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ParthKohli

  • one year ago

How many four digit numbers are there such that the digits add up to 10 assuming that all digits are allowed to be zero?

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  1. amistre64
    • one year ago
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    less than 9999 but more than 1000 :)

  2. amistre64
    • one year ago
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    1117 1171 1711 7111 1126 1162 1261 1621 2611 1216 1612 2161 6121 6211 2116 6112 etc

  3. ParthKohli
    • one year ago
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    A more mathematical approach, probably?

  4. mathslover
    • one year ago
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    1000A + 100B + 10C + D = 10 100A + 10B + C + D/10 = 1

  5. mathslover
    • one year ago
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    if the number is ABCD ^

  6. amistre64
    • one year ago
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    pfft, that is the most basic mathematical appraoach i know :)

  7. ParthKohli
    • one year ago
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    I liked what @mathslover said, but that doesn't resolve it. Sorry for taking the medal back. :-(

  8. mathslover
    • one year ago
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    1000A + 100B + 10C + D = 10 ---- (1) A + B/10 + C/100 + D/1000 = 1/100 ---- (2)

  9. ParthKohli
    • one year ago
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    But we must have four equations, no?

  10. mathslover
    • one year ago
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    No problem :) Parth , thanks though..

  11. mathslover
    • one year ago
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    I am going to subtract the above 2 eqns

  12. amistre64
    • one year ago
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    its most likely a counting application with permutations since order matters

  13. amistre64
    • one year ago
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    how many basic ways are there to define 4 digits that add up to ten, and then how many ways can each group be sorted thru

  14. ParthKohli
    • one year ago
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    Can we use combinatorics for this?

  15. mathslover
    • one year ago
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    hmn yes ^^ wait

  16. amistre64
    • one year ago
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    combinatorics brings to mind the nCr stuff; i think the nPr stuff might be more applicable

  17. harsh314
    • one year ago
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    sorry to correct what @mathslover is doing is a bit wrong /////// A+B+C+D=10 is the requirement///////////

  18. amistre64
    • one year ago
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    1117; how many ways are there to permute this? 1126 how many ways? 1135 1144

  19. amistre64
    • one year ago
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    we have to account for 0 i spose, which is added computations

  20. mathslover
    • one year ago
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    OH yes @harsh314 thanks a lot

  21. amistre64
    • one year ago
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    4+12+12+12 +things with zeros

  22. amistre64
    • one year ago
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    118: stuff like 1180, 1108, 1018, but not 0118 127 136 145 19 28 37 46 55

  23. amistre64
    • one year ago
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    so permute the 3 groups and times it by 3

  24. amistre64
    • one year ago
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    1900, 1090, 1009 looks to be 3 times the 2 groups as well except for the 55 :) 5500, 5050, 5005 is just 3

  25. ParthKohli
    • one year ago
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    This is very long. :-|

  26. ParthKohli
    • one year ago
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    Instead of brute-force, can we try something simpler?

  27. mathslover
    • one year ago
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    I got it

  28. amistre64
    • one year ago
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    this isnt brute force anymore, its refining

  29. mathslover
    • one year ago
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    Lemme check my method again

  30. mathslover
    • one year ago
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    Forget it . . . I think I must try it again :(

  31. amistre64
    • one year ago
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    im to focused on my exam in 30 minutes to think thru this clearerer

  32. mathslover
    • one year ago
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    Parth, do you have answer of it ? ?

  33. ParthKohli
    • one year ago
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    Nope.

  34. harsh314
    • one year ago
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    maybe closer the answer

  35. mathslover
    • one year ago
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    84 numbers ... I think

  36. ParthKohli
    • one year ago
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    How?

  37. mathslover
    • one year ago
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    I will put up soln here soon... just thinking of more no.

  38. mathslover
    • one year ago
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    can the digits repeat Parth?

