Problems & Puzzles:
Puzzles
Puzzle 139.
Sdf(x)=p
Gilbert Johnson, a Mathematics professor from New
Mexico, sent (May/01) the following question:
Let Sdf(n) be the
Smarandache double factorial function, i.e. the smallest integer
such that Sdf(n)!! is divisible by n, where:
m!! = 1x3x5x...xm if m is odd
m!! = 2x4x6x...xm if m is even.
Solve the diophantine equation: Sdf(x)
= p,
where p is a prime number (that is
to say find all the solutions for any p)
References:
C. Dumitrescu, V. Seleacu, "Some Notions and Questions in
Number Theory", Xiquan Publishing House, 1993. See his book online http://www.gallup.unm.edu/~smarandache/SNAQINT.txt
F. Smarandache, "Collected Papers", Vol. II, Kishinev
University Press, 1997.
Felice Russo’s book (studying the Smarandache double factorial
function among others) also online at http://www.gallup.unm.edu/~smarandache/FeliceRussobook1.pdf
Solution
The first solution came form the Russian graduate
student Ivan M. Godunov (21/5/01).To make crystal
clear his answer he uses as example p=101:
"Let's consider all odd numbers from 3 to 99:
S={3,5,7,9,11,13,15,...,97,99}.
(all) the solutions are:
 first 101;
 next 101 times one number from S: 101x3,
101x5, 101x7, ..., 101x97, 101x99;
 101 times two numbers from S: 101x3x5, 101x3x7, ... 101x97x98
(all combinations of two distinct numbers from the set S);
 101 times three distinct numbers from S: 101x3x5x7, 101x3x5x9,
... (all combinations of three distinct numbers from the set S);
 and so on, until we get 101 times all the numbers from the set
S: 101x3x5x7x9x...x99 which is 101!!"
Regarding the number of solutions Ivan wrote:
" For Sdf(x)=101 there will be:
number of solutions = 1 + 1C49 + 2C49 + 3C49 +...+ 49C49 = 2^49
because we have 49 odd numbers in the set S={3,5,7,9,...,99} and
we combine 101 times no number from
S, then 101 times one number from S, then 101 times two numbers from S,
and so on then 101 times all the numbers from S.
I think that generally the number of solutions will be 2^h, where
h=(p3)/2."
***
Ivan also takes the opportunity to make another
question around the same Smarandache double factorial function. The
question is this: If p is
an odd prime number of the form p = 2k+1, calculate Sdf( p^(k+2) )
***
Other people that solved the puzzle 139, at least
partially, was Kim Ngan from Saigon, Vietnam, and  independently  Giuseppe
DiCaprio from Milano, Italy.
***
Felice Russo sent (21/5/2001) the following answer
to the Ivan's question: "Sdf(p^(k+2))=p^2 if p=2*k+1",
but what is the formal proof of this answer? Russo himself wrote the day
after:
"the proof of the Ivanov's question is very easy.
In fact for any prime p=2*k+1 we have excatly k+1 odd multiple of p between
1 and p^2 and then k+2 times p. For example for p=7, k=3 and then between 1
and 49 we have 4 odd multiples of 7, that is: 7, 21, 35, 49 that means
1*3*5*7*11*13.......*49=1*3*5*7.....*(3*7).........*(5*7)*...........*(7*7)=
m*7*7*7*7*7=m*7^5 where m is an integer. Then by definition of Sdf function
it is abvious that Sdf(P^(k+2))=P^2."
***
By his side Kim Ngan wants the complete solution
the equation Sdf(x)=m, where m can be any integer (not necessarily prime).
***
Regarding this last question, Z. Dong, from
Singapore, wrote:
"I want to tell you a solution of the Kim
Ngan's problem Sdf(x)=m, where Sdf is the Smarandache double factorial
function. If is a product of primes to the first power only (square free
numbers), then we replace m by the maximum of these primes, and the problem
becomes equivalent to Gilbert Johnson's Sdf(x)=p=prime... I am not able to
solve the problem when m is not square free. I feel that it needs a research
article."
***
One surprise!... the solution to the original problem and
given by Ivan M. Gudonov has been corrected by Carlos Gustavo
Moreira from Río de Janeiro, Brazil, according to an email sent by Gudonov
himself:
" (for) Sdf(x)=p=prime, the number of
solutions is <= 2^k, where k=(p3)/2. In
my solution is was said equal, and should be less than or equal. The correct
solution (done by Prof. Carlos Gustavo Moreira) of the equation Sdf(x)=p=prime
is pxm, where m is any divisor of (p2)!!
And he give the example for the Smarandache double factorial function
Sdf(x)=17. The solutions are 17xm, where m is any divisor of (172)!! which
is equal to 3x5x7x9x11x13x15=(3^4)x(5^2)x7x11x13 which has
(4+1)x(2+1)x(1+1)x(1+1)x(1+1)=120.
The number is 2^7=128 because some solutions were counted twice, for
example: 17x3x5 is the same as 17x15 or 17x3x15 is the same as 17x5x9.
***
L. Kuciuk, from University of New Mexico, wrote:
...One can generalize this puzzle (see [1]) in the following way:
Let n and k be positive integers, with 1 <= k <= n1.
As an extension of the factorial and double factorial is defined the
Smarandache kfactorial of n as the below product of all possible strictly
positive factors:
SKF(n) = n(nk)(n2k)… .
Thus S1F(n) is just the wellknown factorial of n, i.e. n! =
n(n1)(n2)… 1, and S2F(n) is the wellknown double factorial of n, i.e.
n!! = n(n2)(n4)…, while S3F(n) is the triple factorial of n, i.e. n!!!
= n(n3)(n6)… .
Therefore some Smarandache type functions, such as the classical
Smarandache function, double factorial function, ceil functions, etc. are
extended using this kfactorial definition.
Hence, the Smarandache kfactorial function, Skf(n), is defined as the
smallest positive integer such that SKF(n) is divisible by n.
Some examples:
a) S3f(6)=6 because S3F(6)=6(63)=18 is divisible by 6 and it is the
smallest one;
S3f(16)=8 because S3F(8)=8(83)(86)=8(5)(2)=80 which is divisible by
16, and 8 is the smallest one with this property (I mean S3F(7), S3F(6),
and so on are not divisible by 16).
b) S8f(60)=18 because S8K(18)=18(188)(1816)=18(10)(2)=360 which is
divisible by 60, and S8K(17), S8K(16), etc. are not divisible by 60.
c) If p is prime, then all Skf(p)=p. Nice result!
Also Skf(n) <= n for any n.
Then Puzzle 139 becomes: Determine the number of positive integers x
such that Skf(x)=p, where p is a prime number.
Reference:
[1] M. Perez, The Smarandache kfactorial,
http://www.gallup.unm.edu/~smarandache/SKF.htm
***
