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

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Manager
Joined: 08 Jul 2006
Posts: 90
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Of the three-digit integers greater than 700, how many have [#permalink]  07 Sep 2006, 15:31
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?
Intern
Joined: 02 Aug 2006
Posts: 23
Location: NYC
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Re: Counting Integers [#permalink]  07 Sep 2006, 16:35
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..
CEO
Joined: 20 Nov 2005
Posts: 2919
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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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
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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