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June 12th, 2014, 07:57 AM   #1
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Problem 3

Thanks from raul21
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June 12th, 2014, 05:38 PM   #2
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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
Would you like to add the sequence?
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June 12th, 2014, 06:55 PM   #3
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Well, don't know what's being asked...

However, for [(n+1)(n+2)....(n+500)] / 500!
solution is:
(n + 500)! / (500! * n!)
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June 13th, 2014, 03:17 AM   #4
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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?
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June 13th, 2014, 10:51 AM   #5
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Quote:
Originally Posted by Hoempa View Post
If n = 2 then
$\displaystyle [(n+1)(n+2)....(n+500)] / 500! = 125751$
(n + 500)! / (500! * n!)
= (2 + 500)! / (500! * 2!)
= 502! / (500! * 2)
= 125751

All I was doing is providing an equation to get the TOTAL result...
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June 14th, 2014, 07:31 AM   #6
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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
The output of this program was 503, so it seems that this is some kind of solution. Althought I'm not quite convinced about it. For example, what Hoempa showed earlier, there are much much smaller numbers that could satisfy the condition. For example

$\displaystyle \binom{n + 5}{5}$

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?
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