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dan815
 one year ago
Modified P.E question:
Working from lefttoright 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
dan815
 one year ago
Modified P.E question: Working from lefttoright 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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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0By your modified definition, is the first bouncy number 11 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0nope, i think the question is number system sensitive

dan815
 one year ago
Best ResponseYou've already chosen the best response.3hey look at this pattern for the number of none bouncy numbers from 10 digit to 1 digit it follows this patternn

dan815
 one year ago
Best ResponseYou've already chosen the best response.3for 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
 one year ago
Best ResponseYou've already chosen the best response.3and then we got this general formula a1+a2+....+an = k a_m >=1 number of solutions to this must result from 1+1+...+(1+kn)=k #of Solutions=kn+n1 choose n1 =k1 choose n1

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0lavosh Danial will see in morning .

dan815
 one year ago
Best ResponseYou've already chosen the best response.3oh 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\sum_{b=9}^{10} \sum_{m=1}^{b} (10m)\sum_{k=1}^{m1} \binom{m1}{k}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.3I am gonna say that for any base 'B' it would be \[\sum_{b=B1}^{B} \sum_{m=1}^{b} (Bm)\sum_{k=1}^{m1} \binom{m1}{k}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.3oh 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

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0the length of number does not matter?

dan815
 one year ago
Best ResponseYou've already chosen the best response.3it does the number of non bouncy numbers gets lower and lower for higher digits

dan815
 one year ago
Best ResponseYou've already chosen the best response.3its kind of long but this is pretty much closed form \[\sum_{b=9}^{10} 10*(2^B1)(B1)*2^B 1 10B + B*(B+1)/2\]
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