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x and k are positive integers and x= 5*10^27- 27*10^(5k). If

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Kudos [?]: 178 [0], given: 0

x and k are positive integers and x= 5*10^27- 27*10^(5k). If [#permalink] New post 22 Sep 2006, 09:31
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x and k are positive integers and x= 5*10^27- 27*10^(5k). If s and S are repsectively the smallest and largest possible sums of the digits of the decimal notation of x, then S-s=

(A) 124 (B) 134 (C) 144 (D) 154 (E) none of these
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Kudos [?]: 17 [0], given: 0

 [#permalink] New post 22 Sep 2006, 13:12
Is it E? :shock: May be wrong, though.

For the greatest value of x (and also greatest value of sum of digits), k=1.
x= 5*10^27- 27*10^5 = (10^5) *(5*10^22 - 27)
The decimal notation should be 4(twenty 9s)7300000.
Hence sum of digits is 4 + 180 + 10 = 194

Similarly, for the smallest positive value of x (not looking for negative; assuming that we are looking for the smallest positive value, else it would be infinity)
k=5
x= 5*10^27- 27*10^25 = (10^25) *(500 - 27) = 473(twenty five 0s)
Hence sum of digits is 14.

Hence the difference S-s=180 => Answer E.
  [#permalink] 22 Sep 2006, 13:12
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x and k are positive integers and x= 5*10^27- 27*10^(5k). If

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