GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Nov 2019, 20:24 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Of the three-digit integers greater than 700, how many have two digits

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59039
Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

10
180 00:00

Difficulty:   95% (hard)

Question Stats: 49% (02:38) correct 51% (02:38) wrong based on 2262 sessions

### HideShow timer Statistics

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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59039
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

90
83
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.

_________________
Director  Joined: 22 Mar 2011
Posts: 584
WE: Science (Education)
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

13
1
4
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.

_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
##### General Discussion
SVP  P
Joined: 24 Jul 2011
Posts: 1897
GMAT 1: 780 Q51 V48 GRE 1: Q800 V740 Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

8
2
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.
_________________

Awesome Work | Honest Advise | Outstanding Results

Reach Out, Lets chat!
Email: info at gyanone dot com | +91 98998 31738 | Skype: gyanone.services
Current Student B
Joined: 29 Mar 2012
Posts: 295
Location: India
GMAT 1: 640 Q50 V26 GMAT 2: 660 Q50 V28 GMAT 3: 730 Q50 V38 Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

6
3
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)

Regards,
Math Expert V
Joined: 02 Sep 2009
Posts: 59039
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

5
1
mandrake15 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.

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.

Hope it's clear.
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59039
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

4
3
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59039
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

3
27
Manager  Joined: 12 Mar 2012
Posts: 79
Location: India
Concentration: Technology, Strategy
GMAT 1: 710 Q49 V36 GPA: 3.2
WE: Information Technology (Computer Software)
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

2
1
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.

Please give a kudo if you like my explanation.
Intern  Joined: 15 Dec 2013
Posts: 3
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

2
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.

I just dont't understand why the "3*9*8" can anyone explain me? please
Director  Joined: 07 Aug 2011
Posts: 500
GMAT 1: 630 Q49 V27 Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

2
2
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 .
Magoosh GMAT Instructor G
Joined: 28 Dec 2011
Posts: 4467
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

2
bablu1234 wrote:
Thanks for pointing this out. OG#15 explanation is lengthy and asking user to count numbers between 701 and 999. OG should hire Bunuel .

I do search before posting however I missed for this one.

Thanks again.

Dear bablu1234,
I'm happy to respond, my friend. 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.

Does all this make sense?
Mike _________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Manager  Joined: 28 Sep 2011
Posts: 56
Location: United States
GMAT 1: 520 Q34 V27 GMAT 3: 690 Q47 V38 GPA: 3.01
WE: Information Technology (Commercial Banking)
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

1
2
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

54 + 26 =
Intern  Joined: 14 Apr 2012
Posts: 10
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

1
1
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.
VP  Joined: 02 Jul 2012
Posts: 1099
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42 GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

1
number-system-og-87133.html?fl=similar
_________________
Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types
Manager  Joined: 04 Oct 2013
Posts: 150
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

1
1
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: XYX or 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.

Intern  Joined: 19 Oct 2013
Posts: 16
Location: India
Concentration: General Management, Operations
WE: Engineering (Other)
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

1
1
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]

$$9 + 9 + 8 = 26$$

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!).

e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3138
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

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.

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.

Hi uva,

Zero can definitely be in the units digit.

The hundreds place can take 3 values - 7, 8 or 9

The tens place can take 9 values - From 0 to 9 except the digit taken in the hundreds place

The units digit can take 8 values - From 0 to 9 except the digits taken in the hundreds and tens place.

Hence total possible numbers with distinct digits = 3*9*8 = 216.

Would like to know what made you think that zero can't be at the units place?

Hope it's clear Regards
Harsh
_________________
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3138
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

1
uva wrote:
Hi Harsh,

Question states "Of the three-digit integers greater than 700" so it must be between 701 to 999.

Regards,
Uva

Hi uva,

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.

Regards
Harsh
_________________
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4063
Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

### Show Tags

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 inclusive that 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.

So, our answer is 27+27+26 = 80

RELATED VIDEO

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

Go to page    1   2   3    Next  [ 43 posts ]

Display posts from previous: Sort by

# Of the three-digit integers greater than 700, how many have two digits  