LogFAQs > #986920111

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Topic	A divisibility by 11 problem you've probably never seen.
PurestProdigy 11/02/25 9:38:54 PM #22:	eaglei3 posted... This pattern will occur repeatedly forever base on the TCs statement. The pattern will follow this many required increases to N from the previous hit to find a satisfactory number divisible by 11. +21, +17, +5, +13, +9, +9, +13, +5, +17, +1 <repeat from start> S(21), S(38), S(43), S(56), S(65), S(74), S(87), S(92), S(109), S(110) are all divisible by 11. Now the sequence repeats: (Note S(131) is +21 from 110. S(131), S(148), S(153), S(166), S(175), S(184), S(197), S(202), S(219), S(220) are all divisible by 11. Next repetition starts with S(241). How the fuck do you figure this out without a calculator --- [[[[[[[[[[[[[[[\|\|\|\|\|\|\|\|\|\|\|\|\|]]]]]]]]]]]]]]]] <cite>PurestProdigy posted [p:986920111] in A divisibi... [t:81064497]...</cite> <quote>eaglei3 posted... </cite><quote>This pattern will occur repeatedly forever base on the TCs statement. The pattern will follow this many required increases to N from the previous hit to find a satisfactory number divisible by 11.+21, +17, +5, +13, +9, +9, +13, +5, +17, +1 <repeat from start>S(21), S(38), S(43), S(56), S(65), S(74), S(87), S(92), S(109), S(110) are all divisible by 11.Now the sequence repeats:(Note S(131) is +21 from 110.S(131), S(148), S(153), S(166), S(175), S(184), S(197), S(202), S(219), S(220) are all divisible by 11.Next repetition starts with S(241).</quote>How the fuck do you figure this out without a calculator --- [[[[[[[[[[[[[[[\|\|\|\|\|\|\|\|\|\|\|\|\|]]]]]]]]]]]]]]]]</quote> ... Copied to Clipboard!
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