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Of the three-digit integers greater than 660, how many have two digits 10 Mar 2012, 04:45
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Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?

(A) 47
(B) 60
(C) 92
(D) 95
(E) 96
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D
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Bunuel
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Of the three-digit integers greater than 660, how many have two digits 10 Mar 2012, 05:36
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enigma123
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96
Show more
Three digit number can have only following 3 patterns:

A. all digits are distinct;
B. all three digits are alike;
C. two digits are alike and third digit is different.

We need to calculate C.

C = Total - A - B

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);

A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;

B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Answer: D.
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Of the three-digit integers greater than 660, how many have two digits 10 Mar 2012, 09:40
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my way of thinking is not elegant :) but anyways , see what I did -
we have the range 661-998

1. two 1st digits are the same
then we have-
661-669 (8 options (exclude 666))
770-779(9 options (exclude 777))
880-889 (9 options (exclude 888))
990-998 (9 options (exclude 999))
total=9*3+8=35

2.the 1st and the last are the same
then we have
6*6 (3 options- 676 686 696))
7*7 (9 options (exclude 777))
8*8 (9 options (exclude 888))
9*9 (9 options (exclude 999))
total=9*3+3=30

3. the 2nd and the 3d terms are the same.
then we have
66x- none option, since our first two digits are already the same, and the 3d cant be the same
7xx -9 options (exclude 777)
8xx- 9 options (exclude 888)
9xx 9 options (exclude 999)
total=9*3=27

30+35+27=92

cant find the rest 3 numbers to get 95. I will appreciate if someone helps me :)
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Of the three-digit integers greater than 660, how many have two digits 10 Mar 2012, 09:53
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LalaB
my way of thinking is not elegant :) but anyways , see what I did -
we have the range 661-998

1. two 1st digits are the same
then we have-
661-669 (8 options (exclude 666))
770-779(9 options (exclude 777))
880-889 (9 options (exclude 888))
990-998 (9 options (exclude 999))
total=9*3+8=35

2.the 1st and the last are the same
then we have
6*6 (3 options- 676 686 696))
7*7 (9 options (exclude 777))
8*8 (9 options (exclude 888))
9*9 (9 options (exclude 999))
total=9*3+3=30

3. the 2nd and the 3d terms are the same.
then we have
66x- none option, since our first two digits are already the same, and the 3d cant be the same
7xx -9 options (exclude 777)
8xx- 9 options (exclude 888)
9xx 9 options (exclude 999)
total=9*3=27

30+35+27=92

cant find the rest 3 numbers to get 95. I will appreciate if someone helps me :)
Show more

You are missing 677, 688 and 699.
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Of the three-digit integers greater than 660, how many have two digits 10 Mar 2012, 09:56
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gosh, only I can do such stupid mistakes! all efforts are in vain, if u come to the wrong answer at last.
thnx Bunuel. appreciate ur help. next time I will be more attentive (at least I hope so)
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Of the three-digit integers greater than 660, how many have two digits 13 Aug 2015, 03:15
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enigma123
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?
Show more
from 661 to 699
66_ ( you have 8 numbers )
6_6 ( you have 3 numbers )
_66 ( you have 3 numbers)
Now from 700 to 799
There are 99 numbers in all. In this 72 numbers (9x8) do not satisfy the conditions
therefore, 99-72 =27 numbers satisfy the conditions
Hence, 27x3+14 = 95 is the answer.
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Of the three-digit integers greater than 660, how many have two digits 24 Jan 2016, 00:28
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I tried to use Bunuel's method, but it was hard for me to escape minor mistakes, consequently my answer wasn't correct.

I find the following way is easier to remember:

600 700 800 900
abc 3 9 9 9
abc 8 9 9 9
abc 3 9 9 9

9*10=90-1+6=95
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Of the three-digit integers greater than 660, how many have two digits 25 Aug 2017, 07:05
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Bunuel
enigma123
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. all three digits are alike;
C. two digits are alike and third digit is different.