  39. ParthKohli
    • one year ago
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    Yes.

  40. ParthKohli
    • one year ago
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    Well, a number still has four digits if the digits repeat. :-)

  41. mathslover
    • one year ago
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    I found this on net : http://answers.yahoo.com/question/index?qid=20080330145752AA36Ac2 Now, I am thinking can we use this method to find the answerr?

  42. ParthKohli
    • one year ago
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    We can think of the ones with zeroes like this: three digit numbers that add up to 10 * 4! two digit numbers that add up to 10 * 4!/2 there can't be 3 or 4 zeroes.

  43. harsh314
    • one year ago
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    5193 is the answer or am i wrong as usual

  44. ParthKohli
    • one year ago
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    Hey wait, did I mention that the first digit can be zero?

  45. amistre64
    • one year ago
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    1117 is unique, 4 11xy has 12, and there are 3 ways to chose xy, and no zero groups 4(1)+3(12) 118 is unique, there are 3 ways, and 3 zero groups 1xy has 6 ways, and there are 4 ways to choose xy, and 3 zero groups 3(3)+3(6(4)) 55 is unique, theee is ony 1 way, and 3 zero groups xy 2 ways, there are 4 ways to choose xy, and 3 zero groups 3(1) + 3(2(4)) is there a flaw here?

  46. mathslover
    • one year ago
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    Oh! Is it allowed to have 0910 as the number ?

  47. ParthKohli
    • one year ago
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    Yes.

  48. amistre64
    • one year ago
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    the first digit cant be zero and it still be a 4-digit number tho

  49. harsh314
    • one year ago
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    no but a number with first digit as 0 does not make sense look at my answer plz i think it is right

  50. mathslover
    • one year ago
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    ^ yeah after that 0910 is not a four digit number , it will be three digit number ..

  51. amistre64
    • one year ago
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    have fun, i gotta run ;)

  52. ParthKohli
    • one year ago
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    I know... the question is making such an assumption, so...

  53. mathslover
    • one year ago
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    @amistre64 ;)

  54. mathslover
    • one year ago
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    well there can't be more than 186 numbers.. I think.

  55. mathslover
    • one year ago
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    ^ 196

  56. mathslover
    • one year ago
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    oh sorry 186

  57. ParthKohli
    • one year ago
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    Is it 186?

  58. harsh314
    • one year ago
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    more than 577

  59. mathslover
    • one year ago
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    yes it is 186 @ParthKohli

  60. ParthKohli
    • one year ago
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    The answer is 186?

  61. Meepi
    • one year ago
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    (0..9).each do |a| (0..9).each do |b| (0..9).each do |c| (0..9).each do |d| count = count + 1 if (a + b + c + d) == 10 end end end end puts count gave me 282 as answer, I don't know a mathematical approach though :)

  62. mathslover
    • one year ago
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    yes @ParthKohli

  63. ParthKohli
    • one year ago
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    186 when 0 is allowed or not?

  64. mathslover
    • one year ago
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    allowed

  65. ParthKohli
    • one year ago
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    How? :S

  66. ParthKohli
    • one year ago
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    282 and 186 are both not right :-|

  67. harsh314
    • one year ago
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    i think i got the answer.............

  68. Meepi
    • one year ago
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    282 was with leading zeros, without it would be 219

  69. harsh314
    • one year ago
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    408 without the zeros

  70. harsh314
    • one year ago
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    468 with zeroes

  71. harsh314
    • one year ago
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    plz do reply if wrong.......

  72. harsh314
    • one year ago
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    @ParthKohli

  73. ParthKohli
    • one year ago
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    I'd check

  74. ParthKohli
    • one year ago
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    Done! 13 C 3 = 286

  75. harsh314
    • one year ago
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    yes i got this too a silly error how did you do it............

  76. harsh314
    • one year ago
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    my method is tooo long.......

  77. harsh314
    • one year ago
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    @ParthKohli

  78. ParthKohli
    • one year ago
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    ... 286 is the answer

  79. harsh314
    • one year ago
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    i too got the same but by long method how did you get it .........

