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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 Jun 2011, 21:05
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A
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Question Stats:

56% (02:51) correct 44% (01:12) wrong based on 27 sessions
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

OPEN DISCUSSION OF THIS QUESTION IS HERE: of-the-three-digit-integers-greater-than-700-how-many-have-135188.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Jul 2013, 06:06, edited 1 time in total.
Renamed the topic and edited the question.
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Re: From MitDavidDv: Problem Solving Question-OG 12 [#permalink] New post 18 Jun 2011, 21:17
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Range of Consecutive Integers:
1-x Principle
From 701 to 999: 999-701+1=299 numbers, subtract 3 numbers (777, 888, 999)
299-3= 297.
Fundamental Counting Principle ( To determine how many numbers have zero digits in common)
3 possibilities for the 100's digit
9 possibilities for the 10's digit
8 possibilities for the unit's digit
3*9*8=216 numbers with no digits in common
296-216=80 numbers that have exactly two digits in common.
The answer is C.
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Re: From MitDavidDv: Problem Solving Question-OG 12 [#permalink] New post 18 Jun 2011, 21:37
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case 1:XXY
so ,for first digit 3 options as the no should be in range of 700 and 1000 .second digit has only one option and for third digit 9 options available.
3*1*9=27
case 2:XYX
3*9*1=27
case 3:
YXX
3*9*1=27

total =81 - 1 because it includes 700 .Hence OA is 80.

Any other alternative to this one.pls share.

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Re: From MitDavidDv: Problem Solving Question-OG 12 [#permalink] New post 18 Jun 2011, 22:04
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MitDavidDv 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


The options we have are

XYX
XXY
YXX

lets take 900 series
XYX ( 909,919.................989) ........................... 9 terms
XXY ( 990,991..................998)............................9 terms
YXX ( 900, 911,922............988)........................... 9 terms

total 27 terms
8 series will also have 27 terms
7 series will have 27 -1 term ( 700 cant be taken) = 26 terms

hence total 27+27 +26 = 80 terms
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Re: From MitDavidDv: Problem Solving Question-OG 12 [#permalink] New post 07 Sep 2011, 07:34
Quote:
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


three-digit integers greater than 700: 701 to 999, inclusive.
possible values for hundreds-digit--> 7,8,9
possible values for tens-digit and ones-digit --> 0, 1,2,3,4,5,6,7,8,9

when hundreds-digit and tens-digit are the same: (3x1x10)-3=27 ---> we minus three to exclude 777, 888 and 999
when hundreds-digit and ones-digit are the same: (3x10x1)-3=27 ---> we minus three to exclude 777, 888 and 999
when tens-digit and hundreds-digit are the same:[(3x10x1)-3]-1=26 ---> we minus three to exclude 777, 888 and 999; we minus one to exclude 700

27+27+26 = 80
Answer: C

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Re: From MitDavidDv: Problem Solving Question-OG 12   [#permalink] 07 Sep 2011, 07:34
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