June 12th, 2014, 07:57 AM |
#1 |

Member Joined: Jun 2014 From: Math Forum Posts: 67 Thanks: 4 | Problem 3 |

June 12th, 2014, 05:38 PM |
#2 |

Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 |
Interesting question! It asks for n > 1 such that $\displaystyle {n + 500 \choose 500}$ has no primefactors less than 500. Values of n such that $\displaystyle {n \choose k}$ has no primefactors less than k aren't as such in OEIS yet. Here are values for k = 1 to 30: Code: 1,2 2,3 3,7 4,5 5,23 6,7 7,143 8,44 9,159 10,11 11,47 12,13 13,2239 14,239 15,719 16,17 17,5849 18,19 19,2099 20,43196 21,14871 22,23 23,35423 24,193049 25,2105 26,36287 27,1119 28,29 29,240479 30,31 |

June 12th, 2014, 06:55 PM |
#3 |

Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 |
Well, don't know what's being asked... However, for [(n+1)(n+2)....(n+500)] / 500! solution is: (n + 500)! / (500! * n!) |

June 13th, 2014, 03:17 AM |
#4 |

Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 |
Maybe with some examples you see it; If n = 1, then $\displaystyle [(n+1)(n+2)....(n+500)] / 500! = {501 \choose 500} = 501$ Factoring 501 gives 3*167, which has primefactors less than 500. If n = 2 then $\displaystyle [(n+1)(n+2)....(n+500)] / 500! = 125751$ Factoring 125751 gives 3 * 167 * 251 which has primefactors less than 500. ... If n = m then $\displaystyle [(m+1)(m+2)....(m+500)] / 500! = {m + 500\choose 500}$. Factoring $\displaystyle {m + 500\choose 500}$ gives $\displaystyle \cdots$ which has no primefactors less than 500. What is the least possible positive integer value of m? |

June 13th, 2014, 10:51 AM |
#5 |

Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 | |

June 14th, 2014, 07:31 AM |
#6 |

Senior Member Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics |
All right. I'm definitely not a person who knows the number theory. But I searched this via google, and I was able to find this .pdf article. On the page 3 (or 205) of the paper, the writer says that if $\displaystyle n = s + m\cdot \left( s! \right)^2$, where $\displaystyle m$ is integer, then $\displaystyle \binom{n} {s}$ has not prime factors less than $\displaystyle s$. So. In case of $\displaystyle \binom{n+500}{500}$ if one choose $\displaystyle n = \left( 500! \right)^2$ then the condition should apply. To test this, I wrote a short python program that prints the smallest prime factor of the binomial coefficient: Code: def binomc(n, k): #binomial coefficient n over k result = 1 for i in range(1, k+1): result = result * (n-i+1) / i return result def smallestfact(l): #finds the smallest factor of l fac = 2 while True: if l % fac == 0: #if l is divisable by fac return fac else: fac += 1 def fact(l): #factorial of l f = 1 vv = 2 while vv <= l: f *= vv vv += 1 return f v = fact(500) v = v*v b = binomc(v+500,500) a = smallestfact(b) print a $\displaystyle \binom{n + 5}{5}$ for which Hoempa found $\displaystyle n = 18$ instead of this approach $\displaystyle n = 14400$. Perhaps someone who knows this kind of things better could clarify? |