# Super Star Numbers

A Super Star Number is a positive integer N, such that the 21 vertices of the super star below can be labelled with different positive integers so that the product of the three numbers in any of its 14 straight lines is precisely N.

• a) Find the first two consecutive numbers each of which is a Super Star Number.

• b) Are there arbitrarily long sets of consecutive numbers all of which are Super Stars? Visitors to A Magic Super Star have found a few Super Stars, but so far no two which are consecutive.

There are 75,863 Super Star Numbers not greater than one million. Within this range there are 237 pairs of consecutive Super Star Numbers, yet no triplets.

The first pair, with solutions shown below, is

70,784 and 70,785

Super Star Number: 70,784 Super Star Number: 70,785 Another noteworthy pair is

999,999 and 1,000,000

Just after posting, I notice that my program has reported the first triplet of Super Star Numbers:

1,447,550
1,447,551
1,447,552

Given this, it seems likely that there are longer sets of consecutive Super Star Numbers.

Given four distinct primes p,q,r,s and a positive integer X, the number $$p^{12}q^{12}r^3s^3X$$ is superstar by the following construction:

                                        2 6 2 0
p q r s

7 1 0 2    3 5 0 2
p q r s X  p q r s

3 5 1 1    2 6 1 1    7 1 1 1
p q r s    p q r s    p q r s X
1 7 2 0                                       5 3 2 0
p q r s X                                     p q r s
8 0 0 2                 8 0 1 1    4 4 1 1    0 8 1 1                 0 8 0 2
p q r s                 p q r s    p q r s X  p q r s                 p q r s
3 5 2 0                                       7 1 2 0
p q r s                                       p q r s X
1 7 1 1    6 2 1 1    5 3 1 1
p q r s X  p q r s    p q r s

5 3 0 2    1 7 0 2
p q r s    p q r s X

6 2 2 0
p q r s


Given positive n choose 4n distinct primes $$p_1,q_1,r_1,s_1,p_2,q_2, ..., r_n,s_n$$ and use the Chinese remainder theorem to find positive N such that $$N\equiv-j \mod p_j^{12}q_j^{12}r_j^3s_j^3$$ for $$1\le j\le n$$. Then $$N+1,N+2,...,N+n$$ are all superstar.

• Why $p^{12}q^{12}r^3s^3X$? Was that just the construction you had? Would CRT work just as well with $p^2q^2rsX$ or $p^3q^2r^2X$? Apr 22 at 10:46
• @DanielMathias The problem is that we cannot make many assumptions about the X the CRT is going to give us, for example, it may not be coprime with p,q,r,s. So we need a way to make sure that no matter what the X the 21 numbers will still be guaranteed to be distinct. One easy way (certainly not the only) is to fix the number of prime factors (with multiplicity). You can check that the sum of exponents of p,q,r,s is always 10 so if X is not 1 and you compare a number with an X in it and one without they cannot be the same because the one with X has more prime factors. Apr 22 at 16:07
• I see. This guarantees the existence of a solution without relying on additional proof that a solution exists for any $N$ whose prime factorization meets certain criteria. Apr 22 at 21:04
• @DanielMathias No, I think there is a misunderstanding. The argument goes: Why is N+j superstar? Because by consruction of N we have $N\equiv -j \mod p_j^{12}q_j^{12}r_j^3s_j^3$ which is tantamount to saying there exists $X_j>0$ such that $N + j = p_j^{12}q_j^{12}r_j^3s_j^3X_j$ and now even though we know almost nothing about $X_j$ we can conclude that $N+j$ is superstar. Apr 22 at 21:31
• No misunderstanding. We can conclude that $N+j$ is superstar because we already have a construction that incorporates an arbitrary $X$ value. If we instead found $N\equiv-j \mod p_j^2q_j^2r_js_j$, we would need to prove that all multiples of a superstar number are themselves superstar numbers in order to make the same conclusion. Apr 22 at 22:52