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

Question

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

Bunuel · Accepted Answer

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.

Bunuel · Answer

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BrentGMATPrepNow · Answer

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

Answer:  C

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EvaJager · Answer

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
eq 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

cyberjadugar · Answer

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)

Answer (C).

Regards,

LalaB · Answer

999-700=299 (total numbers)

299-3 =296 (3 means 777/888/999)

3*8*9=216 (doing so, we found total numbers,which have different digits)
296-216=80

ans is C
l

GyanOne · Answer

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.

Abhii46 · Answer

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.

mandrake15 · Answer

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

Bunuel · Answer

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.

arunspanda · Answer

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.    & Z is not equal to X & Y :   XX  Z  or  YY  Z   
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)

Case II.    & Y is not equal to X & Z:   X  Y  X    or  Z  Y  Z  
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)

Case III.   & X is not equal to Y & Z :  X  YY  or  X  ZZ  
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)

Therefore, total numbers of possible digits   = 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.

Answer:  (C)

NoHalfMeasures · Answer

Bunuel, what an elegant solution as always! Please can you provide more such problems for practice? I often struggle with this type. Thanks!

Lucky2783 · Answer

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 .

uva · Answer

Hi Harsh,

Thanks for your reply.

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

Regards,
Uva

EgmatQuantExpert · Answer

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

BrainLab · Answer

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)

bablu1234 · Answer

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.

mikemcgarry · Answer

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 :-)

SVaidyaraman · Answer

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.

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

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GMAT Problem Solving (PS) Questions

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

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