Talk:Steady squares: Difference between revisions
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And there comes the solution:By prepending a digit to n, than n*n must end in n to be a steady square.<BR>
So only solutions of one digit before can be solutions.
<
function CalcSquare(n:LongInt;Pot10:byte);
//pot10 is 10^(count of digits of n -1)
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// so n*n must end in n, to be a steady square
end;
</syntaxhighlight>
:Note also that it must end in 1,5 or 6 (not 0 as any number ending in 0 squared ends with twice as many zeros).--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 12:46, 21 December 2021 (UTC)
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--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 16:11, 3 January 2022 (UTC)
: Yes, the n.th digits sum up to 9 in [[Steady_Squares#Some_Examples]] for 80 digits without the last digit --[[User:Horst.h|Horst.h]] ([[User talk:Horst.h|talk]]) 16:41, 3 January 2022 (UTC)
<br>
: For each digit of the 5-series and 6-series (for the same number of digits), the first through the next-to-last digits (when present) sum up to 9, the final (ones) digit sums to 11.
: Also,
: for single digit, (10^1): 5 + 6 = 11
: for double digits, (10^2): 25 + 76 = 101
: for triple digits, (10^3): 625 + 376 = 1001
: ...pattern continues...
: So the 6-series (for any number of digits) can be obtained by subtracting the 5 series result (for that number of digits) from 10^(number of digits), then adding 1.
: For four digits (and a few others) the leading digit of the 5-series is a zero, so the 6-series result is 10001 - 0625 = 9376.
: When the first digit of the 5-series result is 9 (such as 90625), the corresponding 6-series result is a repeat of the previous number of digits (such as 9376), since the leading digit of the 6-series is zero.
--[[User:Enter your username|Enter your username]] ([[User talk:Enter your username|talk]]) 05:39, 5 January 2022 (UTC)
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