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# Of the three-digit integers greater than 800

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Manager
Joined: 13 Jun 2012
Posts: 205
Location: United States
WE: Supply Chain Management (Computer Hardware)
Of the three-digit integers greater than 800  [#permalink]

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02 Aug 2017, 08:55
5
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Difficulty:

95% (hard)

Question Stats:

29% (02:19) correct 71% (02:26) wrong based on 107 sessions

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Of the three-digit integers greater than 800, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 36
(B) 51
(C) 53
(D) 54
(E) 60
Senior Manager
Joined: 24 Apr 2016
Posts: 327
Of the three-digit integers greater than 800  [#permalink]

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02 Aug 2017, 10:05
3
With the given conditions, three types of numbers are possible

Type 1: AAB -- First and Second Digits are same, third digit is different
Type 2: ABA -- First and Third Digits are same, Second digit is different
Type 3: BAA -- Second and third Digits are same, First digit is different

Type 1: First and Second Digit can be 8 and 9, Third Digit can be 0,1,2,3,4,5,6,7,8/9. (If first and second digit is 8, then third digit cannot be 8 and if first and second digit is 9, then third digit cannot be 9).

Therefore total no. of combinations for Type 1 = 2 *9 = 18

Type 2: First and Third Digit can be 8 and 9, Second Digit can be 0,1,2,3,4,5,6,7,8/9. (If first and third digit is 8, then Second digit cannot be 8 and if first and third digit is 9, then second digit cannot be 9).

Therefore total no. of combinations for Type 2 = 2 *9 = 18

Type 3: If First Digit is 8, then second and third digit can be 1,2,3,4,5,6,7,9. If First Digit is 9, then second and third digit can be 0,1,2,3,4,5,6,7,8.

Therefore total no. of combinations for Type 3 = 8+9 = 17

Total number of combinations = 18 + 18 + 17 = 53

Math Expert
Joined: 02 Sep 2009
Posts: 56366
Re: Of the three-digit integers greater than 800  [#permalink]

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02 Aug 2017, 23:11
1
Turkish wrote:
Of the three-digit integers greater than 800, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 36
(B) 51
(C) 53
(D) 54
(E) 60

This is a copy of the following OG question: https://gmatclub.com/forum/of-the-three ... 35188.html

Similar questions:
https://gmatclub.com/forum/of-the-three ... 27390.html
https://gmatclub.com/forum/how-many-odd ... 71394.html

Hope it helps.
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Joined: 16 Feb 2017
Posts: 19
Re: Of the three-digit integers greater than 800  [#permalink]

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03 Aug 2017, 02:52
3
1
Turkish wrote:
Of the three-digit integers greater than 800, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 36
(B) 51
(C) 53
(D) 54
(E) 60

Three digit numbers have only 3 patterns:
1) all digits are alike
2) only 2 digits are alike
3) all digits are unique

Here, we shall calculate Desired (2 above) = Total - Undesired (1+3 above)

First let's find total digits between 801 <--> 999
Total Digits = 999-801+1 = 199

Now, digits distinct between 801 and 999 --> 2X9X8 = 144
AND, all three alike --> only 2 = 888 and 999

So, our desired answer is = total - undesired
= 199 - 144 - 2 = 199 - 146 = 53

Hope this helps!

Current Student
Joined: 18 Aug 2016
Posts: 617
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
Re: Of the three-digit integers greater than 800  [#permalink]

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03 Aug 2017, 03:15
1
Turkish wrote:
Of the three-digit integers greater than 800, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 36
(B) 51
(C) 53
(D) 54
(E) 60

8xx
tens digit 8 = 9 numbers
unit digit 8 = 9 numbers
unit and tens digit same = 8 (cannot take 8 or 0)

9xx
tens digit 9 = 9 numbers
unit digit 9 = 9 numbers
unit and tens digit same = 9 (cannot take 9)

Total 9*5 + 8 = 53
C
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Luckisnoexcuse
Senior Manager
Joined: 09 Feb 2015
Posts: 352
Location: India
Concentration: Social Entrepreneurship, General Management
Schools: Booth '21 (D)
GMAT 1: 690 Q49 V34
GMAT 2: 720 Q49 V39
GPA: 2.8
Re: Of the three-digit integers greater than 800  [#permalink]

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10 Aug 2017, 11:12
Total number of 3 digit numbers = 199
Total number of 3 digit numbers with all digits same = 2
Total number of digits will all digits different greater than 800 = 9 * 8= 72
Total number of digits will all digits different greater than 900 = 9*8 =72
199 -(2+72*2)=53_
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Re: Of the three-digit integers greater than 800  [#permalink]

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24 Aug 2018, 00:10
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Re: Of the three-digit integers greater than 800   [#permalink] 24 Aug 2018, 00:10
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