Talk:Sub-unit squares: Difference between revisions

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Revision as of 13:13, 27 September 2022 (view source) Nigel Galloway (talk \| contribs) (An alternative method) ← Older edit		Latest revision as of 13:25, 27 September 2022 (view source) Nigel Galloway (talk \| contribs) m (→‎An alternative method)
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Line 21: ==An alternative method== I am going to start by solving a relaxation of the task description. For ana z digit number find perfect squares (gg) ending in 5 such that (gg)+10<sup>z</sup>/9 is also a perfect square. g will be the sequence of integers between ⌈√10<sup>nz</sup>⌉ and ⌊√10<sup>nz+1</sup>-1-10<sup>nz</sup>/9⌋ ending in 5. For 4 digit numbers the candidates (g) are 25 35 45 55 65 75 85; For a given g I form the next perfect square by adding 2g+1. So for g=45 I can form a series of perfect squres: Line 27: g<sup>2</sup>+91 g<sup>2</sup>+91+93 g<sup>2</sup>+91+93+95 ..... so for 45*45<sup>2</sup>+1111 to be a sub-unit square 91+93+95+... must equal 1111. 91 184 279 376 475 576 679 784 891 1000 1111 ...