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 A productive s.o.d. (Posted on 2012-09-19)
Let us denote f(n) = 12...12 (repeated n times) and g(n) = 3...3 (repeated n times). For example: f(3) = 121212 and, g(3) = 333.

(i) Determine the distinct digits in the base ten expansion of f(n)*g(n) whenever n ≥ 3

(ii) Express s.o.d (f(n)*g(n)) in terms of n

*** s.o.d (x) denotes the sum of digits in the base ten expansion of x

 No Solution Yet Submitted by K Sengupta No Rating

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 No Subject | Comment 1 of 5

10      for N=1 to 27
20        Twlve=Twlve*100+12
30        Three=Three*10+3
40        print N,Twlve*Three,fnSod(Twlve*Three)
50      next N
60   end
70      fnSOD(X)
80        Sod=0
90        S=cutspc(str(X))
100        for I=1 to len(S)
110          Sod=Sod+val(mid(S,I,1))
120        next
130      return(Sod)

finds

` n f(n)*g(n)     sod(f(n)*g(n)) 1   36      9 2   39996   36 3   40363596        36 4   40399999596     72 5   40403636359596          63 6   40403999999959596       108 7   40404036363635959596    90 8   40404039999999995959596         144 9   40404040363636363595959596      117 10  40404040399999999999595959596   180 11  40404040403636363636359595959596        144 12  40404040403999999999999959595959596     216 13  40404040404036363636363635959595959596          171 14  40404040404039999999999999995959595959596       252 15  40404040404040363636363636363595959595959596    198 16  40404040404040399999999999999999595959595959596         288 17  40404040404040403636363636363636359595959595959596      225 18  40404040404040403999999999999999999959595959595959596   324 19  40404040404040404036363636363636363635959595959595959596        252 20  40404040404040404039999999999999999999995959595959595959596     360 21  40404040404040404040363636363636363636363595959595959595959596        279 22  40404040404040404040399999999999999999999999595959595959595959596     396 23  40404040404040404040403636363636363636363636359595959595959595959596  306 24  40404040404040404040403999999999999999999999999959595959595959595959596         432 25  40404040404040404040404036363636363636363636363635959595959595959595959596      333 26  40404040404040404040404039999999999999999999999999995959595959595959595959596   468 27  40404040404040404040404040363636363636363636363636363595959595959595959595959596  360 There seems to be:`

floor((n-1)/2) occurrences of "40" at the beginning of the number,

followed by

if n is odd: the first n digits of 36...36
if n is even: "3", then n+1 occurrences of "9"

followed by

floor((n-1)/2) occurrences of "59"

followed by

6

The portion of the s.o.d. that's common to the odd and even n is then:

18*floor((n-1)/2) + 6

For even n, we add 9*(n+1) + 3

That makes the total for even n: 18*(n/2 - 1) + 6 + 9*(n+1) + 3, which simplifies to:

18*n

For odd n, it would seem to depend on whether n is congruent to 1 or 3 mod 4.  However, a look at the calculated sod's above shows that some of the irregularities cancel each other. The result looks like

27*(n-1)/2 + 9

 Posted by Charlie on 2012-09-19 16:46:19

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