My Birthday Problem

[Robert Morewood]

Every year on my birthday, I contemplate my age (in years) trying to find something interesting and preferably unique. This year's contemplation has led to problems requiring only basic tools to comprehend:  The definition of prime divisor along with multiplication, addition, and subtraction.  However, larger numbers appear rather quickly and leading to what may be a very difficult problem.

Since my age is a (larger) prime number, the expression:
n!+1 = (n)*(n-1)*(n-2)*...*(4)*(3)*(2) + 1

came to mind.  It has no non-trivial divisors less than or equal to "n",

hence shows that this IS a prime number larger than "n", for any "n".

Indeed,  my age does divide n!+1 for a particular positive integer "n"

(with "n" less than my age).  In fact, it is the smallest prime divisor.

But that is really not very interesting: 

Every prime number satisfies that condition for some "n".

(A consequence of Wilson's theorem.)

However, changing add to subtract and multiply to add,

my age is also the largest prime divisor of:

(n)+(n-1)+(n-2)+...+(4)+(3)+(2) - 1 = (₏₊₁C₂) - 2; 

(using the same value for "n").

Hence the problem:

My age is (for some "n") both the smallest prime divisor of:

n! + 1 = (n)*(n-1)*(n-2)* ... *(4)*(3)*(2) + 1

and also (with the same "n") the largest prime divisor of:

(₏₊₁C₂) - 2 = (n)+(n-1)+(n-2)+ ... +(4)+(3)+(2) - 1

and it is the smallest such prime.  How old am I?

( This is easily solved:

(  A determined numerical search will get it quick enough. 

(  Spreadsheets or graphing calculators can do it immediately

(  by seeking the greatest common divisor and checking.

I not yet been able to find a second number satisfying both conditions.
However,  searching the greatest common divisors of n!+1 and (₏₊₁C₂)-2,

I have found precisely one other value not equal to 1:  My mother's age!

(Her age is the SECOND smallest prime divisor of that n!+1).

Are there any other solutions to:  GCD[ n!+1, (₏₊₁C₂)-2 ] ≠ 1 ?

[David Ash]

( This is easily solved:

(  A determined numerical search will get it quick enough. 

(  Spreadsheets or graphing calculators can do it immediately

(  by seeking the greatest common divisor and checking.

It can, indeed, be solved quickly using the above approach.

But it can be solved even faster by noting two facts that you have provided: that you are a former participant in Competition Corner, and your age is a prime number. Given the limited time period over which Competition Corner ran, those two facts pretty much narrow it down :)

[Robert Morewood]

Nice to hear from a fellow Waterloo Workshop participant!

David observed that on the problem of my age:

it can be solved even faster by noting two facts that you have provided: that you are a former participant in Competition Corner, and your age is a prime number. Given the limited time period over which Competition Corner ran, those two facts pretty much narrow it down :)

Of course my students will not know, without considerable research, that I was a participant in the Competition Corner nor in what years it ran!

Nonetheless, I am rather more interested in the question:

Is  GCD[ n!+1, (₏₊₁C₂)-2 ]  = 1 for sufficiently large integers "n" ?

I can rule out (infinitely) many primes factors, but not all... 

Is my age the ONLY number which is both the smallest prime divisor of n!+1 while simultaneously being the largest prime divisor of (₏₊₁C₂)-2?