Concatenating two consecutive integers
183 and 184
we get a six-digit number
183184, equalling
square of 428.List similar pairs, below
500,000 (final product).
How about the same task with a pair of a consecutve numbers
in a decreasing order?
Please provide a similar list (same constraints).
(In reply to
Analytic Solution (ascending) by Brian Smith)
Awesome proof.
This explains a lot of patterns in https://oeis.org/A030467
(they leave out the leading zero solutions)
If 10^D+1 is prime, there are no solutions. Since 11 and 101 are prime, the first solutions have three digit values of S.
If N is the number of prime factors, there will be 2^N-2 solutions
(again, unless some have leading zeros) Thus there are 6 solutions for D=3: 1001=7*11*13
But only 2 for each for D=4, 5, 6, 7, 8
But for D=9, 10^D+1=7*11*13*19*52579 so we should see 30 solutions. Interestingly there are only 20. There must be a lot with leading zeros. (S<sqrt(10^(9-1))=10000)
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Posted by Jer
on 2021-08-20 09:24:16 |
re: Analytic Solution (ascending)
