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Digits: Of the 3-digit integers greater than 700, how many h

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Digits: Of the 3-digit integers greater than 700, how many h

by II » Mon Mar 17, 2008 5:45 am

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Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

I am keen to understand different ways of answering this question.
Thanks in advance.
II
Last edited by II on Mon May 05, 2008 1:54 am, edited 1 time in total.
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by mnjoosub » Mon Mar 17, 2008 6:30 am
I don't know whether there is a proper mathematical way to solve this problem but I did it this way.

I tabulate is as follows: see attachment

from the table we can see that the total repeated Nos for 700 = 30
So for 800 and 900 inclusive = 30 *3 = 90
We should less repeated Nos (3*3 = 9) = 81

Remember the question states more than 700 , so less 1 more = 80.

Ans = C
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by mnjoosub » Mon Mar 17, 2008 6:34 am
Sorry I forgot the attachment. :wink:
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Ans.xls
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help!

by sofia cuevas » Wed Aug 06, 2008 2:10 pm
is there any other way to solve this problem? perhaps a faster method?
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by parallel_chase » Wed Aug 06, 2008 3:07 pm
Here is a super fast way of solving this question as compared to above method.

For first digit we have 3 letters to play with (7,8,9)
Next two digits we can have any letter (0,1,2,3,4,5,6,7,8,9)

CASE I (ABB)
3*9*1 = 27

CASE II (BAB)
3*9*1 = 27

CASE III (BBA)
3*1*9 =27

27+27+27 = 81

subtract 1 because we want the number to be greater than 700, the above combination include 700

81-1=80

Once you are comfortable with permutations & combination of digits this thing will take you less than 10 secs and I mean this.

Let me know if you have any doubts.
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by guru_1971 » Fri Dec 10, 2010 1:02 am
The correct answer is 36

3 digit integers greater than 700.

2 cases

case 1: when first digit is 7 then using counting priniciple 1st digit has an option of 1 , and hence since two digits are equal the 2nd digit also is 1 third digit is between 1- 9 ( number greater than 700= 701 ) since the first 2 digits have taken integer 7, third digit will have (9-1) = 8 so the counting principle equation for case 1 is ( 1x1x8)
1 1 8 ( 9-1)
--- X ----------------- X ---------------------
1st digit is 7 2nd digit is same as 1st 3rd digit is 1-9 = 9
case 2

1st digit is not 7, so for the 1 and 2 digit, options are 8, 9 so 2 options for digit 1 and 2 options for digit 2 ( as per the condition of the problem ) now for 3 digit, since the number is greater than 700 digit will be from 1 - 9= and the 1st 2 digits have taken 7 and 8, so third digit will be 9-2= 7 so counting principle equation becomes ( 2 x2x7)

2 2 7
---------- X --------------------- X ---------------------------------
available same as 1st digit here 8 and 9 are taken by 1st 2 digits so 9 -2=7
option is
8 or 9


Solution thus becomes ( 1x1x8)+(2x2x7) = 36
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by diebeatsthegmat » Fri Dec 10, 2010 5:26 am
II wrote:Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

I am keen to understand different ways of answering this question.
Thanks in advance.
II
here the answer i find is also C 80
whats the answer?
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by kevincanspain » Fri Dec 10, 2010 5:29 am
Parallel chase's method is excellent!

Also, there are 299 3-digit integers greater than 700. How many of these do not have exactly two digits equal to each other?
There are 3 such integers that have three digits equal to each other (777,888,999) and 3 x 9 x 8 =216 that feature three distinct digits. Thus there are 299 - 3 - 216 = 80 such integers that have exactly two digits equal to each other
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by kevincanspain » Fri Dec 10, 2010 5:32 am
guru_1971 wrote:The correct answer is 36

3 digit integers greater than 700.

2 cases

case 1: when first digit is 7 then using counting priniciple 1st digit has an option of 1 , and hence since two digits are equal the 2nd digit also is 1 third digit is between 1- 9 ( number greater than 700= 701 ) since the first 2 digits have taken integer 7, third digit will have (9-1) = 8 so the counting principle equation for case 1 is ( 1x1x8)
1 1 8 ( 9-1)
--- X ----------------- X ---------------------
1st digit is 7 2nd digit is same as 1st 3rd digit is 1-9 = 9
case 2

1st digit is not 7, so for the 1 and 2 digit, options are 8, 9 so 2 options for digit 1 and 2 options for digit 2 ( as per the condition of the problem ) now for 3 digit, since the number is greater than 700 digit will be from 1 - 9= and the 1st 2 digits have taken 7 and 8, so third digit will be 9-2= 7 so counting principle equation becomes ( 2 x2x7)

2 2 7
---------- X --------------------- X ---------------------------------
available same as 1st digit here 8 and 9 are taken by 1st 2 digits so 9 -2=7
option is
8 or 9


Solution thus becomes ( 1x1x8)+(2x2x7) = 36
Are you counting possibilities such as 707?
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by BestGMATEliza » Wed Jul 09, 2014 9:51 pm
you only have 3 possibilities for the hundreds digit: 7, 8 or 9.

