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dan815

  • one year ago

a+b+c+d = k a<b<c<d everything is an integer can you find the number of solutions as a function of K

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  1. dan815
    • one year ago
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    @TrojanPoem

  2. TrojanPoem
    • one year ago
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    Solve this in front of me. And give me another example

  3. dan815
    • one year ago
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    okay umm lemme show u why i think this relation is important one

  4. dan815
    • one year ago
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    okay ill show u all my work so far, its not much lol

  5. TrojanPoem
    • one year ago
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    :)

  6. dan815
    • one year ago
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    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

  7. dan815
    • one year ago
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    im looking at only the increasing case, and the decreasing case will be the same numbers

  8. dan815
    • one year ago
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    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

  9. dan815
    • one year ago
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    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

  10. TrojanPoem
    • one year ago
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    123 increasing 321 decreasing 456 inc 654 dec ( non bouncy) all )

  11. dan815
    • one year ago
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    yes that is correct

  12. dan815
    • one year ago
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    okay now lets move to the 9 digit case

  13. dan815
    • one year ago
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    here is where we can see the pattern

  14. TrojanPoem
    • one year ago
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    123456789 987654321

  15. dan815
    • one year ago
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    leets only look at the increasing for now

  16. TrojanPoem
    • one year ago
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    Ok.

  17. dan815
    • one year ago
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    dont think about the decreasing its just a mirror and repeated

  18. TrojanPoem
    • one year ago
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    Yeah, you just turn left to right and so on.

  19. dan815
    • one year ago
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    okay umm wait this is important first

  20. dan815
    • one year ago
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    if u look at the 10 digit case

  21. dan815
    • one year ago
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    |dw:1435690693214:dw|

  22. dan815
    • one year ago
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    there is 1 difference between all the digits,

  23. dan815
    • one year ago
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    the sum of the difference have to add upto 9, and all of them have to be more than 0

  24. dan815
    • one year ago
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    so u can see there is no way to arrange the 1s to get anything new that tells us there is infact only 1

  25. dan815
    • one year ago
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    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

  26. TrojanPoem
    • one year ago
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    What do you mean by all of them have to be more than 0 ? why not 0 ?

  27. dan815
    • one year ago
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    which is 1, not 0 anymore, but anything more than 1 means we will fail

  28. dan815
    • one year ago
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    because if the difference is 0 then we have 1 1 1 1 1 which is bouncy

  29. dan815
    • one year ago
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    12345 all have diff 1 for their digits

  30. dan815
    • one year ago
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    following so far?

  31. TrojanPoem
    • one year ago
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    You mean the difference between the numbers can't be zero. Fine I understood all.

  32. dan815
    • one year ago
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    okay now look at this

  33. TrojanPoem
    • one year ago
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    The sum of difference 12345 is 4

  34. dan815
    • one year ago
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    |dw:1435690935221:dw|

  35. dan815
    • one year ago
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    so for the 9 digit case we know that 1 is the biggest number

  36. dan815
    • one year ago
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    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

  37. dan815
    • one year ago
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    when you pick 1, the difference have to add up to 8 and we have 8 places so everything is 1

  38. dan815
    • one year ago
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    but when you pick 0, the differences can add up to 9 and we have 8 places

  39. TrojanPoem
    • one year ago
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    But if we used 2, won't the number above 23456789--- will be bouncy if we added any other digit ?

  40. dan815
    • one year ago
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    yes that is why 2 is not possible

  41. dan815
    • one year ago
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    are u good upto here?

  42. dan815
    • one year ago
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    this is where the parition question starts

  43. TrojanPoem
    • one year ago
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    Yeah. But you must give me 1 or 2 easy questions as not to lose the info I gained.

  44. dan815
    • one year ago
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    okay i think we should just work on this simpler problem

  45. dan815
    • one year ago
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    a+b+c+d = k a<b<c<d everything is an integer can you find the number of solutions as a function of K

  46. dan815
    • one year ago
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    lets say K = 7

  47. dan815
    • one year ago
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    a+b+c+d=7 how many solutions are there to this? where a<b<c<d

  48. TrojanPoem
    • one year ago
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    Bouncy or non bouncy numbers ?

  49. dan815
    • one year ago
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    no this is just a separate question

  50. dan815
    • one year ago
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    you have to find hte number of integer solutions to that equation

  51. TrojanPoem
    • one year ago
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    Lol a = 0 , b = 1, c= 2 , d= 4 or a = 0 , b = 0 , c = 0 , d = 7 there is infinity. Clarify your aimbition

  52. dan815
    • one year ago
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    a<b<c<d is a condition

  53. TrojanPoem
    • one year ago
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    Oh

  54. TrojanPoem
    • one year ago
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    a = 4 ,b = 2, c = 1 ,d = 0

  55. dan815
    • one year ago
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    you have to find the number* of integer solutions

  56. TrojanPoem
    • one year ago
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    a+ b = 2 , a < b < solve this.

