[seqfan] Re: N = 3417. Now replace a digit by the "multiply" sign
Nick Matteo
kundor at kundor.org
Wed Aug 28 22:32:03 CEST 2019
Hi Éric,
There are infinitely many hits.
Let me use * to denote multiplication, and concatenation to mean
concatenation of digits.
As Robert Israel points out, if P*Q | PDQ for some digit D,
then P*Q0 divides PDQ0, so there are infinitely many examples that way.
But if we discount multiples of 10 for P and Q, and also discount any
Q with leading zeros,
we can still create infinitely many examples.
Let Q be the 2k+2 digit number consisting of 90 (k times) and ending 91
(e.g. 91, 9091, 909091, 90909091...)
and P be 9Q (so 991, 99091, 9909091, 990909091...)
Then P*Q | P9Q. Notice, since P is 9Q, this number is PP; PP / P =
10...01, where there are 2k+2 zeros.
Such a number factors as 11 * 90...9091, where there are k 90s, i.e. as 11 * Q.
So 11*P*Q = P9Q.
Cheers,
Nick Matteo
On Tue, Aug 27, 2019 at 10:08 AM Éric Angelini <bk263401 at skynet.be> wrote:
>
> Hello SeqFans,
>
> Take an integer N (for example N = 3417).
> Erase one of N's digits to form integers P and Q.
> (For N = 3417, erasing 4 produces P = 3 and Q = 17).
> If P*Q divides N, we have a hit.
> (this is the case here as 3*17 = 51 and 3417/51 = 67
> although the other possible erasure doesn't produce
> an integer: 34*7 = 238 and 3417/238 = 14.35714...)
>
> Jean-Marc Falcoz has computed 10,000 such « hits »
> (with N > 99 and Q that cannot have a leading 0,
> except if Q = 0); he noticed that the density of
> hits decreases dramatically when N gets bigger and
> bigger. Which leads to this: is the hit-sequence
> finite? More pictures and data here, for those
> who might be interested:
> https://bit.ly/2Lag2Fh (in French).
> Best,
> É.
>
> --
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