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Diagnostic: Properties of numbers OG 12 PS 11 - Don t agree

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Diagnostic: Properties of numbers OG 12 PS 11 - Don t agree

by Princesse » Thu Nov 29, 2012 6:45 am
Hi all,

While working on the diagnostic test of the OG12, I have found an answer I don t understand:

"
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

The solution is C.
"

There are 3 pairs of hundreds: 701 - 799 / 800 - 899 / 900 - 999
There are 3 possibilities to place the double digits which apply for each pair: tens and ones (9)/ hundreds and tens (9)/ hundreds and ones (9 !!!).

I don t understand how there could be 9 possibilities to get the same hundred and one number on a scale of 701 to 999. IMHO only 707, 808 and 909 match to the restrictions.

Do you have any idea regarding the topic?

Many thanks in advance!
Last edited by Princesse on Fri Nov 30, 2012 12:47 am, edited 1 time in total.
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by GMATGuruNY » Thu Nov 29, 2012 6:55 am
Princesse wrote:Hi all,

While working on the diagnostic test of the OG12, I have found an answer I don t understand:

"
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

The solution is C.
"

There are 3 pairs of hundreds: 701 - 799 / 800 - 899 / 900 - 999
There are 3 possibilities to place the double digits which apply for each pair: tens and ones (9)/ hundreds and tens (9)/ hundreds and ones (9 !!!).

I don t understand how there could be 9 possibilities to get the same hundred and one number on a scale of 701 to 999. IMHO only 707, 808 and 909 match to the restrictions.

Do you have any idea regarding the topic?

Many thanks in advance!
Hundreds and tens digit the same, units digit different:
Number of choices for hundreds digit = 3 (we can use 7, 8, or 9)
Number of choices for tens digit = 1 (must be the same as the hundreds digit)
Number of choices for units digit = 9 (we can use any digit other the digit already used)
Multiplying the number of choices for each digit, we get 3*1*9 = 27 integers.

Hundreds and units digit the same, tens digit different:
Number of choices for hundreds digit = 3 (we can use 7, 8, or 9)
Number of choices for units digit = 1 (must be the same as the hundreds digit)
Number of choices for tens digit = 9 (we can use any digit other the digit already used)
Multiplying the number of choices for each digit, we get 3*1*9 = 27 integers.

Hundreds digit different, tens and units digit the same:
Number of choices for hundreds digit = 3 (we can use 7, 8, or 9)
Number of choices for tens digit = 9 (we can use anything other than the digit already used)
Number of choices for units digit = 1 (must be the same as the tens digit)
Multiplying the number of choices for each digit, we get 3*9*1 = 27 integers.
But these 27 integers include 700, which we cannot include, leaving us 27-1 = 26 integers.

Total number of possible integers = 27+27+26 = 80.

The correct answer is C.
Last edited by GMATGuruNY on Sun Jan 20, 2013 12:33 pm, edited 1 time in total.
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by GMATGuruNY » Thu Nov 29, 2012 7:04 am
Princesse wrote:Hi all,

While working on the diagnostic test of the OG12, I have found an answer I don t understand:

"
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

The solution is C.
"

There are 3 pairs of hundreds: 701 - 799 / 800 - 899 / 900 - 999
There are 3 possibilities to place the double digits which apply for each pair: tens and ones (9)/ hundreds and tens (9)/ hundreds and ones (9 !!!).

I don t understand how there could be 9 possibilities to get the same hundred and one number on a scale of 701 to 999. IMHO only 707, 808 and 909 match to the restrictions.

Do you have any idea regarding the topic?

Many thanks in advance!
An alternate approach:

Integers with exactly 2 digits the same = Total integers - Integers with all 3 digits the same - Integers with all 3 digits different.

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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by Brent@GMATPrepNow » Thu Nov 29, 2012 7:04 am
Princesse 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
In the future, please use the spoiler function to hide the answer so that others can try the question without seeing the answer.

One approach is to list the 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, there are 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
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by Princesse » Fri Nov 30, 2012 12:49 am
Thank you very much for the explanation !

PS I have added the spoiler to the initial post ;)
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