Problems & Puzzles:
Problems
Problem 19. Divisors (I): The least number N(d)
with d number of divisors.
N(2) = 2 = is the least number with 2 divisors (1
& 2)
N(3) = 4 = is the least number with 3 divisors (1, 2,
& 4)
N(4) = 6 = is the least number with 4 divisors (1, 2, 3,
& 6)
a) Find N(4000), N(4001) and N(4002),
respectively.
b) Can you explain the
general algorithm to calculate N(d)?
a) Jud McCranie, T.W.A.
Baumann & Enoch Haga, solved independently
the part a) of this puzzle:
N(4000) =
2^4*3^4*5^4*7*11*13*17*19 = 261891630000
N(4001) = 2^4000
N(4002) = 2^28*3^22*5^2*7 = 1474163083033391923200
b) Jud
McCranie, T.W.A. Baumann & Enoch Haga sent
basically the same procedure to find N(d) for a given d:
1. Factorize d as a
product of his prime divisors:
d = p1^a1*p2^a2*p3^a3...
2. convert this factorization in another arithmetically
equivalent factorization, composed of nonpowered
monotonically decreasing & not necesarilly prime
factors....(uf!...)
d = p1^a1*p2^a2*p3^a3... = b1*b2*b3... ; such that
b1=>b2=>b3...
You must realize that for every given d, there are
several arithmetically equivalent factorizations that can
be done: by example:
if d = 16 =2^4 then there are 5
equivalent factorizations:
d = 2*2*2*2 = 4*2*2 = 4*4 = 8*2 = 16
3. N is the minimal number resulting of computing
2^(b11)*3^(b21)*5^(b31)*... for all the equivalent
factorizations of d. Working the same example:
N(16) = the minimal of
these{2*3*5*7, 2^3*3*5, 2^3*3^3, 2^7*3, 2^15}= 2^3*3*5 =
120
***
I have the 'feeling' that a more detailed analysis
can be done to guide &/or
optimize the execution of the steps 2 & 3 of the
above procedure. As far as I have been thinking about it,
I believe the following:
From all the
possible factorizations of d you need only consider those
which the sum of the factors is equal to the sum of all
the nonpowered prime factors.
If this is truth,
this also means that the minimal N is only one of some
particular of all the possible factorizations of the
product monotonically decreasing of all the nonpowered
odd primes of d by 2^z, z=>0, that is to say that you
need to consider only all the possible minimal
factorizations of 2^z.
Let's work another example: if d = 240 = (5*3)*(2*2*2*2)
then you need to consider only the different minimal
factorizations of 2*2*2*2; so the only factorizations of
d = 240 to consider are 5*3*2*2*2*2, 5*4*3*2*2 &
5*4*4*3.
Then, N(240) = the minimal of these{2^4*3^2*5*7*11*13,
2^4*3^3*5^2*7*11, 2^4*3^3*5^3*7^2}= 2^4*3^2*5*7*11*13 = 72072
Of course that if d is odd, the factorization of d that
produces the minamal value of N is simply that obtained
by the product of all the non powered prime factors of d
in monotonically decreasing order.
I bold the fact that I don't know if all this is
absolutely true. All I have is some empirical evidence
confirming this conjecture.
Can you
simplify/profundize this?
***
On March 08 Christian Bau
wrote:
The solution so far
on the problem page is the following: Find all
possible ways to write the integer d as the product of factors >= 2
which are decreasing in size, that is d = n1 * n2 * n3 ... * nk, where
n1 >= n2 >= n3 >= ... >= nk >= 2. For each such product, calculate the
value
x = 2^(n1  1) * 3^(n2  1) * 5^(n3  1) * ... * pk ^ (nk 1).
Then N is the smallest of all these values x.
For small d, like d = 20, all possible ways of writing d as a product
are easily found, but this method is quite unusable for large d, like
d = 100! because of the enormous number of possible candidates. We
need to find severe restrictions on the possible factorisations of d
for such a large number.
Assume we found an x as above, that is x = 2^(n1  1) * 3^(n2  1) *
5^(n3  1) * ... * pk ^ (nk 1). n1 is either the largest prime factor
of d, or a composite number. Assume that n1 is not a prime, so we can
write n1 = a*b, where a >= b >= 2. Assume that p is the first prime
number that isn't used in the product that gives x. We can find
another number y that has exactly the same number of divisors as x by
letting
y = 2^(a1) * 3^(n2  1) * 5^(n3  1) * ... * pk ^ (nk 1) * p^(b1).
If x were the smallest number with d divisors, then we have
necessarily x < y and therefore 2^(ab1) < 2^(a1) * p^(b1). This is
equivalent to 2^(a(b1)) < p^(b1) or 2^a < p.
This is a severe restriction: In a concrete example, if p = 103 it
means that 2^a < 103 or a <= 6. So ab must not be the product of two
numbers with the larger one >= 7. This leaves ab = 2*2, 3*2, 4*2, 5*2,
6*2, 3*3, 5*3, or 5*5 (not for example 4*4, because that is equal to
8*2 with a factor greater than 6). In this example, if n1 is composite
then it is at most 25. So the possible candidates for n1 are either
the largest prime factor of d, or those of the numbers n1 = 25, 15,
12, 10, 9, 8, 6, 4 that are greater than the largest prime factor.
We turn this now into an algorithm. Given is the number d. We want to
find the numbers n1, n2, n3 etc. in the factorisation N = 2^(n11) *
3^(n22) * 5^(n31) etc. of the smallest number N with exactly d
divisors.
Let i = 1 and D = d. Repeat the following steps as long as D > 1:
1. Find the complete prime factorisation of D. Let pmax be the largest
prime factor of D. Let k be the total number of prime factors of D.
Let p be the (k + i  1)th prime number, and let pi be the ith prime
number. Let m = floor (log (p) / log (pi)). The first possible
candidate for ni is the prime number pmax. Further candidates are all
the products a * b, where b is a prime number 2 <= b <= m which is one
of the factors of D, a is an integer in the range b <= a <= m which
has no prime factors less than b, with the further property that a*b
is a factor of D, and a*b > pmax. Finally, if i > 1 then a*b must not
be greater than the previous exponent n(i1).
2. For each of the possible numbers ni: Choose ni  1 as the power for
the prime number pi in the product for x. Divide D by ni, and let i = i
+1.
As a concrete example that works out very well, let d = 100! The
factorisation of d is d = 2^97 * 3^48 * 5^24 ... * 83 * 89 * 97. There
are exactly 239 factors, and the 239th prime is 1493. floor (log
(1493) / log (2)) = 10, so either n1 = 97 or n1 is a product with the
largest factor up to 10. The largest number with no factor greater
than 10 is 49, therefore n1 = 97. In the same way, floor (log (1493) /
log (3)) = 6, so either n2 = 89 or n2 is composite with no factor
greater than 6, which makes n2 = 89, then n3 = 83 and so on.
As another example, let d = 2^238 *5. Again we have 239 factors, the
239th prime is 1493, and either n1 = 5 or n1 is composite with no
factor greater than 10. Here the possible composite values for n1 are
4*2, 5*2, 8*2 and 10*2 which we have to try. We will again find that
n2 should have no factor greater than 6 if it is composite, so for n2
the choices will be n2 = 5, 8 or 10, and the choices for n3 and n4
etc. are even further reduced.
***
