Problems & Puzzles: Puzzles

 Puzzle 599. Weird numbers Adrian Chow sent the following puzzle. I got a result about Weird Number. P*N is a weird number if: 1. N is a weird number and  2. P is a prime larger than the sum of divisors of N. Example 70 is the smallest weird number and  70 * 149 = 10430 is also a weird number. 10430 * 21601 = 225298430 is also a weird number. But I don't know why? Q. Can you provide an explanation of find a counterxample?

Contributions came from Jan van Delden, Fred Schneider, Paul Schmidt, Nick McGrath, Antoine Verroken & Emmanuel Vantieghem.

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Jan, Fred, Paul & Antoine, basically sent the same proof for the explanation asked by Adrian:

P*N is a weird number if:

1. N is a weird number and
2. P is a prime larger than the sum of divisors of N.

Weird numbers, by definition are abundant but not semiperfect.

Since P is prime and greater than N, the factors of P*N are the factors of N and P times each of the factors of N

e.g.

Factors of 70 = 1,2,5,7,10,14,35,70
Factors of 70*149 = 1,2,5,7,10,14,35,70,149,298,745,1043,1490,2086,5215,10430

The proper factors of P*N are the proper factors of N, N, and P times each of N

First, P*N must be abundant.

The sum of the proper divisors of P*N is F + N + P*F
F will denote the sum of all the proper divisors of N
F>N by definition of abundance
P*F > P*N
F + N + P*F > P*N
so P*N is abundant

P*N is semiperfect if the following two conditions hold:

F1 + N + P*F2 <> P*N   and   F1 + P*F2 <> P*N

Where F1 and F2 are equal to the sum of all or some of the proper divisors of N
F1 <> N and F2 <> N by definition if N is weird.
F will denote the sum of all the proper divisors of N

Let's solve F1 + N + P*F2 <> P*N first

Assume they are equal

F1 + N + P*F2 = P*N
(F1 + N + P*F2)/P = N
(F1 + N)/P + F2 = N
N and F2 are integers, so (F1 + N)/P must be an integer.
But, from the definition of the problem, P > (F+N),
F1 <= F, so F1+N < P
N > 0 and F1 >= 0 so (F1+N)/P > 0
therefore (F1+N)/P < 1 and is not an integer and
F1 + N + P*F2 <> P*N

The second equation: F1 + P*F2 <> P*N

Assume they are equal

F1 + P*F2 = P*N
F1/P + F2 = N
F1/P < 1 also, but, if F1=0, then F1/P is an integer
0/P + F2 = N
F2 = N but this contradicts the definition of F2 which must not equal N for N to be weird
therefore
F1 + P*F2 <> P*N

Both equations cannot equal the number P*N, so P*N is not semiperfect.

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Nick wrote:

Let's assume that P*N is not weird.

Let S be the set of divisors of N. Then the set of divisors of N*P are (S) U (S)*P, (where U = union)

Let s and t be any subsets of S.

P*N is not weird therefore there must be subsets s and t such that  s + P*t = P*N

If s is the empty set then  P*t = P*N   or  t = N.  Contradiction because N is weird.

Otherwise,  s/P + t = N   or   N - t = s/P

N - t   is an integer but  s/P is not because P>S  and s is a subset of S. Another contradiction.

Therefore our assumption that P*N is not weird is false.

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Emmnanuel wrote:

The result is not new.  It occurs without proof in
(page 5, last frame).
What follows now is my proof :
Let  N  be a weird number and let { d(0), d(1), ..., d(k) }  be the set of its divisorsBy definition of weirdness, s = d(0)+d(1)+...+d(k) > 2N  and no partial sum of the  d(i)  equals  N.

Let  P  be a prime number, > s.  Clearly, P is not a divisor of  N, whence the set of divisors of  P N  is

{d(0), d(1), ..., d(k), P d(0), P d(1), ..., P d(k) }.

The sum of all the divisors of  P N  is thus equal to  s + P s, >  2 P N. (***)

Now, assume  P N = d(i_1)+...+d(i_g)+P ( d(j_1)+...d(j_h) )  is a sum of some of its divisors.  Clearly, P  divides  d(i_1)+...+d(i_g). Since such a sum is smaller than  s, and  s  is smaller than  P,  that sum must be zero.  Thus, P N = P ( d(j_1)+...d(j_h) ), which implies that  N  is a sum of some of its divisors, impossible.  Thus, no partial sum of divisors of  P N  can equal  P N, which together with (***) implies that  P N  is a weird number-

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A. Verroken wrote:

Please will find in annex solution of p.599a ( Vantieghem ).

A weird number Q is abundant,sum of proper divisors > Q and no subset of the proper divisors sums to Q.

Q = W*P^n      W is a weird number            P is prime > sigma (W )          n > 0

1. because W is weird and a multiple of a abundant number is abundant ,Q is abundant
2. abundant : sigma(Q) – 2*Q > 0  à sigma(Q) – Q > Q , thus sum of proper divisors is bigger than Q.

3. no subset of the proper divisors of Q sums to Q.

a.       sigma(W) = A = d(0) + d(1) + … + d(k) ;  A < P  ;  A maximum = P – 1

b.      sigma(Q)  = B = A + A*P + A*P^2 + A*P^n because  B = A * ( 1 + P + .. P^n ) = A * [ P^(n+1)-1 ] / [ P – 1 ] ; with A, max. P – 1 , B max. = P^(n+1) – 1  à P^(n+1) > B max.
c.       s(‘): sum of some of the divisors of A
z’ : sum of some of the proper divisors of A
C :  sum of subset of proper divisors of Q
C = s’ + s’’*P + s’’’*P^2 + s’’’’P^3 + ...+ s# *P^(n-1) + z’*P^n
C/P^n = D = s’/P^n + s’’/P^(n-1) + … + s#/P + z’
s(‘) max. = P – 1
D max. = (P – 1 ) * [ 1/P^n + 1/P^(n-1) … 1/P ] + z’
D max. = [ P^(n+1) – 1 ] / P^(n+1) + z’ = E + z’
E max. is < 1 and z’ <> W à D <> W
Thus Q is a weird number if  W is a weird number and P(prime ) > sigma ( W )

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