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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

Diagnostic Test Question: 11 Page: 22 Difficulty: 650

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 number can have only following 3 patterns: A. all digits are distinct; B. two digits are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

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02 Jul 2012, 02:49

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Bunuel 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

Diagnostic Test Question: 11 Page: 22 Difficulty: 650

The first digit can be 7, 8 or 9, so three possibilities. If the first digit is repeated, we have \(2 * 9\) possibilities, because the numbers can be of the form \(aab\) or \(aba\), where \(b\neq a\). If the first digit is not repeated, the number is of the form \(abb\), for which there are 9 possibilities. This gives a total of 27 choices for a given first digit. Altogether, we will have \(3 * 27 = 81\) choices from which we have to eliminate one choice, 700 (because the numbers should be greater than 700). We are finally left with 80 choices.

Answer: C
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Re: Of the three-digit integers greater than 700, how many have [#permalink]

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02 Jul 2012, 05:33

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Hi,

To satify the given condition, required no. of cases = total numbers - numbers with all digits different - numbers when all three digits are same, number greater than 700; total numbers = 1*10*10 = 100 numbers with all digits different = 1*9*8 = 72 numbers when all three digits are same (777) = 1 req. = 100- 72 - 1 = 27 considering the numbers between 700 & 999 = 27*3=81 Answer is 80 ('cause 700 can't be included)

Re: Of the three-digit integers greater than 700, how many have [#permalink]

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02 Jul 2012, 06:34

There are 2 numbers that have the same two digits for each of the 10's (700-709, 710-719, 720-729, etc.), with the exception for where the tens digit is the same as the hundreds (770-779), in this scenario there are 9 digits that have the same two digits.

So for 700-799 inclusive there are (9 * 2) + 9 - 1 = 26 numbers that the same two digits. Subtract one cause 700 shouldn't be included. For 800 - 999 the calculation simply becomes 27 * 2 = 54

Re: Of the three-digit integers greater than 700, how many have [#permalink]

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02 Jul 2012, 12:22

Total Number of 3 digit number greter than 700 = 299 In 700 series - The total number of number with distinct digit = 9*8 So from 700 till 999 we will have - 3*9*8 = 216 So total number of numbers with duplicate digit = 299-216 = 83 Need to Subtract 777,888 & 999 so answer is 80.

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 number can have only following 3 patterns: A. all digits are distinct; B. two digits are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

Re: Of the 3digit nos greater than 700, how many have 2 digits [#permalink]

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15 Feb 2013, 23:41

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For 3 digit numbers greater than 700, the first digit has to be 7, 8, or 9. The last two digits can be anything except 00.

There can be three possible cases for such numbers:

Case 1- first two digits same. The last digit has to be different from the first two. As the first digit is constrained to be 7,8,or 9, total number of such numbers = 3*9 = 27

Case 2 - first and third digits same. By the same logic as above, total number of such numbers = 27

Case 3 - second and third digits same. Total number of such numbers = (3*9)-1 = 26 (we need to subtract one to remove the 00 case).

Therefore total = 27 + 27 + 26 = 80 numbers (option C)

We have assumed here that numbers where all three digits are the same are not allowed.
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Re: Of the three-digit integers greater than 700, how many have [#permalink]

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12 Mar 2013, 21:42

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From 701 to 799 we will have: 711, 722.... 799( excluding 777 ) = 8; From 770 to 779 we have 9 numbers ( excluding 777 ). We will also have numbers such as 707, 717, 727.... 797, total 9 numbers ( excluding 777 ) Similarly from 800 to 899 we will have 9 numbers ( we will exclude 888 ) From 880 to 899 we will once again have 9 numbers We will also have numbers such as 808, 818.... 898, total 9 numbers ( excluding 888 ) Similarly for numbers greater than equal to 900 and till 999 we will have 27 such numbers. Total = 26(>700) + 27(>=800) + 27(>=900) = 80 such numbers.

Re: Of the three-digit integers greater than 700, how many have [#permalink]

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16 Dec 2013, 14:57

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Bunuel wrote:

SOLUTION

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 number can have only following 3 patterns: A. all digits are distinct; B. two digits are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.

I just dont't understand why the "3*9*8" can anyone explain me? please

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 number can have only following 3 patterns: A. all digits are distinct; B. two digits are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.