  80. Meepi
    • one year ago
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    I still find it odd that my bruteforce method which calculates the sum of the digits from 0 upto 9999 is off by 4

  81. ParthKohli
    • one year ago
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    We wanted the number of non negative numbers \(a_1,a_2,a_3,a_4\) such that \(a_1 + a_2 + a_3 + a_4 = 10\). A formula says that it would be \(\binom{10 + 4 - 1}{4 - 1}\)

  82. harsh314
    • one year ago
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    oh so no logic behind it as such..............

  83. Meepi
    • one year ago
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    But bruteforcing every digit from 0 to 9999 gives 282 though, shouldn't they be the same?

  84. ParthKohli
    • one year ago
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    I don't know. :-|

  85. Meepi
    • one year ago
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    Weird :l

  86. harsh314
    • one year ago
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    my method was long and a bit calculation mistake.............. case1:::'1" as the first digit ,,,,,, the following subcases arise (a) .....with 2 zeroes ........9 has to be the sum of the remaining three numbers so total possible combinations are 3. (b) .......with 1 zero we have 810,720,630,540......all these have 6 combinations each so total 24 (c) .......with no zeroes.....711(3combinations),621(6),531(6),522(3),432(6),441(3),333(1) so total of 28 combinations hence for the 1st case we have 45 combinations similarly we can have others tooo..........and then add all........we get 286

  87. harsh314
    • one year ago
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    but it would be great if any einstein of the present century could explain how @ParthKohli 's method came into being.............

  88. harsh314
    • one year ago
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    @satellite73

  89. amistre64
    • one year ago
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    works in google chrome :) <html> <body> <input type=button value="AddUp" onClick="qwe()"><br> 0000: <input id="txtOut0"><br> 1000: <input id="txtOut1"> </body> <script language=javascript> var a=0,b=0,c=0,d=0 var count=0 function qwe() {for (var x=1000; x<=10000; x=x+1) {if (a+b+c+d == 10){count=count+1} d=d+1; if (d>10) {d=0; c=c+1} if (c>10){c=0; b=b+1} if (b>10){b=0; a=a+1} if (a >9){x=10001} } document.getElementById("txtOut0").value=count //resets a=1,b=0,c=0,d=0, count=0 for (var x=1000; x<=10000; x=x+1) {if (a+b+c+d == 10){count=count+1} d=d+1; if (d>10) {d=0; c=c+1} if (c>10){c=0; b=b+1} if (b>10){b=0; a=a+1} if (a >9){x=10001} } document.getElementById("txtOut1").value=count } </script> </html>

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  90. amistre64
    • one year ago
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    i should prolly include an output such that we can verify its a+b+c+d parts to make sure its operating as expected

  91. Meepi
    • one year ago
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    I'm working on one :)

  92. amistre64
    • one year ago
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    gotta weed out a bug :) 0,0,0,10 0,0,1,9 0,0,2,8 0,0,3,7 0,0,4,6 0,0,5,5 0,0,6,4 0,0,7,3 0,0,8,2 0,0,9,1 0,0,10,0 0,1,0,9 0,1,1,8

  93. amistre64
    • one year ago
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    if abcd = 10 then reset to 0, not if greater than 10 .... lol

  94. amistre64
    • one year ago
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    im missing 9001, 9010, 9100 from the end of my 0xxx for some reason 279+3 = 282 otherwise its 219 if we use actual 4-digit numbers :)

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  95. Meepi
    • one year ago
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    282 is correct :)

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  96. amistre64
    • one year ago
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    yay!! :)

  97. amistre64
    • one year ago
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    ... brute math, the easy way :)

  98. Meepi
    • one year ago
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    Computers made these problem a lot easier :)

  99. Meepi
    • one year ago
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    Unfortunately Parth has 286 as answer though :(

  100. amistre64
    • one year ago
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    parth is counting somethings more than once then. i adjusted my loops to 0 to 10000 and it caught the last 3 ;)

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