dan815
  • dan815
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
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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ganeshie8
  • ganeshie8
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Astrophysics
  • Astrophysics
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ganeshie8
  • ganeshie8
By your modified definition, is the first bouncy number 11 ?

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dan815
  • dan815
yep that is true
Astrophysics
  • Astrophysics
Binary?
ganeshie8
  • ganeshie8
nope, i think the question is number system sensitive
dan815
  • dan815
hey look at this pattern for the number of none bouncy numbers from 10 digit to 1 digit it follows this patternn
dan815
  • dan815
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} . . .
dan815
  • dan815
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
ikram002p
  • ikram002p
lavosh Danial will see in morning -.-
dan815
  • dan815
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
dan815
  • dan815
|dw:1435712976484:dw|
dan815
  • dan815
\[\sum_{b=9}^{10} \sum_{m=1}^{b} (10-m)\sum_{k=1}^{m-1} \binom{m-1}{k}\]
dan815
  • dan815
I am gonna say that for any base 'B' it would be \[\sum_{b=B-1}^{B} \sum_{m=1}^{b} (B-m)\sum_{k=1}^{m-1} \binom{m-1}{k}\]
dan815
  • dan815
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
dan815
  • dan815
|dw:1435714315667:dw|
xapproachesinfinity
  • xapproachesinfinity
*
xapproachesinfinity
  • xapproachesinfinity
the length of number does not matter?
dan815
  • dan815
it does the number of non bouncy numbers gets lower and lower for higher digits
dan815
  • dan815
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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