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Summing Ones (Posted on 2013-02-27) Difficulty: 3 of 5
Find the sum
1+11+111+.......+111..111(n digits)

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution solution | Comment 2 of 4 |

10   T=1:Tot=T:Proto=123456790//999999999
20   for N=2 to 45
30     T=T*10+1:Tot=Tot+T
35     Approx=int(10^N*Proto)
40      print N,Tot,Approx-Tot
41     Prev=Approx-Tot
50   next

shows sums for n=2 through 45 together with the difference between this and an approximation based on a previous run of a simpler version of the program:

 2       12      0
 3       123     0
 4       1234    0
 5       12345   0
 6       123456          0
 7       1234567         0
 8       12345678        1
 9       123456789       1
 10      1234567900      1
 11      12345679011     1
 12      123456790122    1
 13      1234567901233   1
 14      12345679012344          1
 15      123456790123455         1
 16      1234567901234566        1
 17      12345679012345677       2
 18      123456790123456788      2
 19      1234567901234567899     2
 20      12345679012345679010    2
 21      123456790123456790121   2
 22      1234567901234567901232          2
 23      12345679012345679012343         2
 24      123456790123456790123454        2
 25      1234567901234567901234565       2
 26      12345679012345679012345676      3
 27      123456790123456790123456787     3
 28      1234567901234567901234567898    3
 29      12345679012345679012345679009   3
 30      123456790123456790123456790120          3
 31      1234567901234567901234567901231         3
 32      12345679012345679012345679012342        3
 33      123456790123456790123456790123453       3
 34      1234567901234567901234567901234564      3
 35      12345679012345679012345679012345675     4
 36      123456790123456790123456790123456786    4
 37      1234567901234567901234567901234567897   4
 38      12345679012345679012345679012345679008          4
 39      123456790123456790123456790123456790119         4
 40      1234567901234567901234567901234567901230        4
 41      12345679012345679012345679012345679012341       4
 42      123456790123456790123456790123456790123452      4
 43      1234567901234567901234567901234567901234563     4
 44      12345679012345679012345679012345679012345674    5
 45      123456790123456790123456790123456790123456785   5
 
 The discrepancy increases by 1 every ninth value of n. This is shown farther down the list by:


   10   T=1:Tot=T:Proto=123456790//999999999
   20   for N=2 to 200
   30     T=T*10+1:Tot=Tot+T
   35     Approx=int(10^N*Proto)
   40     if Approx-Tot<>Prev then print N;Approx-Tot
   41     Prev=Approx-Tot
   50   next

 8  1
 17  2
 26  3
 35  4
 44  5
 53  6
 62  7
 71  8
 80  9
 89  10
 98  11
 107  12
 116  13
 125  14
 134  15
 143  16
 152  17
 161  18
 170  19
 179  20
 188  21
 197  22
 
The reduced value of proto is actually 10/81.

The initial approximation is [10^n * 10/81] where the square brackets indicate the floor function. But to take into consideration the growing discrepancy from the approximation, we need to subtract [(n+1)/9].

Putting the two together and simplifying we get:

[10^(n+1) / 81] - [(n+1)/9]

As verification, this:

10   T=1:Tot=T:Proto=123456790//999999999
20   for N=2 to 2000
30     T=T*10+1:Tot=Tot+T
35     Approx=int(10^(N+1)//81)-int((n+1)//9)
40     if Approx-Tot<>Prev then print N;Approx-Tot
41     Prev=Approx-Tot
50   next

produces no output, as Prev starts out and stays zero.


  Posted by Charlie on 2013-02-27 13:23:06
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