@TrojanPoem
TrojanPoem
  • TrojanPoem
Solve this in front of me. And give me another example
dan815
  • dan815
okay umm lemme show u why i think this relation is important one

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dan815
  • dan815
okay ill show u all my work so far, its not much lol
TrojanPoem
  • TrojanPoem
:)
dan815
  • dan815
okay so i started working backwards, now i allowed 0 to be in the first place of the digit, we can take care of the problems of that later on
dan815
  • dan815
im looking at only the increasing case, and the decreasing case will be the same numbers
dan815
  • dan815
so for 10 digit numbers we have 1 non bouncy number that is strictly increasing 0123456789 and 1 non bouncy number that is strickly decreasing 9876543210
dan815
  • dan815
if there were 2 strickly increasing then there have to be 2 strictly decreasing so you can see why we only care about one half
TrojanPoem
  • TrojanPoem
123 increasing 321 decreasing 456 inc 654 dec ( non bouncy) all )
dan815
  • dan815
yes that is correct
dan815
  • dan815
okay now lets move to the 9 digit case
dan815
  • dan815
here is where we can see the pattern
TrojanPoem
  • TrojanPoem
123456789 987654321
dan815
  • dan815
leets only look at the increasing for now
TrojanPoem
  • TrojanPoem
Ok.
dan815
  • dan815
dont think about the decreasing its just a mirror and repeated
TrojanPoem
  • TrojanPoem
Yeah, you just turn left to right and so on.
dan815
  • dan815
okay umm wait this is important first
dan815
  • dan815
if u look at the 10 digit case
dan815
  • dan815
|dw:1435690693214:dw|
dan815
  • dan815
there is 1 difference between all the digits,
dan815
  • dan815
the sum of the difference have to add upto 9, and all of them have to be more than 0
dan815
  • dan815
so u can see there is no way to arrange the 1s to get anything new that tells us there is infact only 1
dan815
  • dan815
now if u look at the 9 digit, case we have 8 difference now there is a max number we can pick for the first place
TrojanPoem
  • TrojanPoem
What do you mean by all of them have to be more than 0 ? why not 0 ?
dan815
  • dan815
which is 1, not 0 anymore, but anything more than 1 means we will fail
dan815
  • dan815
because if the difference is 0 then we have 1 1 1 1 1 which is bouncy
dan815
  • dan815
12345 all have diff 1 for their digits
dan815
  • dan815
following so far?
TrojanPoem
  • TrojanPoem
You mean the difference between the numbers can't be zero. Fine I understood all.
dan815
  • dan815
okay now look at this
TrojanPoem
  • TrojanPoem
The sum of difference 12345 is 4
dan815
  • dan815
|dw:1435690935221:dw|
dan815
  • dan815
so for the 9 digit case we know that 1 is the biggest number
dan815
  • dan815
now if you pick 1, then all the following differences have to be 1, so we know starting with 1 there is only 1 number 123456789
dan815
  • dan815
when you pick 1, the difference have to add up to 8 and we have 8 places so everything is 1
dan815
  • dan815
but when you pick 0, the differences can add up to 9 and we have 8 places
TrojanPoem
  • TrojanPoem
But if we used 2, won't the number above 23456789--- will be bouncy if we added any other digit ?
dan815
  • dan815
yes that is why 2 is not possible
dan815
  • dan815
are u good upto here?
dan815
  • dan815
this is where the parition question starts
TrojanPoem
  • TrojanPoem
Yeah. But you must give me 1 or 2 easy questions as not to lose the info I gained.
dan815
  • dan815
okay i think we should just work on this simpler problem
dan815
  • dan815
a+b+c+d = k a
dan815
  • dan815
lets say K = 7
dan815
  • dan815
a+b+c+d=7 how many solutions are there to this? where a
TrojanPoem
  • TrojanPoem
Bouncy or non bouncy numbers ?
dan815
  • dan815
no this is just a separate question
dan815
  • dan815
you have to find hte number of integer solutions to that equation
TrojanPoem
  • TrojanPoem
Lol a = 0 , b = 1, c= 2 , d= 4 or a = 0 , b = 0 , c = 0 , d = 7 there is infinity. Clarify your aimbition
dan815
  • dan815
a
TrojanPoem
  • TrojanPoem
Oh
TrojanPoem
  • TrojanPoem
a = 4 ,b = 2, c = 1 ,d = 0
dan815
  • dan815
you have to find the number* of integer solutions
TrojanPoem
  • TrojanPoem
a+ b = 2 , a < b < solve this.
dan815
  • dan815
there is only 1 solution to that
dan815
  • dan815
oh and a b c d > 0 ofcourse
TrojanPoem
  • TrojanPoem
Ok .. because only when a = 2 But what about this a + b + c = 50 a < b
dan815
  • dan815
or equal to 0
dan815
  • dan815
yeah thats the question im asking u lol
dan815
  • dan815
50 is a big one try just 7 for now
TrojanPoem
  • TrojanPoem
Ok a = 4, b = 2, c = 1, d = 0 (1) a = 5, b= 2, c =1 , d= =0 Maybe 2 or so
dan815
  • dan815
i think a logical way to break this question down is to look at the differences
dan815
  • dan815
a+b+c=7 okay say a=0 this means we have a difference of 7 to distribute
TrojanPoem
  • TrojanPoem
Ok , so failed because it doesn't satisfy the condition.
