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(a)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.
