## Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining di

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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 di

by BTGModeratorVI » Sun Jul 19, 2020 1:36 pm

00:00

A

B

C

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E

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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?

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

Source: Official guide

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### Re: Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remainin

by [email protected] » Tue Jul 21, 2020 4:55 am
BTGModeratorVI wrote:
Sun Jul 19, 2020 1:36 pm
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

Source: Official guide
Let's count the numbers from 701 to 799; then from 800 to 899 and then from 900 to 999.

• 701 to 799:

-From 711, 722, 733, 744, 755, 766, 788, and 799, there are 8 such numbers, excluding 777.

-Now let's take numbers starting with 77, i.e. 77_. The units digits can be any from 0 to 9, except 7. So, there are 9 such numbers.

-Now let's take numbers as 7_7. The tens digits can be any from 0 to 9, except 7. So, there are 9 such numbers.

Total such numbers = 8 + 9 + 9 = 26

• 800 to 899:

The count would be 26 + 1 = 27 as we have to include 800, too.

• 900 to 998:

The count would be 26 + 1 = 27 as we have to include 900, too.

Total count = 26 + 27 + 27 = 80.

Hope this helps!

-Jay
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### Re: Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remainin

by [email protected] » Tue Jul 21, 2020 7:48 am
BTGModeratorVI wrote:
Sun Jul 19, 2020 1:36 pm
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

Source: Official guide
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

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### Re: Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remainin

by [email protected] » Sat Jul 24, 2021 3:40 am
BTGModeratorVI wrote:
Sun Jul 19, 2020 1:36 pm
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