  57. dan815
    • one year ago
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    there is only 1 solution to that

  58. dan815
    • one year ago
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    oh and a b c d > 0 ofcourse

  59. TrojanPoem
    • one year ago
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    Ok .. because only when a = 2 But what about this a + b + c = 50 a < b <c

  60. dan815
    • one year ago
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    or equal to 0

  61. dan815
    • one year ago
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    yeah thats the question im asking u lol

  62. dan815
    • one year ago
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    50 is a big one try just 7 for now

  63. TrojanPoem
    • one year ago
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    Ok a = 4, b = 2, c = 1, d = 0 (1) a = 5, b= 2, c =1 , d= =0 Maybe 2 or so

  64. dan815
    • one year ago
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    i think a logical way to break this question down is to look at the differences

  65. dan815
    • one year ago
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    a+b+c=7 okay say a=0 this means we have a difference of 7 to distribute

  66. TrojanPoem
    • one year ago
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    Ok , so failed because it doesn't satisfy the condition.

  67. dan815
    • one year ago
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    0 , 1, 6 0,2,5 0,3,4 no other solutions with a = 0

  68. dan815
    • one year ago
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    hmm okay u think about it for a bit, i am gonna look at something

  69. TrojanPoem
    • one year ago
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    Damn -_- I saw the condition a > b > c >d

  70. dan815
    • one year ago
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    thats okay too same thing

  71. TrojanPoem
    • one year ago
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    But In your solution you made a = 0 and other number = 0 this is not possible you have broken the condition right ?

  72. dan815
    • one year ago
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    xD dont get caught on details lol

  73. dan815
    • one year ago
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    we need to work faster rn

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

  75. TrojanPoem
    • one year ago
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    are*

  76. dan815
    • one year ago
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    i found 11 increasing non bouncy numbers for digit 9 , im gonna work on 8 now

  77. dan815
    • one year ago
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    try this problem instead ditch the older one a + b+ c = 7 and a,b,c >0 and

  78. dan815
    • one year ago
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    so we are allowing the equal case now

  79. TrojanPoem
    • one year ago
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    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

  80. TrojanPoem
    • one year ago
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    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.

  81. dan815
    • one year ago
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    yes thats where i am right now

  82. dan815
    • one year ago
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    now instead of this, people have found a formula to do that counting for the number of solutions

  83. dan815
    • one year ago
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    which is our goal

  84. TrojanPoem
    • one year ago
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    AND YOU LEFT ME CALCULATING !!!!!!!!!!!!!!!!! >:(

  85. TrojanPoem
    • one year ago
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    But at least I learnt the basics :P

  86. dan815
    • one year ago
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    this stuff is an intro to something called paritioning theorems in mathematics

  87. dan815
    • one year ago
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    people have been working on approximations formulas for this count

  88. TrojanPoem
    • one year ago
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    approximations ? So answers are not accurate ?

  89. dan815
    • one year ago
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    the degree of error is quite low, so for the number we are dealing with the error will not effect it

  90. TrojanPoem
    • one year ago
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    What's that formula ?

  91. dan815
    • one year ago
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    in other words its like people are finding formulas that are able to solve higher and higher values for k a+b+c=k

  92. dan815
    • one year ago
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    i dont know, there are a lot apparently

  93. dan815
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    okay but theres still work to be done here before the formula

  94. dan815
    • one year ago
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    there are more simpler ways to count

  95. TrojanPoem
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    Like ?

  96. dan815
    • one year ago
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    okay like take this case for 9 digit

  97. dan815
    • one year ago
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    the difference if you pick 0 as the start

  98. dan815
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    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

  99. dan815
    • one year ago
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    the other 2 cases are when all the differences are 1 which gives us 2 more cases so a total of 9

  100. dan815
    • one year ago
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    now for the 8 digit case, we can use some combinatorics to count faster

  101. dan815
    • one year ago
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    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

  102. dan815
    • one year ago
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    and so on for 7

  103. TrojanPoem
    • one year ago
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    explain " there is one 2" we can have a 3 somewhere or a 2 somewhere

  104. dan815
    • one year ago
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    if u look at the differences

  105. dan815
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    when u have 0 as a start

  106. dan815
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    |dw:1435693429832:dw|

  107. dan815
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    and we can have 2 and 2

  108. dan815
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    how many ways to distribute 2, 2s in 8 positions

  109. dan815
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    it would be 8*7/2

  110. TrojanPoem
    • one year ago
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    I am dumb lol. |dw:1435693617081:dw|

  111. dan815
    • one year ago
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    oops lol it should be 1 1, 2

  112. dan815
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    |dw:1435693673194:dw|

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

  114. dan815
    • one year ago
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    what do u mean?

  115. TrojanPoem
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    1 2 4 56789 ?

  116. dan815
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    no look at the differences

  117. dan815
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    we are trying to find the unique differences and see the number of permutatiosn each have

  118. dan815
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    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

  119. dan815
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    and that completes the case for a=0 for the 8 digit case

  120. dan815
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    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

  121. dan815
    • one year ago
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    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

  122. TrojanPoem
    • one year ago
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    Give me simple exercise

  123. TrojanPoem
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    a simple*

  124. dan815
    • one year ago
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    a+b+c=5 a,b,c>0

  125. dan815
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    find a nice solution to that and we will solve this whole quetion in like 0.0000000000001 computation time hehe

  126. dan815
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    a+b+c=5 a,b,c>0 Total number of solutions = ?