We need to calculate C. C = Total - A - B

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Answer: D.
Show more


Hello Bunuel, Considering this logic , can we say there would 9*8*7 three digit numbers with distinct digits
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Of the three-digit integers greater than 660, how many have two digits 25 Aug 2017, 08:36
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Arsh4MBA
Bunuel
enigma123
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. all three digits are alike;
C. two digits are alike and third digit is different.

We need to calculate C. C = Total - A - B

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Answer: D.


Hello Bunuel, Considering this logic , can we say there would 9*8*7 three digit numbers with distinct digits
Show more

There are 9*9*8 3-digit numbers with distinct digits.

9 options for the first digit (from 1 to 9 inclusive).
9 options for the second digit (from 0 to 9 inclusive minus the one we used for the first digit ).
8 options for the third digit (from 0 to 9 inclusive minus 2 digits we used for the first and the second digits)
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Of the three-digit integers greater than 660, how many have two digits 25 Aug 2017, 08:51
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LalaB
my way of thinking is not elegant :) but anyways , see what I did -
we have the range 661-998

1. two 1st digits are the same
then we have-
661-669 (8 options (exclude 666))
770-779(9 options (exclude 777))
880-889 (9 options (exclude 888))
990-998 (9 options (exclude 999))
total=9*3+8=35

2.the 1st and the last are the same
then we have
6*6 (3 options- 676 686 696))
7*7 (9 options (exclude 777))
8*8 (9 options (exclude 888))
9*9 (9 options (exclude 999))
total=9*3+3=30

3. the 2nd and the 3d terms are the same.
then we have
66x- none option, since our first two digits are already the same, and the 3d cant be the same
7xx -9 options (exclude 777)
8xx- 9 options (exclude 888)
9xx 9 options (exclude 999)
total=9*3=27

30+35+27=92

cant find the rest 3 numbers to get 95. I will appreciate if someone helps me :)
Show more


700 , 800 , 900 are missing you have to consider these as well



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Of the three-digit integers greater than 660, how many have two digits 12 Nov 2017, 13:03
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Bunuel
enigma123
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. all three digits are alike;
C. two digits are alike and third digit is different.

We need to calculate C. C = Total - A - B

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Answer: D.
Show more


Hi Bunuel can you kindly elaborate on the first part of the "all the digits are distinct" I didn't quite understand the 3*8, however, I do understand the second part.

Thanks!
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Of the three-digit integers greater than 660, how many have two digits 12 Nov 2017, 21:13
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Salsanousi
Bunuel
enigma123
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. all three digits are alike;
C. two digits are alike and third digit is different.

We need to calculate C. C = Total - A - B

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Answer: D.


Hi Bunuel can you kindly elaborate on the first part of the "all the digits are distinct" I didn't quite understand the 3*8, however, I do understand the second part.

Thanks!
Show more

Three-digit numbers greater than 660 with the hundreds digit of 6: 6XY - X can take only 3 values (7, 8, or 9) and Y can take 8 values (Out of 10 digits, not 6 and not the onte we used for X).
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Of the three-digit integers greater than 660, how many have two digits 12 Nov 2017, 21:39
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Let us first take 700-799

With tens digit 7 we have 9 cases. With units digit 7 we have 9 cases. With tens and units digits same we have 9 cases for a total of 27 cases. Similarly for 800-899 and 900-999 for a total of 81 cases.

Let us now take 661-699

With tens digit 6 , we have 8 cases. With units digit 6, we have 3 cases and with units and tens digit same. we have 3 cases for a total of 14 cases.

So for numbers greater than 660 we have 81+14=95 cases
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Of the three-digit integers greater than 660, how many have two digits 14 Nov 2017, 05:27
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Bunuel Thanks!
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Of the three-digit integers greater than 660, how many have two digits 26 Jan 2019, 11:08
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Let's breakdown the problem into 2 sections:
i) 661 - 699:
case I: x x y
- - - x -> takes only one value i.e., 6 and y takes 9 digits (0,1,2,3,4,5,7,8,9. Since we are already using up 6) It becomes 1*1*9=9
case II: x y x
Applying the same rule as above, we again get 9.
Hence 9+9 numbers between 661-699

ii) Consider 700-999

X X Y => X can take 3 numbers (7,8,9) and Y can take 9 numbers, giving 3*1*9 = 27
X Y X and Y X X=> Same logic as above we get 27*3=81

i) + ii) = 18+81 = 99

Now out of 99, we have 666,777,888,999 that can not be considered.