700s

you can have 7xx (ex: 722), for this there are 8 possibilities for digits (0 doesn't count because it must be greater than 700, neither does 7)
there is also 7x7 (ex: 702), for this there are 9 possibilities (only 7 doesn't count)
then there is 77x (ex:771), for this there are also 9 possibilities

800s

8xx- 9 possibilities, because you can include 0 so 0-9, not including 8
8x8- 9 possibilities
88x- 9 possibilities

900s

9xx- 9 possibilities, because you can include 0 so 0-9, not including 8
9x9- 9 possibilities
99x- 9 possibilities

Add them all up and you get 80 (C)
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by GMATGuruNY » Thu Jul 10, 2014 2:52 am
II wrote:Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
Integers with exactly 2 digits the same = Total integers - Integers with all 3 digits the same - Integers with all 3 digits different.

Total integers:
To count consecutive integers, use the following formula:
Number of integers = biggest - smallest + 1.
Thus:
Total = 999 - 701 + 1 = 299.

Integers with all 3 digits the same:
777, 888, 999.
Number of options = 3.

Integers with all 3 digits different:
Number of options for the hundreds digit = 3. (7, 8, or 9)
Number of options for the tens digit = 9. (Any digit 0-9 other than the digit already used.)
Number of options for the units digit = 8. (Any digit 0-9 other than the two digits already used.)
To combine these options, we multiply:
3*9*8 = 216.

Thus:
Integers with exactly 2 digits the same = 299-3-216 = 80.

The correct answer is C.
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by Brent@GMATPrepNow » Thu Jul 10, 2014 6:20 am
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
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 = [spoiler]80 = C[/spoiler]

Cheers,
Brent
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by GMATinsight » Fri Jul 11, 2014 9:49 am
Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
Between 700 to 799

770, 771, 771, 773, 774, 775, 776, 778, 779 = 9 Numbers
707, 717, 727, 737, 747, 757, 767, 787, 797 = 9 Numbers
711, 722, 733, 744, 755, 766, 788, 799 = 8 Numbers

Total Such Numbers = 9+8+9 = 26

Between 800 to 899

880, 881, 881, 883, 884, 885, 886, 887, 889 = 9 Numbers
808, 818, 828, 838, 848, 858, 868, 878, 898 = 9 Numbers
800, 811, 822, 833, 844, 855, 866, 877, 899 = 9 Numbers

Total Such Numbers = 9+9+9 = 27

Between 900 to 999

990, 991, 992, 993, 994, 995, 996, 997, 998 = 9 Numbers
909, 919, 929, 939, 949, 959, 969, 979, 989 = 9 Numbers
900, 911, 922, 933, 944, 955, 966, 977, 988 = 9 Numbers

Total Such Numbers = 9+9+9 = 27

Total Numbers = 26+27+27 = 80 Answer
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by Anaira Mitch » Mon Nov 07, 2016 6:36 am
GMATGuruNY wrote:
II wrote:Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
Integers with exactly 2 digits the same = Total integers - Integers with all 3 digits the same - Integers with all 3 digits different.

Total integers:
To count consecutive integers, use the following formula:
Number of integers = biggest - smallest + 1.
Thus:
Total = 999 - 701 + 1 = 299.

Integers with all 3 digits the same:
777, 888, 999.
Number of options = 3.

Integers with all 3 digits different:
Number of options for the hundreds digit = 3. (7, 8, or 9)
Number of options for the tens digit = 9. (Any digit 0-9 other than the digit already used.)
Number of options for the units digit = 8. (Any digit 0-9 other than the two digits already used.)
To combine these options, we multiply:
3*9*8 = 216.

Thus:
Integers with exactly 2 digits the same = 299-3-216 = 80.

The correct answer is C.


Amazing solution Mitch. Thanks for your guidance. Official Guide explanation is quite complex to understand.
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by Scott@TargetTestPrep » Thu Apr 12, 2018 3:55 pm
II wrote:Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(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.

Answer: C

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