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# OG 11- Quant Sample- Question 11

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Senior Manager
Joined: 25 Nov 2006
Posts: 332
Schools: St Gallen, Cambridge, HEC Montreal
OG 11- Quant Sample- Question 11 [#permalink]

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19 Mar 2007, 17:10
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hello,

Would anybody have a formula/explanation for this:
"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?"
Senior Manager
Joined: 11 Feb 2007
Posts: 351

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19 Mar 2007, 17:38
This isn't a formula or a mathematical explanation but I would just go with the primitive method (I won't be able to come up with a formula in 2 minutes anyway... )

8xx where x = 0~7, 9 - 9 of them
8x8 where x = 0~7, 9 - 9 of them
88x where x = 0~7, 9 - 9 of them
total 27

since it would be the same for the others in 700s and 900s

27 x 3 = 81

however, since it has to be greater than 700, subtract the case 7xx where x = 0.

81 - 1 = 80

What is the OA?
(I might have missed out on something...)

This might not be what you wanted..but...
Senior Manager
Joined: 25 Nov 2006
Posts: 332
Schools: St Gallen, Cambridge, HEC Montreal

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19 Mar 2007, 17:47
Thanks.

Answer is 80, so you're right.

Could you explain this more? I am not sure to get it

8xx where x = 0~7, 9 - 9 of them
8x8 where x = 0~7, 9 - 9 of them
88x where x = 0~7, 9 - 9 of them
Senior Manager
Joined: 11 Feb 2007
Posts: 351

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19 Mar 2007, 18:13
lumone wrote:
Thanks.

Answer is 80, so you're right.

Could you explain this more? I am not sure to get it

8xx where x = 0~7, 9 - 9 of them
8x8 where x = 0~7, 9 - 9 of them
88x where x = 0~7, 9 - 9 of them

8xx where x = 0~7, 9 - 9 of them
-> Keep the 8 fixed and make the ten's and unit's digits the same. For example, 800, 811, 822, etc. However, you cannot have 8xx where x = 8 since then, there would be 3 (not 2) identical numbers.

8x8 where x = 0~7, 9 - 9 of them
-> Keep the hundred's and unit's digits fixed and rotate the middle digits.

88x where x = 0~7, 9 - 9 of them
-> Keep the first two digits fixed and rotate the last digits.

Manager
Joined: 12 Feb 2007
Posts: 167

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19 Mar 2007, 18:32
What am I doing wrong here?

I used a brute force method,

700
707
711
717

so every change in the tens digit, there are 2 possibilities, except for 77x where there are 10 possibilities, so therefore there are 26 possibilities for the 700's, same of 800's and the 900's, so

26*3=78
Senior Manager
Joined: 11 Feb 2007
Posts: 351

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19 Mar 2007, 19:00
Tuneman wrote:
What am I doing wrong here?

I used a brute force method,

700
707
711
717

so every change in the tens digit, there are 2 possibilities, except for 77x where there are 10 possibilities, so therefore there are 26 possibilities for the 700's, same of 800's and the 900's, so

26*3=78

Hi Tuneman,

You shouldn't include 700 cuz it said that the number is greater than 700. More important, 77x does not have 10 possibilities because 777 would have three identical numbers instead of two.
CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

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07 Jan 2008, 14:09
ricokevin wrote:
This isn't a formula or a mathematical explanation but I would just go with the primitive method (I won't be able to come up with a formula in 2 minutes anyway... )

8xx where x = 0~7, 9 - 9 of them
8x8 where x = 0~7, 9 - 9 of them
88x where x = 0~7, 9 - 9 of them
total 27

since it would be the same for the others in 700s and 900s

27 x 3 = 81

however, since it has to be greater than 700, subtract the case 7xx where x = 0.

81 - 1 = 80

What is the OA?
(I might have missed out on something...)

This might not be what you wanted..but...

_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Re:   [#permalink] 07 Jan 2008, 14:09
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