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Re: Of the three-digit integers greater than 700, how many have two digits
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30 May 2015, 10:49

1

uva wrote:

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) ??

You are right about the range of the numbers. In the solution presented we are subtracting from total possible numbers (i.e. 299) the numbers which have all the 3 digits same and the numbers which have all the three digits distinct. In case of numbers with distinct digits between 701 and 999 both inclusive we can have numbers like 710, 720... etc. So zero can be there at the units place.

Alternatively in case of numbers with exactly two similar digits we can have numbers like 800,900 or 770, 880, 990 which have zero at their units place.

Hope it's clear . Let me know if you are having trouble at any point of the explanation.

Re: Of the three-digit integers greater than 700, how many have two digits
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25 Jul 2015, 04:22

Here we can use a counting method in which we have 3 cases (XX means the same number)

Case1: XXY First: We can choose 7,8 or 9 --> 3 options Second: --> here we have only 1 option (see XX) Third: We can pick any number except the number picked for the first digit --> 9 => 3*1*9=27

CASE2: XYX Same as above we have 3*9*1=27

CASE3: YXX Same logic as above --> 3*8*1=24 (we can not choose 700 here because of the restriction >700) but in this case we eliminate options 800 and 900 which meets the requirement of the restriction >700 --> So we have here 24+2=26

Overall = 27+27+24=80 (C)

Method 2

#of arrangements of xxy =3!/2! 1st digit: 3 choices (7,8 or 9) 2nd digit: let's say same as 1st digit - 1 choice 3rd: any number other then the first 2 - 9 choices

--> 3!/2!*3*1*9=81 but we must substract 1 because we have included 700 in this calculation => 81-1=80 Answer (C)
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Re: Of the three-digit integers greater than 700, how many have two digits
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07 Dec 2015, 18:04

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.

Wasnt sure initially why Bunuel multiplied 9 instead of 10 digits for the second place. But got it after a second thought - One way to write the same equation to understand better - (1*9*8) + (1*9*8) + (1*9*8) = 3*9*8 accounts of 1st digit as 7 + 1st digit as 8 + 1st digit as 9 Since one digit is already taken - we are left with 9 and then for last digit eventually with 8
_________________

Re: Of the three-digit integers greater than 700, how many have two digits
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10 Jun 2016, 22:47

bablu1234 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

Whats the purpose of this type of absurd question? There is no logic in this question other than remembering that 800 and 900 should also be counted but not 700. Its quite foolishness to count all those combinations between 700-999.

Topic Merged. It is not an absurd question. Refer to the above discussions.

Concentration: General Management, Entrepreneurship

GMAT 1: 510 Q22 V19

GPA: 3.41

Re: Of the three-digit integers greater than 700, how many have two digits
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11 Jun 2016, 11:11

Vyshak wrote:

bablu1234 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

Whats the purpose of this type of absurd question? There is no logic in this question other than remembering that 800 and 900 should also be counted but not 700. Its quite foolishness to count all those combinations between 700-999.

Topic Merged. It is not an absurd question. Refer to the above discussions.

I believe you already have gotten an good explanation of this particular math question. Bunuel is indeed a gifted teacher.

I want to point out something to you about the OG. The questions in the OG are released GMAT question: these questions are among the most rigorously examined questions on the entire planet. They went through several rounds of testing before they ever were on the GMAT, and they generated buckets of data while they were on the GMAT. By the the time any questions gets released from the GMAT and into an OG, it has a mountain of data behind its validity. By contrast, the explanations in the OG are only written when the book needs to be published, and when GMAC introduce additional questions in a new edition, some poor grad student somewhere has to produce new explanations. It's not clear to me that these explanations go through any feedback process at all. Some of the explanations are good, and some are simply atrocious. A few say things that are blatantly wrong. Most experts here on GMAT Club can give much better explanations than those that appear in the OG.

If the questions in the OG are like the crown jewels of Britain, the OG explanations are like costume jewelry one buys at a discount. If the questions in the OG are a world-class meal specially prepared by the finest chefs on earth, the OG explanations are like fast food. The difference in quality is mindboggling, and many students are entirely unaware of this difference because they are all bound in the same book with the same font.

Don't be fooled. Rely on the questions in the OG, but don't rely on the explanations. Get higher quality explanations here on GMAT Club, from Bunuel, Souvik, myself, and others.

