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Of the three-digit integers greater than 800 [#permalink]

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02 Aug 2017, 08:55

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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?

Of the three-digit integers greater than 800 [#permalink]

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02 Aug 2017, 10:05

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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

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?

Re: Of the three-digit integers greater than 800 [#permalink]

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03 Aug 2017, 02:52

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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

Re: Of the three-digit integers greater than 800 [#permalink]

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03 Aug 2017, 03:15

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)

Concentration: Social Entrepreneurship, General Management

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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