Problems & Puzzles: Puzzles

Puzzle 784. Find some more solutions Claudio Meller in his entry 1392, poses a challenge from Eric Angelini. Write infinite copies of an integer n, C= n.n.n.n.... Obtain the successive products of the digits of C from left to right, P1, P2, ... Are there n integers such that n is equal to one of these products? Example: 735 = 7*3*5*7 Only two  more solutions are known: 18432 = 1* 8 * 4 * 3 * 2 * 1 * 8 * 4 * 3 442368 = 4 * 4 * 2 * 3 * 6 * 8 * 4 * 4 * 2 * 3 (Reported to Claudio, by Vicente Izquierdo) Q. Find more solutions.

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Contributions came from Jan van Delden, Emmanuel Vantieghem and Giovanni Resta

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Jan wrote:

The digits in our number n have divisors {1,2,3,5,7}.
If n is a product of digits, as specified, it must be 7-smooth.

Clearly we can’t have 0 as a digit.

 

Moreover:

if n is odd the prime-divisors must be in the set {3,5,7}.
if n is even the prime-divisors must be in the set {2,3,7}.

 

First construct a sieve using these primes, say between 10^(k*10) and 10^((k+1)*10).

The  maximum number of smooth numbers in such a set,

if n is odd:    789

if n is even: 1733.

 

I found no more solutions having 10000 digits or less, except the trivial solutions: 0..9.
However, I can’t think of a reason that no more solutions are possible.
I only suspect (I didn’t test this) that the probability that a 0 emerges approaches 1 (fast).

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Emmanuel wrote on "Tue, Apr 28, 2015 at 10:51 AM"

To find solutions, I am inspired by David Wilson's appraoch in his mail to the SeqFan's mail list (see : http://list.seqfan.eu/pipermail/seqfan/2015-April/014748.html ).

There, he observes that a solution of the problem must be an integer of the form  (2^a)*(3^b)*(5^c)(7^d)  with no zero among its digits.
 

I made the list of all  m <= 2^300  (and > 1) of such numbers, eliminating those  m  that have both the digits 2 and 5. My list has  1809  integers.

Only one extra solution appeared :

3682784876146817236992 = 3*6*8*2*7*8*4*8*7*6*1*4*6*8*1*7*2*3*6*9*9*2*3*6*8*2*7*8*4*8*7*6

=(2^37)*(3 ^13)*(7^5)

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Giovanni wrote on "Wed, Apr 29, 2015 at 9:03 AM"

up to 10^100 there is only another solution, apart the little ones
(735,18432,442368), i.e.

3682784876146817236992 = (3*6*8*...*9*2) * (3*6*8*2*7*8*4*8*7*6).

If we allow to multiply digits from both ends
of the number, like in

4794391461888 = 8*8*8*(4*7*9*4*3*9*1*4*6*1*8*8*8)*4*7,

then the solutions up to 10^100 are
128, 175, 384, 735, 1296, 18432, 34992, 442368, 4128768, 13395375,
13436928, 161243136, 1269789696, 4161798144, 149824733184,
611784327168, 4794391461888, 2877833474998272, 3682784876146817236992.

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Later, on May 2, Jan wrote again:

"I checked and my routine had a flaw. I agree with the other results."

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