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Products of digits. (Posted on 2012-12-05) Difficulty: 3 of 5
Let P(n) be the product of non-zero digits of the number n.Find the value of P(1)+P(2)+P(3).......P(2012)

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution analytic and computer solutions Comment 1 of 1

The sum of the first 9 digits is 45 and this would be the sum for P(1) to P(9).

For 2-digit numbers, 10 through 99, the 45 is multiplied by the 9th triangular number, i.e. that same 45, but also add in another 45 for the 2-digit numbers with zero in the units position, so the 2-digit numbers account for 45*45+45=2070.

Similarly, 3-digit numbers account for 45*(45+45*45+45)+45 = 95220 as all the 1-digit and 2-digit numbers are repeated with 100, 200, ... added and those with two trailing zeros account for the additional 45.

The 1 thousands will result in 45*(45+2070+95220) plus 1 more for 1000 itself, making 97336.

Then 2000 through 2009 account for 2+2*45 = 92.

Finally 2010, 2011 and 2012 account for 2+2+4 = 8.

       45
     2070 
    95220   
    97336
       92
        8
  -------
   194771


  
computer method:  
  
list
   10   for N=1 to 2012
   20     D=cutspc(str(N))
   30     P=1
   40     for I=1 to len(D)
   50        if mid(D,I,1)>"0" then P=P*val(mid(D,I,1))
   60     next
   70     Tot=Tot+P
   80   next
   90   print Tot
OK
run
 194771
OK  


  Posted by Charlie on 2012-12-05 17:33:44
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