dan815
  • dan815
0 , 1, 6 0,2,5 0,3,4 no other solutions with a = 0
dan815
  • dan815
hmm okay u think about it for a bit, i am gonna look at something
TrojanPoem
  • TrojanPoem
Damn -_- I saw the condition a > b > c >d
dan815
  • dan815
thats okay too same thing
TrojanPoem
  • TrojanPoem
But In your solution you made a = 0 and other number = 0 this is not possible you have broken the condition right ?
dan815
  • dan815
xD dont get caught on details lol
dan815
  • dan815
we need to work faster rn
TrojanPoem
  • TrojanPoem
Lol, details is important. But what about a+b+c+d = 10^9 won't I die before completing it ?
TrojanPoem
  • TrojanPoem
are*
dan815
  • dan815
i found 11 increasing non bouncy numbers for digit 9 , im gonna work on 8 now
dan815
  • dan815
try this problem instead ditch the older one a + b+ c = 7 and a,b,c >0 and
dan815
  • dan815
so we are allowing the equal case now
TrojanPoem
  • TrojanPoem
Ok first a =0 b + c = 7 b = 0 c = 7 b =2 c = 5 b = 1 c = 6 b = 3 c = 4 b = 4 c = 3 b = 5 c = 2 b = 6 c = 1 Ok I got it , but still there is a lot a lot a lot of solutions
TrojanPoem
  • TrojanPoem
I got a scheme on how to do it first a = 0 b +c = 7 ( b from 0 to 7) and calculate c then a = 1 b+c = 6 (b from 0 to 6) and calculate c ... until the end.
dan815
  • dan815
yes thats where i am right now
dan815
  • dan815
now instead of this, people have found a formula to do that counting for the number of solutions
dan815
  • dan815
which is our goal
TrojanPoem
  • TrojanPoem
AND YOU LEFT ME CALCULATING !!!!!!!!!!!!!!!!! >:(
TrojanPoem
  • TrojanPoem
But at least I learnt the basics :P
dan815
  • dan815
this stuff is an intro to something called paritioning theorems in mathematics
dan815
  • dan815
people have been working on approximations formulas for this count
TrojanPoem
  • TrojanPoem
approximations ? So answers are not accurate ?
dan815
  • dan815
the degree of error is quite low, so for the number we are dealing with the error will not effect it
TrojanPoem
  • TrojanPoem
What's that formula ?
dan815
  • dan815
in other words its like people are finding formulas that are able to solve higher and higher values for k a+b+c=k
dan815
  • dan815
i dont know, there are a lot apparently
dan815
  • dan815
okay but theres still work to be done here before the formula
dan815
  • dan815
there are more simpler ways to count
TrojanPoem
  • TrojanPoem
Like ?
dan815
  • dan815
okay like take this case for 9 digit
dan815
  • dan815
the difference if you pick 0 as the start
dan815
  • dan815
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
dan815
  • dan815
the other 2 cases are when all the differences are 1 which gives us 2 more cases so a total of 9
dan815
  • dan815
now for the 8 digit case, we can use some combinatorics to count faster
dan815
  • dan815
for a starting value of 1 we have 8 cases for a starting value of 0 we can have a 3 somewhere or a 2 somewhere so 16 more ways from that
dan815
  • dan815
and so on for 7
TrojanPoem
  • TrojanPoem
explain " there is one 2" we can have a 3 somewhere or a 2 somewhere
dan815
  • dan815
if u look at the differences
dan815
  • dan815
when u have 0 as a start
dan815
  • dan815
|dw:1435693429832:dw|
dan815
  • dan815
and we can have 2 and 2
dan815
  • dan815
how many ways to distribute 2, 2s in 8 positions
dan815
  • dan815
it would be 8*7/2
TrojanPoem
  • TrojanPoem
I am dumb lol. |dw:1435693617081:dw|
dan815
  • dan815
oops lol it should be 1 1, 2
dan815
  • dan815
|dw:1435693673194:dw|
TrojanPoem
  • TrojanPoem
And you didn't tell me where the 3 gone ?
dan815
  • dan815
what do u mean?
TrojanPoem
  • TrojanPoem
1 2 4 56789 ?
dan815
  • dan815
no look at the differences
dan815
  • dan815
we are trying to find the unique differences and see the number of permutatiosn each have
dan815
  • dan815
there is 1 way to arrange all 1s there are 8 ways to arrange one 2 and rest 1 there are 8 ways to arrange one 3 and rest 1 there are 8*7/2 ways to arrange two 2s and rest 1
dan815
  • dan815
and that completes the case for a=0 for the 8 digit case
dan815
  • dan815
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
dan815
  • dan815
so everything is basic computation including the combinatorics we can make computer do, the one part we need to figure out is the unique solutions
TrojanPoem
  • TrojanPoem
Give me simple exercise
TrojanPoem
  • TrojanPoem
a simple*
dan815
  • dan815
a+b+c=5 a,b,c>0
dan815
  • dan815
find a nice solution to that and we will solve this whole quetion in like 0.0000000000001 computation time hehe
dan815
  • dan815
a+b+c=5 a,b,c>0 Total number of solutions = ?