I believe you already have gotten an good explanation of this particular math question. Bunuel is indeed a gifted teacher.

I want to point out something to you about the OG. The questions in the OG are released GMAT question: these questions are among the most rigorously examined questions on the entire planet. They went through several rounds of testing before they ever were on the GMAT, and they generated buckets of data while they were on the GMAT. By the the time any questions gets released from the GMAT and into an OG, it has a mountain of data behind its validity. By contrast, the explanations in the OG are only written when the book needs to be published, and when GMAC introduce additional questions in a new edition, some poor grad student somewhere has to produce new explanations. It's not clear to me that these explanations go through any feedback process at all. Some of the explanations are good, and some are simply atrocious. A few say things that are blatantly wrong. Most experts here on GMAT Club can give much better explanations than those that appear in the OG.

If the questions in the OG are like the crown jewels of Britain, the OG explanations are like costume jewelry one buys at a discount. If the questions in the OG are a world-class meal specially prepared by the finest chefs on earth, the OG explanations are like fast food. The difference in quality is mindboggling, and many students are entirely unaware of this difference because they are all bound in the same book with the same font.

Don't be fooled. Rely on the questions in the OG, but don't rely on the explanations. Get higher quality explanations here on GMAT Club, from Bunuel, Souvik, myself, and others.

Re: Of the three-digit integers greater than 700, how many have two digits
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27 Nov 2016, 05:57

Marcos Ramalho wrote:

I tried to solve the aforementioned problem but i reached 82, which is different from (supposedly) the correct answer (80).

"Of the three-digit integers greater than 700, hoe many have two digits that are equal to each other and the remaining digit different from the other two?"

Digits that i computed that are within the problem criteria:

Re: Of the three-digit integers greater than 700, how many have two digits
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27 Nov 2016, 10:44

1

Top Contributor

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

One approach is to start LISTING numbers and look for a PATTERN.

Let's first focus on the numbers from 800 to 899 inclusive. We have 3 cases to consider: 8XX, 8X8, and 88X

8XX 800 811 822 . . . 899 Since we cannot include 888 in this list, there are 9 numbers in the form 8XX

8X8 808 818 828 . . . 898 Since we cannot include 888 in this list, there are 9 numbers in the form 8X8

88X 880 881 882 . . . 889 Since we cannot include 888 in this list, there are 9 numbers in the form 88X

So, there are 27 (9+9+9) numbers from 800 to 899 inclusive that meet the given criteria.

Using the same logic, we can see that there are 27 numbers from 900 to 999 inclusive that meet the given criteria.

And there are 27 numbers from 700 to 999 inclusivethat meet the given criteria. HOWEVER, the question says that we're looking at numbers greater than 700, so the number 700 does not meet the criteria. So, there are actually 26 numbers from 701 to 799 inclusive that meet the given criteria.

Re: Of the three-digit integers greater than 700, how many have two digits
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21 May 2017, 19:01

Top Contributor

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

1. When tens digit equal to the hundreds digit, the number of possibilities for the units digit is 9 , for each possibility of hundreds digit for a total of 3*9=27 3. When units digit equal to the hundreds digit, the number of possibilities for the tens digit is 9 , for each possibility of hundreds digit for a total of 3*9=27 4. When tens digit is equal to units digit, there are 9 possibilities for each value of hundreds digit, for a total of 27,

So numbers, satisfying the constraints is (3*27)-1=80. 1 has to be subtracted because 700 is excluded.
_________________

Re: Of the three-digit integers greater than 700, how many have two digits
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16 Feb 2018, 17:39

bunuel can you expalin the goic for A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9), its not quite clear. Thanks in advance

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 three-digit integers greater than 700, how many have two digits
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16 Feb 2018, 20:05

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

pattern can be aab, aba, or abb aab: d1=3/3; d2=1/10; d3=9/10 aba: d1=3/3; d2=9/10; d3=1/10 abb: d1=3/3; d2=9/10; d3=1/10 3*(3*9*1)=81 81-1(for excluded 700)=80 C

Re: Of the three-digit integers greater than 700, how many have two digits
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12 Apr 2018, 16:30

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

We can solve this problem by analyzing the digits (0 to 9) that are being “doubled”.

If 0 is doubled, then the numbers can only be 800 and 900. So we have 2 numbers.

If 1 is doubled, then the numbers can only be 711, 811 and 911. So we have 3 numbers.

