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

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Joined: 04 May 2006
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of the three-digit integers greater than 700, how many have [#permalink]  28 Apr 2008, 22:20
of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit diffferent from the other two?

A. 90
B. 82
C. 80
D. 45
E. 36
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Joined: 17 Jan 2008
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Re: Interesting question! [#permalink]  28 Apr 2008, 22:42
sondenso wrote:
of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit diffferent from the other two?

A. 90
B. 82
C. 80
D. 45
E. 36

Not sure what the mathematical way of doing this is but I got 80 by counting out the number between 701 and 800 and then multiplying by 3. There are 27 between 701 and 800. 27*3=81 but 1000 doesn't count since it's a four-digit number so -1. 81-1=80.

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Re: Interesting question! [#permalink]  29 Apr 2008, 06:46
1
KUDOS
Between 701 and 800

707, 717... for a total of 9 (not including 777)
Similarly, 770, 771 for a total of 9 (not incl 777)
Also, you have 711, 722, 733 for a total of 9 (incl 800)

Multiplying this by 3, gives 81.. This included 1000 so correct answer is C.
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Joined: 06 Jan 2008
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Re: Interesting question! [#permalink]  29 Apr 2008, 07:38
sondenso wrote:
of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit diffferent from the other two?

A. 90
B. 82
C. 80
D. 45
E. 36

7_ _ ( 700 < Total number of digits <=799 is 99)
8_ _ ( 800 <= Total number of digits <=899 is 100 )
9_ _ ( 900 <= Total number of digits <=999 is 100 )

Consider 7 _ _
Lets calculate number of three digit numbers greater than 700 which have all three DIFFERENT numbers.
7_ _
second digit can take ONLY 9 digits. (becoz we need to exclude 7)
third digit can take ONLY 8 digits. (bcoz we need to exclude 7 and the above number that we already selected.)
So total = 9 * 8 = 72
Now TOTAL - ALL Different numbers = numbers which have same digits in it
99 - 72 =27
But wait this includes 777, so 27 - 1 = 26

Similarly for 8 _ _,
100 - 72 = 28
But wait this includes 888, so 28 - 1 = 27

Similarly for 9 _ _,
100 - 72 = 28
But wait this includes 999, so 28 - 1 = 27

Total = 26 + 27 +27 =80
C
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Re: Interesting question! [#permalink]  29 Apr 2008, 08:50
yellowjacket wrote:
Between 701 and 800

707, 717... for a total of 9 (not including 777)
Similarly, 770, 771 for a total of 9 (not incl 777)
Also, you have 711, 722, 733 for a total of 9 (incl 800)

Multiplying this by 3, gives 81.. This included 1000 so correct answer is C.

Good explanation. +1
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Re: Interesting question! [#permalink]  30 Apr 2008, 12:17
Fastest way --

We need to fill 3 digits
_ _ _

Two cases
Case 1. Leftmost and another digit same -
Leftmost digit can be filled in 3 ways (7,8,9). Of the two digits, one is same as leftmost, and other can be filled in 9 ways. So, total = 9*2 = 18.
Now, second and third digit can be interchanged to give a new number. Hence 18*2 = 36

Case 2. Second and third digit same
Leftmost digit can be filled in 3 ways (7,8,9). Other two digits are same and can be filled in 9 ways
3*9=27

Total numbers = 54+27 = 81.
This includes 700 as well. So total number of numbers greater than 700 is 80.
Re: Interesting question!   [#permalink] 30 Apr 2008, 12:17
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