I just dont't understand why the "3*9*8" can anyone explain me? please

A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);

Since given three-digit integers must be greater than 700, then the first digit can take three values: 7, 8, or 9. The second digit can take 9 values: 10 digits minus one digit we've already used for the first digit. Similarly, the third digit can take 8 values: 10 digits minus two digits we've already used for the first and second digits.

Re: Of the three-digit integers greater than 700, how many have [#permalink]

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25 Jan 2014, 21:19

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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

Let the three digit number be represented as X Y Z.

There are 3 cases:

Case I. [ X=Y ] & Z is not equal to X & Y : XXZ or YYZ X can be either 7, 8 or 9, so digit at X can be chosen in 3 ways After X is chosen, Y can be chosen in 1 way After X & Y are chosen, Z can be chosen in 9 ways Thus, possible No of digits = (3 ways) * (1 way) * (9 ways) = 27 ....(1) [example numbers: 774,779,882,993 etc]

Case II. [ X=Z ] & Y is not equal to X & Z: XYXor ZYZ X can be either 7, 8 or 9, so digit at X can be chosen in 3 ways After X is chosen, Z can be chosen in 1 way After X & Z are chosen, Z can be chosen in 9 ways Thus, possible No of digits = (3 ways) * (9 ways) * (1 way) = 27 ....(2) [example numbers: 747,797,828,939 etc]

Case III. [ Y =Z ]& X is not equal to Y & Z :XYY or XZZ X can be either 7, 8 or 9, so digit at X can be chosen in 3 ways After X is chosen, Y can be chosen in 9 ways After Y is chosen, Z can have 1 way Thus, possible No of digits = (3 ways) * (1 way) * (9 ways) = 27 ....(3) [example numbers: 744,799,822,933 etc]

Therefore, total numbers of possible digits [sum of Case (1), (2) & (3) above] = 27 + 27 + 27 - 1 = 80 One digit is subtracted from total number of possible digits to eliminate one possibility of XYZ = 700 to satisfy the condition that digit > 700.

Re: Of the three-digit integers greater than 700, how many have [#permalink]

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19 May 2014, 09:28

Bunuel, what an elegant solution as always! Please can you provide more such problems for practice? I often struggle with this type. Thanks!
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My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

Re: Of the three-digit integers greater than 700, how many have [#permalink]

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21 May 2014, 02:25

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

Consider numbers from 701 to 799

Case 1: Numbers such as 707, 717..... 797 (Except 777)

\(1*9*1 = 9\) ---> (1)

Case 2: Numbers such as 770, 771..... 779 (Except 777)

\(1*1*9 = 9\) ---> (2)

Case 3: Numbers such as 711, 722...... 799 (Except 777)

\(1*8*1 = 8\) ---> (3) [NOTE: Did not consider 0 & 7 for the unit's and ten's digit]

Adding (1), (2) & (3)

\(9 + 9 + 8 = 26\)

This can be done for 801 to 899 and 901 to 999.

So, we get \(26 * 3 = 78\)

We also need to consider \(800\) and \(900\) (But NOT \(700\))

So, \(78 + 2 = 80\)

Can an expert evaluate the method. I understand that this method is long and could take time. But I want to be sure that I considered the cases correctly (In case, it comes to this during the exam!).

Re: Of the three-digit integers greater than 700, how many have [#permalink]

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28 Mar 2015, 01:23

Bunuel 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

Diagnostic Test Question: 11 Page: 22 Difficulty: 650

lets count the number for one range 700-799 and we will multiply it by 3 .

77_= 9 7_7= 9 7__= 9 so total 27 numbers are there starting with 7XX . total 81 such numbers will be there in 700-999 inclusive .

as the question asks for all such numbers which are greater than 700 so 81 -1 = 80 Answer D .
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Re: Of the three-digit integers greater than 700, how many have [#permalink]

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30 May 2015, 10:19

Hi All,

With the below explanation,

Quote:

three digit number can have only following 3 patterns: A. all digits are distinct; B. two digits are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.

I am not able to understand this line, A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); shouldn' t it be 3*8*8 (since zero cant be there in units digit right) ??

Regards, Uva.

gmatclubot

Re: Of the three-digit integers greater than 700, how many have
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30 May 2015, 10:19

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