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Of the three-digit integers greater than 700, how many have

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Of the three-digit integers greater than 700, how many have [#permalink] New post 18 Feb 2006, 14:44
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Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A. 90
B. 82
C. 80
D. 45
E. 36
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 [#permalink] New post 18 Feb 2006, 16:43
C=80?

Three types of numbers:

XYX -> 3*9*1 = 27 (Egs: 717, 858.....)
XXY -> 3*1*9 = 27-1 = 26(need to subtract 1 because 700 is not allowed) (Egs: 772, 884....)
XYY -> 3*9*1 = 27 (Egs:755, 866.....)

Sum = 80

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 [#permalink] New post 18 Feb 2006, 19:14
Numbers with first digit as 7 and next two digits as same = 8 ( exclude 7,0 )
Numbers with first digit as 7 and next two digits as same = 9 ( exclude 8 )
Numbers with first digit as 7 and next two digits as same = 9 ( exclude 9 )

Numbers with first and second digits are 7 = 9 ( last digit can't be 7)
Numbers with first and third digits are 7 = 9 ( second digit can't be 7)

Similary we get 9+9+9+9 for second and third digits as 8,9 when the first digit is 8 or 9.

Total = 8 + 9*8 = 80.
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 [#permalink] New post 19 Feb 2006, 16:27
OA: C :)
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 [#permalink] New post 19 Feb 2006, 17:25
For hundreds digit 7 + 2 other digits
Remaining digits - 11,22,33,44,55,66,88,99
Total: 8

For hundreds digit 7 + 7 in ones position
Remaining digts - 17,27,37,47,57,67,87,97,07
Total: 9

For hundreds digit7 + 7 in tens position
Remaining digits - 70,71,72,73,74,75,76,78,79
Total: 9

For hundreds digit 8,9 + other digits
Remaining digits: 00,11,22,33,44,55,66,77,88/99
Total: 9*2 = 18

For hundreds digit 8,9 + 8/9 in ones position
Remaining digits: 18,28,38,48,58,68,78,98,08 (19,29,39,49,59,69,79,89,09)
Total: 9*2 = 18

For hundreds digit 8,9 + 8/9 in tens position
Remaining digits: 80,81,82,83,84,85,86,87,89 (90,91,92,93,94,95,96,97,98)
Total: 9*2 = 18

Combines total = 80
  [#permalink] 19 Feb 2006, 17:25
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