Problems & Puzzles: Puzzles

 Puzzle 830. Entry 1439 from Claudio Meller's site. Claudio Meller asked to find on integer N than can be decomposed in n factors (not prime factors!) such that increasing on unit to each factor of N you get another integer M such that M=2016*N Meller allowed to use unitary factors for N. Accordingly a Puzzler, Mmonchi sent the following valid solution. 1x1x1x1x1x1x1x1x1x2x2x2x6=48=N 2x2x2x2x2x2x2x2x2x3x3x3x7=M=48x2016=N*2016 I proposed to solve the same puzzle but avoiding to use unitary factors for N. Again Mmonchi sent the following solution: 2*2*2*2*2*2*2*2*2*2*2*2*3*3*3*3*3*3*3*3*3*6=N 3*3*3*3*3*3*3*3*3*3*3*3*4*4*4*4*4*4*4*4*4*7=2016*N Q1. Do you devise a strategy/algorithm in order to get these last kind of solutions for any M=K*N condition? Q2. Send your minimal N solutions for K= 2012, 2013, ... 2017.

Contribution came from Emmanuel Vantieghem

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Emmanuel wrote:

Here is my answer to question 2 :

K                                               N
2012   2*2*2*2*2*2*2*2*2*2*3*3*3*3*3*3*3*3*3*3*4*4*4*250*502 = 485664325632000
2013   2*2*2*2*2*2*2*2*2*2*2*3*3*3*3*3*3*3*3*3*4*4*10*60 = 386983526400
2014   2*2*2*2*2*2*2*2*2*2*2*2*3*3*3*3*3*3*3*3*3*12*18*52 = 905541451776
2015   2*2*2*2*2*2*2*2*2*2*2*3*3*3*3*3*3*3*3*3*4*4*12*30 = 232190115840
2016   2*2*2*2*2*2*2*2*2*2*2*2*3*3*3*3*3*3*3*3*3*6 = 483729408
2017   2*2*2*2*2*2*2*2*2*2*2*2*3*3*3*3*3*3*3*3*3*6*2016 = 975198486528

I'm not 100% sure  N  is minimal.

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