Problems & Puzzles:
2013 and 2014
reported in his entry
1061 that 2013 & 2014 share a property:
2013 = 3 x 11 x 61 & 2013
+ 3 + 11 + 61 = 2088
2014 = 2 x 19 x 53 & 2014
+ 2 + 19 + 53 = 2088
Can you get the smallest triplet of this type*?
* Perhaps this is a cousin-problem of the
unsolved problem posed by Shyam Sunder Gupta: "find three
consecutive numbers whose sum of divisors is same". As a
matter of fact this will be the matter of the next
Contributions came from Emmanuel Vantieghem & Jan van Delden.
Emmanuel Vantieghem wrote:
A triplet must be of the form 8m+5, 8m+6,
There is no triplet whose members are less
This is my proof :
I restricted my attention to squarefree
It is then clear that a triple of this type must be
of the form 4m+1, 4m+2, 4m+3.
Asume 4m+1 = p*q*r, 4m+2 = 2*s*t, 4m+3 = u*v*w are the
prime factorizations. Then, an even number of prime divisors
of 4m+1 are of the form 4k+3. In all cases, p+q+r will be of
the form 4a+3. The condition p*q*r +p+q+r = 2*s*t+2+s+t then
implies p+q+r = 3+s+t, which implies s+t =
4c. Since s and t are odd primes, one must be 4b+1, the
other 4c-1. This happens only when s*t = 4d+3. Thus, 4m+2 =
Jan van Delden wrote:
I searched until 3.1*10^9 but found no triplet.