If 2, 3, 4, 5, or 6 is doubled, then we should have 3 numbers for each case since it’s analogous to 1 being doubled.

If 7 is doubled, then the numbers can only be 770, 771, 772, 773, 774, 775, 776, 778, 779; 707, 717, 727, 737, 747, 757, 767, 787, 797; 877, and 977. So we have a total of 20 numbers.

If 8 or 9 is doubled, then we should have 20 numbers for each case since it’s analogous to 7 being doubled.

Thus, altogether we have 2 + 6 x 3 + 3 x 20 = 2 + 18 + 60 = 80 such numbers.

Re: Of the three-digit integers greater than 700, how many have two digits
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29 Apr 2018, 06:42

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.

pushpitkc hello , can please help me to wrap my mind around Bunuels`s solution

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

I can not understand why second and third digits are 9 and 8 respectively, any idea?

Re: Of the three-digit integers greater than 700, how many have two digits
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29 Apr 2018, 07:16

1

dave13 wrote:

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.

pushpitkc hello , can please help me to wrap my mind around Bunuels`s solution

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

I can not understand why second and third digits are 9 and 8 respectively, any idea?

Hey Dave,

The hundred's digit can be 7 OR 8 OR 9. Thus, there are 3 ways to fill the first place.

While filling the hundred's place, we will have only 9 options with us. The reason that we have 9 options with us is because ALL the digits of the number are distinct. And we have already used one of the number in the hundreds place.

Think about it. Once we put the digit 7 in the hundred's place, we can put 0 OR 1 OR 2 OR 3 OR 4 OR 5 OR 6 OR 8 OR 9 (9 options) in the tens place.

Now suppose you have place 1 in the tens place. So you have used 7 and 1 in the two places, thus, you will be left with 8 options for the units place( 0 OR 2 OR 3 OR 4 OR 5 OR 6 OR 8 OR 9)

Thus, that is the reason, we are writing all the possible cases as 3 x 9 x 8 (3 possible cases for the hundreds place, 9 possible cases for the tens place and 8 possible cases for the units place)

Of the three-digit integers greater than 700, how many have two digits
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12 May 2018, 06:37

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).

Numbers with all distinct digits = 3*9*8=216 (first digit can take 3 values - 7, 8, 9; second digit can take 9 values and the third digit 8 values);

how can these numbers be distinct, if first value is 7 then second digit can be any number from any of these sets: 0, 1, 2, 3, 4, 5, 6 7, 8, or this set( 1, 2, 3, 4, 5, 6 , 7, 8) what i am trying to say is that 7 can still be as second digit as well.... so if second digit take 9 values, 7 still can be among these 9 digits, why ? for example take this set as mentioned above 0, 1, 2, 3, 4, 5, 6 7, 8, there are 9 digits here, but as you see i simply replaced number 9 which is missing in this set with number 7

the same (my logic) applies to third digit...

pushpitkc could you explain why am i wrong ? please

Re: Of the three-digit integers greater than 700, how many have two digits
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06 Sep 2018, 21:45

Im not sure if I'm the only one that's doing this a completely different way, but it gets the answer quite quickly.

So if for one second we assume the question includes the number 700, then the case that's being asked will repeat itself 3 times (for the 700's, 800's and 900's) so the answer will have to be a multiple of 3. Now because we know 700 isnt included then we know the final answer plus 1 has to be a multiple of 3, aka (3x -1 = y, x= # of options for each of 700's, 800's and 900's and y = the answer to the questions)

So if we check all the answers: a) 90 --> 90+1 is not divisible by 3 so its wrong b) 82 --> 82+1 is not divisible by 3 so its wrong c) 80 --> 80 + 1 is divisible by 3 so it could be right d) 45 --> 45 + 1 is not divisible by 3 so its wrong e) 36 --> 36 + 1 is not divisible by 3 so its wrong.

Re: Of the three-digit integers greater than 700, how many have two digits
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28 Apr 2019, 23:35

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.

Hi Bunuel, Can you please explain how you got the highlighted part? Especially, 3*9*8? Hundreds number can be 3 because there are only 3 possibilities, 7,8,9. Tens number can be 0-9 = 10 numbers. But since one of these would be taken up by the 100's number, 10-1 = 9 possibilities. Ones number can be 0-9 = 10 numbers. Since two of these would be taken up by 100's and 10's, 10-2 = 8 possibilities.