Below is an OG-13, diagnostic test question:
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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How to solve it without actually listing the numbers?
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GMATGuruNY
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Integers with exactly 2 digits the same = Total integers - Integers with all 3 digits the same - Integers with all 3 digits different.NaimaB wrote:Below is an OG-13, diagnostic test question:
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
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.
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NaimaB
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GMATGuruNY,
This is an awesome solution. Thank you very much for your quick reply.
I actually listed all the numbers and then counted!!! How stupid I am!
This question's difficulty is at what level? Easy, medium or high?
This is an awesome solution. Thank you very much for your quick reply.
I actually listed all the numbers and then counted!!! How stupid I am!
This question's difficulty is at what level? Easy, medium or high?
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Brent@GMATPrepNow
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One approach is to start LISTING numbers and look for a PATTERN.NaimaB wrote: 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
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 = [spoiler]80 = C[/spoiler]
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
