Of the three-digit integers greater than 800

Question

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

GMATAspirer09 · Accepted Answer

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!

quantumliner · Answer

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

Answer is c

Bunuel · Answer

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.

Luckisnoexcuse · Answer

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

goforgmat · Answer

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_

KarishmaB · Answer

We need numbers of the type AAB or ABA or ABB greater than 800. We note that 800 is of the type ABB so first write "-1" so that you don't forget to exclude 800 from your counting.

AAB/ABA 
Select A in 2 ways (8 or 9) and B in 9 ways. Arrange A and B in 2 ways (AB or BA). 
So total such numbers = 2 * 9 * 2 = 36

ABB 
Select A in 2 ways (8 or 9) and B in 9 ways (A cannot be selected). 
So total such numbers = 2*9 = 18

Total numbers = 36 + 18 - 1 = 53

Answer (C)

Kinshook · Answer

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?

Case I: Of the form AAB
Options for A = {8,9} : 2 options
Options for B = {0,1,2,3,4,5,6,7,8,9} - A : 9 options
Total numbers possible of the form AAB = 2*9 = 18

Case 2: Of the form ABA
Options for A = {8,9} : 2 options
Options for B = {0,1,2,3,4,5,6,7,8,9} - A : 9 options
Total numbers possible of the form AAB = 2*9 = 18

Case 3: Of the form BAA
Options for B = {8,9} : 2 options
Options for A = {0,1,2,3,4,5,6,7,8,9} - A : 9 options
Total numbers possible of the form AAB = 2*9 = 18
But 800 is not possible
Total numbers possible of the form AAB = 18 - 1 = 17

Of the three-digit integers greater than 800, the number of integer which have two digits that are equal to each other and the remaining digit different from the other two = 18+ 18+17 = 53

IMO C

Of the three-digit integers greater than 800

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Of the three-digit integers greater than 800

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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