Unsolved? problem

[Richard Guy]

Dear Subbarao,
               Good to hear from you and
to know that the old brain is still
ticking over.  I think that Murray's
problem goes back quite a bit earlier.

Singmaster's Sources doesn't mention
this particular problem, but has
references to similar ones back to
1854.

On writing it in the form

     9 * (10x + y)  =  y * 10^k + x

we have
          89x = y(10^k - 9)

and  89  is a `semi-long' prime
with the period for  1/89  having
length  (89 - 1)/2 = 44, the number
of digits in the answer.

If you want even more spectacular
examples, look at

    (10^k - 1) * 10^k  -  1

where Murray's problem has  k = 1.

k = 6  and  k = 9  give primes, but
I don't know what their order is,
mod 10.  Best,    R.

On Thu, 28 Apr 2005, msubbara wrote:

> Dear Richard:
> It is indeed a good idea to collect Murray Klamkin problems. I am particularly
> fond of one his problems-but it did not appear in Math journals, but in the
> book by L.A.Graham : Ingenious Mathematical problems and methods, Dover.
> Murray's problem,which appears as #72 in the book,reads: find the smallest
> number such that if the last digit is removed and placed at the  beginning to
> become the  first digit, this new number is nine times the original one.
> The solution  given in the book itself is the astronomical fijure
>
> 10,112,359,550,561,797,752,808,988,764,044,943,820,224,719.
> What would be the solution if we do similar operation with the last two digits
> and ask '99 times the original number", or replace the last three digits
> similarly to   get 999 times the original number etc.
> Subbarao
>
>
>
>
>
>
>
>
>> ===== Original Message From Richard Guy <rkg at cpsc.ucalgary.ca> =====
>> I'm collecting Murray Klamkin
>> problems and solutions and am
>> currently going thru Math Mag.
>>
>> I came across Problem 886,
>> Math Mag 48(1975) 57--58
>> [nothing to do with Murray]
>> which isn't properly stated
>> but should read as in OEIS
>> A003508 :
>>
>> a(n) = a(n-1) + 1 + sum of
>> distinct prime factors of
>> a(n-1) that are < a(n-1).
>>
>> This leads to
>> 1,2,3,4,7,8,11,12,18,24,30,41,42,55,...
>>
>> The original problem asked that
>> if you start elsewhere, e.g.,
>>
>> 5,6,12, ...  or
>> 9,13,14,24, ... or
>> 10,18, ... or
>> 15,24, ...
>>
>> do you always merge with the
>> original sequence?  Evidently
>>
>> 91,112,122,186,... takes a little
>> while.
>>
>> Has anyone ... Can anyone prove
>> Charles Trigg's guess ?    R.
>
>