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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 07 Sep 2006, 16:31
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

Explain your reasoning. Also, can PR or CR be used for this problem...how so?
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Re: Counting Integers [#permalink] New post 07 Sep 2006, 17:35
Rayn wrote:
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

Explain your reasoning. Also, can PR or CR be used for this problem...how so?


I get none of the above answers... :(

The numbers are:

707, 717, 727,...797 - 9 numbers (excluding 777)
711, 722, 733, ...799 - 8 numbers (excluding 700 and 777)

808, 818, 828,...898 - 9 numbers
800, 811, 822,...899 - 9 numbers

909, 919, 929, ...989 - 9 numbers
900, 911, 922, ...988 - 9 numbers

Totals to 53..
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 [#permalink] New post 07 Sep 2006, 20:59
We will count only from 700-799

7XX where X is any digit other than 7. = 9
7X7 where X is any digit other than 7 = 9
77X where X is any digit other than 7 = 9
Total form 700-799 = 27
Similarly there are 27 from 800-899 and 27 from 900-999.

Total = 27+27+27 = 81
but we have to exclude 700.
So answer = 81-1 = 80
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

  [#permalink] 07 Sep 2006, 20:59
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