99-6=95
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Of the three-digit integers greater than 660, how many have two digits 27 Jul 2023, 20:35
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enigma123
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96
I always get stuck on these questions. Any idea what is the concept and how can I solve?
Show more

There are three possibilities:
a) We need the units digit to be 1 and above keeping the other 2 digits 66 - Clearly there are 8 options (661, 662, 663, 664, 665, 667, 668 and 669) ... (i)
b) We need the tens digit to be 7 and above keeping the hundreds digit 6 -
One scenario is that the tens and units digits are same: 677, 688, 699 - 3 options ... (ii)
Another is that the hundreds and units digits are same: 676, 686, 696 - 3 options ... (iii)
c) We need the hundreds digit to be 7 and above (3 options - 7, 8 and 9) -
One scenario is that the tens and units digits are same for each value of the hundreds place: 3 * 9 * 1 = 27 (i.e. 700, 711, ... 799, 800, etc.) - we chose 9 since the digits 7 cannot be used in the tens and units place and similarly for 8 and 9) ... (iv)
Another is that the hundreds and tens places are equal (units is different) - 3 * 1 * 9 = 27 ... (v)
Another is that the hundreds and units places are equal (tens is different) - 3 * 9 * 1 = 27 ... (vi)

Thus, total = 8 + 3 + 3 + 27 + 27 + 27 = 95
Answer D
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Of the three-digit integers greater than 660, how many have two digits 10 Aug 2023, 17:14
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Bunuel
enigma123
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. all three digits are alike;
C. two digits are alike and third digit is different.

We need to calculate C. C = Total - A - B

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Answer: D.
Show more

what about 661 to 699?

where we add that?
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Of the three-digit integers greater than 660, how many have two digits 12 Aug 2023, 02:23
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enigma123
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 47
(B) 60
(C) 92
(D) 95
(E) 96

I always get stuck on these questions. Any idea what is the concept and how can I solve?

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. all three digits are alike;
C. two digits are alike and third digit is different.

We need to calculate C. C = Total - A - B

Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);
A. all digits are distinct = 3*8+3*9*8 = 240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;
B. all three digits are alike = 4 (666, 777, 888, 999).

So, 339-240-4=95.

Answer: D.

what about 661 to 699?

where we add that?
Show more

Your question is not clear. What do you mean by referencing the numbers 661 to 699? These numbers are already included in the total count of numbers (95), which have two digits that are equal to each other and the remaining digit different from the other two. Why should these particular numbers be singled out or treated separately?
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Of the three-digit integers greater than 660, how many have two digits 12 Oct 2024, 00:38
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So if we were calculating 3-digit numbers that have two same and one different digit from 330 to 999:

Assume again three digit number can have only following 3 patterns:
A. all digits are distinct;
B. all three digits are alike;
C. two digits are alike and third digit is different.

And C = Total - A - B

Total numbers from 330 to 999 = 669 (3-digit numbers greater than 330);

A. all digits are distinct = 6*8+6*9*8 = 480.
B. all three digits are alike = {3, 4, 5, 6, 7, 8, 9} = 7
C = 669 - 480 - 7 = 182?
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Of the three-digit integers greater than 660, how many have two digits 12 Oct 2024, 00:46
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felon57
So if we were calculating 3-digit numbers that have two same and one different digit from 330 to 999:

Assume again three digit number can have only following 3 patterns:
A. all digits are distinct;
B. all three digits are alike;
C. two digits are alike and third digit is different.

And C = Total - A - B

Total numbers from 330 to 999 = 669 (3-digit numbers greater than 330);

A. all digits are distinct = 6*8+6*9*8 = 480.
B. all three digits are alike = {3, 4, 5, 6, 7, 8, 9} = 7
C = 669 - 480 - 7 = 182?
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Correct. Good job!
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