Of the threedigit 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
Answer: C
Source: Official guide
Of the threedigit integers greater than 700, how many have two digits that are equal to each other and the remaining di
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Let's count the numbers from 701 to 799; then from 800 to 899 and then from 900 to 999.BTGModeratorVI wrote: ↑Sun Jul 19, 2020 1:36 pmOf the threedigit 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
Answer: C
Source: Official guide
• 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.
Correct answer: C
Hope this helps!
Jay
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One approach is to start LISTING numbers and look for a PATTERN.BTGModeratorVI wrote: ↑Sun Jul 19, 2020 1:36 pmOf the threedigit 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
Answer: C
Source: Official guide
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
Answer: C
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Solution:BTGModeratorVI wrote: ↑Sun Jul 19, 2020 1:36 pmOf the threedigit 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
Answer: C
Source: Official guide
If the number 700 is included, then we should have the same number of integers in the 700s as in the 800s as in the 900s. In other words, if the number 700 is included, the total number of integers should be a multiple of 3. However, since the number 700 is not included, the total number of integers should be 1 less than a multiple of 3. Looking at the choices, we see that only 80 satisfies this criterion. Thus, it’s the right answer.
Answer: C
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