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 Divisibility in a sequence (Posted on 2017-08-12)
Show that in every sequence of 79 consecutive positive integers, there is a positive integer whose sum of digits is divisible by 13.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 Existence proof (lowest pair requiring 79) Comment 2 of 2 |
(In reply to proof (but not a proof that this is the lowest achievable) by Charlie)

The triggering of a change in the higher-order digits' sum mod 13 is ordinarily 1, except when the last digit is 9. In that case the mod-13 value decreases by 8 or increases by 5, again, except when the next digit to the left is again 9, etc.  If x represents a set of digits not ending in 9, then x9 results in +5; x99 represents +9; x999, +0; x9999, +4;x99999,+8; x999999,+12;x9999999,+3;x99999999, +7; what we are looking for.  The last two digits of the first number are 60, with a mod-13 value of 6, while 99999999 has a value of 7, so the value of the last ten digits is 13 or zero and we don't need to add more. So, if I'm right 9999999960 would be the first case where the next sod with a value of 0 mod 13 would be 79 higher, at  10000000039. The sod's of each of these numbers is in fact divisible by 13. Finding the mod values of numbers between, shows

`    n     sod q r    where q is the quotient when sod is divided by 13 and r the remainder9999999960 78 6 09999999961 79 6 19999999962 80 6 29999999963 81 6 39999999964 82 6 49999999965 83 6 59999999966 84 6 69999999967 85 6 79999999968 86 6 89999999969 87 6 99999999970 79 6 19999999971 80 6 29999999972 81 6 39999999973 82 6 49999999974 83 6 59999999975 84 6 69999999976 85 6 79999999977 86 6 89999999978 87 6 99999999979 88 6 109999999980 80 6 29999999981 81 6 39999999982 82 6 49999999983 83 6 59999999984 84 6 69999999985 85 6 79999999986 86 6 89999999987 87 6 99999999988 88 6 109999999989 89 6 119999999990 81 6 39999999991 82 6 49999999992 83 6 59999999993 84 6 69999999994 85 6 79999999995 86 6 89999999996 87 6 99999999997 88 6 109999999998 89 6 119999999999 90 6 1210000000000 1 0 110000000001 2 0 210000000002 3 0 310000000003 4 0 410000000004 5 0 510000000005 6 0 610000000006 7 0 710000000007 8 0 810000000008 9 0 910000000009 10 0 1010000000010 2 0 210000000011 3 0 310000000012 4 0 410000000013 5 0 510000000014 6 0 610000000015 7 0 710000000016 8 0 810000000017 9 0 910000000018 10 0 1010000000019 11 0 1110000000020 3 0 310000000021 4 0 410000000022 5 0 510000000023 6 0 610000000024 7 0 710000000025 8 0 810000000026 9 0 910000000027 10 0 1010000000028 11 0 1110000000029 12 0 1210000000030 4 0 410000000031 5 0 510000000032 6 0 610000000033 7 0 710000000034 8 0 810000000035 9 0 910000000036 10 0 1010000000037 11 0 1110000000038 12 0 1210000000039 13 1 0`

 Posted by Charlie on 2017-08-12 12:36:27

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