Problems & Puzzles:
Puzzles
Puzzle 116. A=B+C, A*B*C is a primorial...
Naohiro
Nomoto asks for numbers A, B & C such that the following four rules
apply:
 A
= B + C
 A*B*C
is a primorial
 A,
B, & C are, each, a product of n of distinct
primes
 gcd(A,B)=gcd(A,C)=1
Nomoto
has found only two examples:
5
= 2 + 3
1105
= 231 + 874 [ 5*13*17 =
3*7*11 + 2*19*23 ]
Question
1: Find the next example
If
the condition 3 is relaxed and we let that A, B & C are the product of
distinct primes, not necessarily the same quantity, example:
10
= 3 + 7 [ 2*5 = 3
+ 7 ]
There are
some more examples, being the largest currently found, the following one:
496961=495726+1235
[
17x23x31x41 = 2x3x7x11x29x37 + 5x13x19 ]
Question
2. Find the next 3 examples
Question
3. Can you argue if this sequence is finite or infinite?
___________
Notes:
All the numbers for the sequence described in the question 2 are a subset
of the
Sloane's integers
sequence A057035.
Hakan Summakoğlu wrote (May
2011)
Q2: 219965 = 213486 +
6479 [ 5x29x37x41 = 2x3x7x13x17x23 + 11x19x31]
***
Jan van Delden wrote (May 2011)
I was wondering about
the number of solutions of Puzzle 116, I thought that could be improved.
Must have missed this one..
My contribution:
Q1/Q2:
We have P:=p[n]#=a.b.c and a=b+c. Given P there are two equations with 3
unknowns. We can therefore (for instance) choose b and a and c follow. From
this one can deduce that if b^2+4P/b is a square, say D^2, we have: a=(D+b)/2
and c=(Db)/2.
Furthermore the minimum number of factors of P to be assigned to a can be
calculated beforehand (using the largest prime factors of P), leading to the
maximum number of factors to be assigned to b. Unfortunately this maximum is
about n/3, which still gives a lot of tests to perform!
I found the following 20 solutions, in order of appearance (primarily Q2):
5=2+3 (Q1)
10=3+7 2x5=3+7
35= 2+33 5x7=2+3x11
22= 7+15 2x11=7+3x5
21=11+10 3x7=11+2x5
55=13+42 5x11=13+2x3x7
221=11+210 13x17=11+2x3x5x7
182=17+165 2x7x13=17+3x5x11
231=10+221 3x7x11=2x5+13x17
442=57+385 2x13x17=3x19+5x7x11
357=247+110 3x7x17=13x19+2x5x11
1430=119+1311 2x5x11x13=7x17+3x19x23
1105=874+231 5x13x17=2x19x23+3x7x11 (Q1)
2958=1495+1463 2x3x17x29=5x13x23+7x11x19
23345=374+22971 5x7x23x29=2x11x17+3x13x19x31
496961=1235+495726 17x23x31x41=5x13x19+2x3x7x11x29x37
219965=6479+213486 5x29x37x41=11x19x31+2x3x7x13x17x23
144210=16523+127687 2x3x5x11x19x23=13x31x41+7x17x29x37
226083=6118+219965 3x11x13x17x31=2x7x19x23+5x29x37x41
106981=57646+49335 7x17x29x31=2x19x37x41+3x5x11x13x23
No further solutions with n<=35
Q3:
I computed the minimum absolute distance between b^2+4P/b and the nearest
square (where no solution was found), let's call this dist(D^2). The
smallest distance 103 was found (n in [12,1426]) at n=18, b=12509182. At
first sight this is large, however a and c fail to be integer by a mere
0.000000133 (=dist(a),dist(b)). However the resulting distance between P and
a.b.c is huge, since a.b.c=(D+b)/2.b.(Db)/2=b.(D^2b^2)/4, this gap
is of the order b.dist(D^2)/4=12509182.103/4.
In general dist(a)=dist(b)=dist(D^2)/(4D) and dist(P)=b.dist(D^2)/4. And
dist(D^2) is maximal D+1/4. More should be known about the distribution of
dist(D^2) to say more.
***
