Re: Initial digits of squares
- From: bert <bert.hutchings@xxxxxxxxxxxxxx>
- Date: Thu, 26 Mar 2009 11:25:00 -0700 (PDT)
On 26 Mar, 17:15, Zak Seidov <zaksei...@xxxxxxxxx> wrote:
Dear you,
is it known (or possible to prove)
that for any set of m digits,
say, m=9, 314159265 (see Pi?),
there is/are square/s n^2
beginning with given set?
In our case, smallest "soln" is
n=177245385, n^2=31415926503798225.
What about way to calculate n for more digits of Pi,
say, m up to 100?
I was all ready to reply that you should just extend
the required leading digits with the same number of
digits 9, then round down its integer square root,
but when I tested this with m = 30, I found that:
56049 91216 39792 86993 11282 43387^2 begins with
31415 92653 58979 32384 62643 38328 ..., while
56049 91216 39792 86993 11282 43386^2 begins with
31415 92653 58979 32384 62643 38326 ..., thus there
is no 60-digit square beginning with the desired
31415 92653 58979 32384 62643 38327 ...
You have to tack on one extra 9-digit, then you can
get a 61-digit square:
1 77245 38509 05516 02729 81674 83341^2 begins with
3 14159 26535 89793 23846 26433 83278 ...
--
.
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