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May 25, 2016 at 21:45 comment added Barry Cipra Really nice! I've added a link to it in an answer to a related question, math.stackexchange.com/questions/1788290/…
May 25, 2016 at 9:35 comment added awllower Thanks for the nice explanations!
May 25, 2016 at 8:41 comment added user284001 @joriki Bravo Sir, bravo!
May 25, 2016 at 8:34 comment added joriki @awllower: I started from both ends, I saw that both squares differ by just one from a multiple of $10$, so that left $70\cdot5^9\le9\cdot8^8$ to prove, then I realised that $5^3\approx2^7$, and that left $70\lt72$.
May 25, 2016 at 8:24 comment added awllower May I ask how did you come up with these inequalities? Thanks!
May 25, 2016 at 8:20 vote accept hola
May 25, 2016 at 8:08 history answered joriki CC BY-SA 3.0