Talk:Minimum multiple of m where digital sum equals m

Observation for bigger m

I am trying to find a way to minimize the range for the search.
The start n is easy to construct

  m   Start/End n       multiples 1..9 of 100/m
                      x1       x2       x3       x4       x5       x6       x7       x8       x9
  41       1463--  2.43902  4.87805  7.31707  9.75610 12.19512 14.63415 17.07317 19.51220 21.95122
           4339
1463*41 = 59,983. 6*10,000-1 = 59,999 has sum of digits = 41 
The factor number is 6 one step therefor 1463/6 = 243.833
But the result n = 4339 / 243.833 ~ 17,8 not integer like  

  42       1666--  2.38095  4.76190  7.14286  9.52381 11.90476 14.28571 16.66667 19.04762 21.42857
           2119
much better to see n start = 8xk and n final is 10xk 
  43       1860-- >2.32558< 4.65116  6.97674  9.30233 11.62791 13.95349 16.27907 !18.60465! 20.93023
           2323
but much better for bigger m here x1000/m
                           x1       x2       x3       x4       x5       x6       x7       x8       x9
 140 428571428571420--  7.14286 14.28571 21.42857 28.57143 35.71429 42.85714 50.00000 57.14286 64.28571
     571427857142857   n start = x6 n final = x8
 141 49645390070921--  7.09220 14.18440 21.27660 28.36879 35.46099 42.55319 49.64539 56.73759 63.82979
     63822694964539    n start = x7 n final = x9
 142 56338028169014--  7.04225 14.08451 21.12676 28.16901 35.21127 42.25352 49.29577 56.33803 63.38028
    140140838028169    n start = x8 n final = x12
 143 62937062937062--  6.99301 13.98601 20.97902 27.97203 34.96503 41.95804 48.95105 55.94406 62.93706
    391606993006993    n start = x9 n final = x 56
 144 69444444444444--  6.94444 13.88889 20.83333 27.77778 34.72222 41.66667 48.61111 55.55556 62.50000
    277777777777777    n start = x1 n final = x4
 145 137931034482758--  6.89655 13.79310 20.68966 27.58621 34.48276 41.37931 48.27586 55.17241 62.06897
     482751724137931    n start = x2 n final = x7
 146 205479452054794--  6.84932 13.69863 20.54795 27.39726 34.24658 41.09589 47.94521 54.79452 61.64384
     471917801369863    n start = x3 n final = x6.89
 147 272108843537414--  6.80272 13.60544 20.40816 27.21088 34.01361 40.81633 47.61905 54.42177 61.22449
     401360544217687    n start = x4 n final = x5.9
 148 337837837837837--  6.75676 13.51351 20.27027 27.02703 33.78378 40.54054 47.29730 54.05405 60.81081
    1081081081081081    n start = x5 n final = x16
 149 402684563758389--  6.71141 13.42282 20.13423 26.84564 33.55705 40.26846 46.97987 53.69128 60.40268
     536912751677851    n start = x6 n final = x8

--Horst.h (talk) 14:50, 2 February 2022 (UTC)

I think for bigger values turning the task into a problem in combinatorics will be better. Taking 140 simplistically the minimum is 5999999999999999, which obviously is not divisible by 140. The solution is 79999899999999980. So I could try 6+15 digits which sum to 134 then 7+15 digits which sum to 133 until I find a number divisible by 140. Let me call this 15 digit number set N. Note that 6+N will have the same N as 15+N, 24+N, 33+N, 42+N, 51+N, and 60+N. If I optimize this for a particular number the prime factors of 140 are 2,2,5 and 7 from which it follows that the final digit must be 0 and the last but 1 must be even. Which makes the search space very small.--Nigel Galloway (talk) 16:21, 2 February 2022 (UTC)