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

Answer: C
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Of the three-digit integers greater than 700, how many have two digits 06 Oct 2019, 05:08
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We don't even have to solve this question. Just use the logic!!

Because of the symmetry, 3 digit integers that have two digits that are equal to each other and the remaining digit different from the other two from 700-799 is equal to that are from 800-899 or 900-999

Hence total 3 digit numbers greater than 700 have two digits that are equal to each other and the remaining digit different from the other two
= 3k-1 (as we have to exclude 700)

Only option C is in the form of 3k-1

Bunuel
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
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Of the three-digit integers greater than 700, how many have two digits 08 Feb 2023, 22:30
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Is this a 650 Difficulty level question?
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Of the three-digit integers greater than 700, how many have two digits 09 Feb 2023, 02:11
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Is this a 650 Difficulty level question?

You can check the difficulty level of a question along with the stats on it in the first post. For this question Difficulty = 700-Level. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.
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Of the three-digit integers greater than 700, how many have two digits 03 Jun 2023, 21:17
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Bunuel
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

Arrangement method
9_ _ =(1*1*9 )*3!/2! = 27
same same different = 3!/2!
similarly for
8_ _ = 27
and 7 _ _ =27 but we want greater than 700 so minus 1 and the 27 will include 700 so minus 1 = 26

so
27+27+26 = 80
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Of the three-digit integers greater than 700, how many have two digits 29 May 2024, 20:51
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Interesting question, way easier to do total minus what can't happen:

­
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Of the three-digit integers greater than 700, how many have two digits 26 Jun 2024, 05:38
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1st- with 7XX numbers
1. 7__ Here with same remaining two digits only possible combinations are (11,22,33,44,55,66,88,99) = 8 combinations
2. 7_7 Here 9 combinations are possible except 777.
3. 77_ Here as well 9 combinations are possible 
Total for 7xx 8+9+9 = 26
Simlary if we check for 8XX and 9XX we will get 27 for each as 800 and 900 is possible as it wasnt in 7xx case because of given condition in question "..greater than 700.."
Total possible cases = 26 + 27 + 27 = 80.
Answer C.
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Of the three-digit integers greater than 700, how many have two digits 28 Jun 2024, 05:38
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Bunuel
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,

Thank you for the solution. Just wanted to check with you if the below formula can be used to solve these types of questions.

{9*n*(10-k)}-1

where n= number of digits (in this case, 3)
k= first digit of the number (in this case, 7)

I understood the concept, just wanted to check the formula as it will save a lot of time.

Thank you!­
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Of the three-digit integers greater than 700, how many have two digits 04 Sep 2024, 03:21
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Bunuel
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.
Could you explain more why A is 3x9x8?
What are the constraints for the second and third digits?

Many thanks!
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Of the three-digit integers greater than 700, how many have two digits 04 Sep 2024, 06:39
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Bunuel
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.
Could you explain more why A is 3x9x8?
What are the constraints for the second and third digits?

Many thanks!
In case A, the digits are distinct. So, for the second digit, we cannot use the digit selected for the first one, and for the third digit, we cannot use the digits used for the first and second ones, hence 9 and 8 options, respectively.

Hope it's clear.
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Of the three-digit integers greater than 700, how many have two digits 16 Feb 2025, 15:00
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I need to understand how my approach is incorrect -
3C1 (Choosing 1 out of 3) * 9C1 (Choosing 1 out of 9) * 2C1 (Choosing one from the before 2)

What did I forget to take into account?
Bunuel
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
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Of the three-digit integers greater than 700, how many have two digits 16 Feb 2025, 20:54
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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?

Let the number be of the form

Case 1: AAB in which A is same and B is different
A = {7,8,9} 3 options
B = 0-9 except A: 9 options
Total cases = 3*9 = 27

Case 2: ABA in which A is same and B is different
A = {7,8,9} 3 options
B = 0-9 except A: 9 options
Total cases = 3*9 = 27

Case 3: BAA in which A is same and B is different
B = {7,8,9} 3 options
A = 0-9 except A: 9 options
Total cases = 3*9 = 27

Total cases = 27+27+27=81

Since the three-digit integers are greater than 700, 700 is excluded from total cases.

Total cases = 81 - 1 = 80

IMO C
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