TrojanPoem
  • TrojanPoem
hahahahahahahahahaaaaaaaaaaaa, Nice solution ? I am using dumbs way.
dan815
  • dan815
here is something that might be useful im not sure yet
dan815
  • dan815
if you take the 2nd differences we can allow 0 so this is great
dan815
  • dan815
then we can use a nice combinatorics trick
dan815
  • dan815
like look at this other question okay where we allow 0
dan815
  • dan815
a+b+c = 10 lets say
dan815
  • dan815
you can think of this problem like this think of the 10 as being 10 ones
TrojanPoem
  • TrojanPoem
|dw:1435694528337:dw|
dan815
  • dan815
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
dan815
  • dan815
will tell us total number of solutions
dan815
  • dan815
1 1 1 1+ 1 1 1 1+ 1 1 this is saying 4 + 4 + 2
TrojanPoem
  • TrojanPoem
Wait wait, here is my problem "put 2 signs to get unique solution"
dan815
  • dan815
there are a total of 12 spots if u count the 1s and + sign as spots so that means 12 choose 2 are the total number of integer solutiosn to a+b+c = 10
TrojanPoem
  • TrojanPoem
I understood !
TrojanPoem
  • TrojanPoem
HAHAHAHA , I UNDERSTOOD Xd
TrojanPoem
  • TrojanPoem
Give me example fast I understood xD
dan815
  • dan815
this is a useful result, now u can answer questions like say you and 2 friends have 10 dollars and only loonies allwoed now u know the total number of ways your money can be split
TrojanPoem
  • TrojanPoem
You were explaining to me combinations and permutations ?
dan815
  • dan815
or something with a huge solution but u can atleast get an expression for it, like a billion dollars is in a town of 10,000 people, you can now answer the total number of ways this money can be split between the people
dan815
  • dan815
the solution to his, if its loonies must be
dan815
  • dan815
billion + 9999 choose 9999
TrojanPoem
  • TrojanPoem
1,000,000,000 P 10,000
dan815
  • dan815
not Permute but choose
dan815
  • dan815
but if u want to allow cents too then its ((billion*100 )+ 9999) choose 9999
TrojanPoem
  • TrojanPoem
But Permutations are right here ?
dan815
  • dan815
when we say chosoe and permute we are actually not thinking about permute and choosing
dan815
  • dan815
in your mind you know the formula you get and u just say was it a choose or permute
TrojanPoem
  • TrojanPoem
But how about the a+b+c+d = k ? I still can't think about it
dan815
  • dan815
okay so here is what im thinking about a way to solve this
TrojanPoem
  • TrojanPoem
When It was 7 you told me write them as 1111111 now 11+111+11 and calculate 2+3+2
dan815
  • dan815
wait i wanna work something out on paper first
TrojanPoem
  • TrojanPoem
Take your time.
dan815
  • dan815
hey man I think I found something very interseting :O but im not quitee sure yet what to do with it!! but taking a look at the 2nd difference is very very very very very interseting
dan815
  • dan815
I feeel like this will eventually lead to our own counting method for this whole thing
TrojanPoem
  • TrojanPoem
What did you find ? :>
dan815
  • dan815
okay okay look at this!! there is only 1 case we even need to lookat!
dan815
  • dan815
okay for example this one a+b+c=7 lets say and a,b,c>= 1
dan815
  • dan815
now we consider the simplst case
dan815
  • dan815
1+1+5
dan815
  • dan815
okay now here is what we do
dan815
  • dan815
|dw:1435695479325:dw|
dan815
  • dan815
now that 2nd difference can be solved with this method
dan815
  • dan815
5 choose 1 ways
dan815
  • dan815
so there msut be 5 solutions to this
dan815
  • dan815
let see if its right!
dan815
  • dan815
|dw:1435695530619:dw|
dan815
  • dan815
this gives us all the solutions for fixing first number
dan815
  • dan815
lets try a bigger example
dan815
  • dan815
|dw:1435695642003:dw|
dan815
  • dan815
i just need to deal with that first digit part lemme see about that one thing hmm
TrojanPoem
  • TrojanPoem
I must give you 100 medal for that work , awesome.
dan815
  • dan815
hahaha itss nothinngg
dan815
  • dan815
this stuff is very interestin :D i feel like we found something very neat!
TrojanPoem
  • TrojanPoem
:D
TrojanPoem
  • TrojanPoem
Oh, before I forget. Dan are you good at partial integration ?
dan815
  • dan815
okay wait here is the bad part
dan815
  • dan815
we solved a slightly different problem
dan815
  • dan815
lol because 112233 is still classified as a increasing lmao
TrojanPoem
  • TrojanPoem
Oo
dan815
  • dan815
but this is kind of good because we found this really cool relationship!
dan815
  • dan815
ill change that question and post it ON OS, let other ppl try it!
TrojanPoem
  • TrojanPoem
As you like.

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