Problem Solving

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

Sorry I forgot the attachment.

is there any other way to solve this problem? perhaps a faster method?

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

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?

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

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?

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)

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

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.