  127. TrojanPoem
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    hahahahahahahahahaaaaaaaaaaaa, Nice solution ? I am using dumbs way.

  128. dan815
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    here is something that might be useful im not sure yet

  129. dan815
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    if you take the 2nd differences we can allow 0 so this is great

  130. dan815
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    then we can use a nice combinatorics trick

  131. dan815
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    like look at this other question okay where we allow 0

  132. dan815
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    a+b+c = 10 lets say

  133. dan815
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    you can think of this problem like this think of the 10 as being 10 ones

  134. TrojanPoem
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    |dw:1435694528337:dw|

  135. dan815
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    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

  136. dan815
    • one year ago
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    will tell us total number of solutions

  137. dan815
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    1 1 1 1+ 1 1 1 1+ 1 1 this is saying 4 + 4 + 2

  138. TrojanPoem
    • one year ago
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    Wait wait, here is my problem "put 2 signs to get unique solution"

  139. dan815
    • one year ago
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    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

  140. TrojanPoem
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    I understood !

  141. TrojanPoem
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    HAHAHAHA , I UNDERSTOOD Xd

  142. TrojanPoem
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    Give me example fast I understood xD

  143. dan815
    • one year ago
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    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

  144. TrojanPoem
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    You were explaining to me combinations and permutations ?

  145. dan815
    • one year ago
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    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

  146. dan815
    • one year ago
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    the solution to his, if its loonies must be

  147. dan815
    • one year ago
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    billion + 9999 choose 9999

  148. TrojanPoem
    • one year ago
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    1,000,000,000 P 10,000

  149. dan815
    • one year ago
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    not Permute but choose

  150. dan815
    • one year ago
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    but if u want to allow cents too then its ((billion*100 )+ 9999) choose 9999

  151. TrojanPoem
    • one year ago
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    But Permutations are right here ?

  152. dan815
    • one year ago
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    when we say chosoe and permute we are actually not thinking about permute and choosing

  153. dan815
    • one year ago
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    in your mind you know the formula you get and u just say was it a choose or permute

  154. TrojanPoem
    • one year ago
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    But how about the a+b+c+d = k ? I still can't think about it

  155. dan815
    • one year ago
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    okay so here is what im thinking about a way to solve this

  156. TrojanPoem
    • one year ago
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    When It was 7 you told me write them as 1111111 now 11+111+11 and calculate 2+3+2

  157. dan815
    • one year ago
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    wait i wanna work something out on paper first

  158. TrojanPoem
    • one year ago
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    Take your time.

  159. dan815
    • one year ago
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    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

  160. dan815
    • one year ago
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    I feeel like this will eventually lead to our own counting method for this whole thing

  161. TrojanPoem
    • one year ago
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    What did you find ? :>

  162. dan815
    • one year ago
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    okay okay look at this!! there is only 1 case we even need to lookat!

  163. dan815
    • one year ago
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    okay for example this one a+b+c=7 lets say and a,b,c>= 1

  164. dan815
    • one year ago
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    now we consider the simplst case

  165. dan815
    • one year ago
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    1+1+5

  166. dan815
    • one year ago
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    okay now here is what we do

  167. dan815
    • one year ago
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    |dw:1435695479325:dw|

  168. dan815
    • one year ago
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    now that 2nd difference can be solved with this method

  169. dan815
    • one year ago
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    5 choose 1 ways

  170. dan815
    • one year ago
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    so there msut be 5 solutions to this

  171. dan815
    • one year ago
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    let see if its right!

  172. dan815
    • one year ago
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    |dw:1435695530619:dw|

  173. dan815
    • one year ago
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    this gives us all the solutions for fixing first number

  174. dan815
    • one year ago
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    lets try a bigger example

  175. dan815
    • one year ago
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    |dw:1435695642003:dw|

  176. dan815
    • one year ago
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    i just need to deal with that first digit part lemme see about that one thing hmm

  177. TrojanPoem
    • one year ago
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    I must give you 100 medal for that work , awesome.

  178. dan815
    • one year ago
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    hahaha itss nothinngg

  179. dan815
    • one year ago
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    this stuff is very interestin :D i feel like we found something very neat!

  180. TrojanPoem
    • one year ago
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    :D

  181. TrojanPoem
    • one year ago
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    Oh, before I forget. Dan are you good at partial integration ?

  182. dan815
    • one year ago
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    okay wait here is the bad part

  183. dan815
    • one year ago
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    we solved a slightly different problem

  184. dan815
    • one year ago
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    lol because 112233 is still classified as a increasing lmao

  185. TrojanPoem
    • one year ago
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    Oo

  186. dan815
    • one year ago
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    but this is kind of good because we found this really cool relationship!

  187. dan815
    • one year ago
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    ill change that question and post it ON OS, let other ppl try it!

  188. TrojanPoem
    • one year ago
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    As you like.

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