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dan815

  • one year ago

Modified P.E question: Working from left-to-right if no digit is exceeded and not equal by the digit to its left it is called an increasing number; for example, 134568. Similarly if no digit is exceeded and not equal by the digit to its right it is called a decreasing number; for example, 65420. We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349. What is the total number of non bouncy numbers? Here is a similar question with more details : https://projecteuler.net/problem=112

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  1. ganeshie8
    • one year ago
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    .

  2. Astrophysics
    • one year ago
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    .

  3. ganeshie8
    • one year ago
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    By your modified definition, is the first bouncy number 11 ?

  4. dan815
    • one year ago
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    yep that is true

  5. Astrophysics
    • one year ago
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    Binary?

  6. ganeshie8
    • one year ago
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    nope, i think the question is number system sensitive

  7. dan815
    • one year ago
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    hey look at this pattern for the number of none bouncy numbers from 10 digit to 1 digit it follows this patternn

  8. dan815
    • one year ago
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    for 10 digit solving the equation a1+a2+a3+....+a9=k , where k E { 9} where a_n >= 1 for 9 digit solving the equation a1+a2+a3+....+a8=k , where k E {9,8,8} for 8 digit solving the equation a1+a2+a3+....+a7=k , where k E {9,8,8,7,7,7} . . .

  9. dan815
    • one year ago
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    and then we got this general formula a1+a2+....+an = k a_m >=1 number of solutions to this must result from 1+1+...+(1+k-n)=k #of Solutions=k-n+n-1 choose n-1 =k-1 choose n-1

  10. ikram002p
    • one year ago
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    lavosh Danial will see in morning -.-

  11. dan815
    • one year ago
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    oh actually it might not be too much work from here to see how many non repeating numbers are there to cover the actual peuler question

  12. dan815
    • one year ago
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    |dw:1435712976484:dw|

  13. dan815
    • one year ago
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    \[\sum_{b=9}^{10} \sum_{m=1}^{b} (10-m)\sum_{k=1}^{m-1} \binom{m-1}{k}\]

  14. dan815
    • one year ago
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    I am gonna say that for any base 'B' it would be \[\sum_{b=B-1}^{B} \sum_{m=1}^{b} (B-m)\sum_{k=1}^{m-1} \binom{m-1}{k}\]

  15. dan815
    • one year ago
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    oh i found the inner one on wiki hehe theres a nice form to it, i think ive worked on the proof for this one too xD

  16. dan815
    • one year ago
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    |dw:1435714315667:dw|

  17. xapproachesinfinity
    • one year ago
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    *

  18. xapproachesinfinity
    • one year ago
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    the length of number does not matter?

  19. dan815
    • one year ago
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    it does the number of non bouncy numbers gets lower and lower for higher digits

  20. dan815
    • one year ago
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    its kind of long but this is pretty much closed form \[\sum_{b=9}^{10} 10*(2^B-1)-(B-1)*2^B -1 -10B + B*(B+1)/2\]

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