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Official GMAT Prep Exam 5 Question

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by 800_or_bust » Fri Jun 17, 2016 12:04 pm
Of the three digit positive integers whose digits are all different and nonzero, how many are odd integers greater than 700?

(a) 84
(b) 91
(c) 100
(d) 105
(e) 243

OA: b
800 or bust!
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by 800_or_bust » Fri Jun 17, 2016 12:10 pm
So I got this as Question #23 on Exam 5 (Exam Pack 2). I missed it on the first go-around. Fell for a nice trap answer they had set up, and assumed I had done the math correctly. Upon review, I was able to recognize my error and generate the correct answer. But I figured I'd post it here and see if there are any alternative means to solve it, and perhaps gain some more insight. Plus, I figured some of the other would-be exam takers might like to see some official practice questions (especially if they didn't purchase the second exam pack).
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by [email protected] » Fri Jun 17, 2016 3:05 pm
Hi 800_or_bust,

To start, you should be careful about posting too many questions from the GMAC CATs - you might inadvertently end up 'ruining' the CAT-taking experience for someone later on (especially if they've seen too many of the questions in advance).

For this question, you might find it useful to break the overall calculation into 'pieces'....

The numbers that were looking for must fit the following restrictions:
1) ODD 3-digit numbers greater than 700.
2) All digits are DIFFERENT and not 0.

Let's start with those between 700 and 800...

The first digit is 'locked in': a 7.
The third digit must be ODD, but CANNOT be 7: 1, 3 5 or 9
The second digit can be anything OTHER than 7, 0 and whatever the third digit is: 7 options
So, there are (1)(4)(7) = 28 numbers in this sub-group

A similar sub-group exists for the numbers between 900 and 1000...

The first digit is 'locked in': a 9.
The third digit must be ODD, but CANNOT be 9: 1, 3 5 or 7
The second digit can be anything OTHER than 9, 0 and whatever the third digit is: 7 options
So, there are (1)(4)(7) = 28 numbers in this sub-group

The numbers between 800 and 900 follow a slightly different pattern though:

The first digit is 'locked in': an 8.
The third digit must be ODD: 1, 3, 5, 7 or 9
The second digit can be anything OTHER than 8, 0 and whatever the third digit is: 7 options
So, there are (1)(5)(7) = 35 numbers in this sub-group

28+28+35 = 91 possible numbers

Final Answer: B

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by GMATGuruNY » Sat Jun 18, 2016 2:11 am
Of three digit positive integers whose three digits are all different and non zero, how many are odd integers greater than 700

1)84
2)91
3)100
4)105
5)243
Case 1: Hundreds digit is 7 or 9
Number of options for the hundreds digit = 2. (7 or 9)
Number of options for the units digit = 4. (Any odd digit but the one in the hundreds place)
Number of options for the tens digit = 7. (Any digit 1-9 but the two already used)
To combine these options, we multiply:
2*4*7 = 56.

Case 2: Hundreds digit is 8
Number of options for the hundreds digit = 1. (Must be 8)
Number of options for the units digit = 5. (Any of the 5 odd digits)
Number of options for the tens digit = 7. (Any digit 1-9 but the two already used)
To combine these options, we multiply:
1*5*7 = 35.

Total options = 56+35 = 91.

The correct answer is B.
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by Matt@VeritasPrep » Thu Jun 23, 2016 4:28 pm
Another approach:

There are three choices for the first integer (7, 8, 9), five choices for the last integer (1, 3, 5, 7, 9), and nine choices (1 through 9) for the middle integer, giving us 3 * 5 * 9 = 135 options.

But every option with repeating digits must be eliminated:

7 7 7: 1 option
7 _ 7 (where the blank is NOT 7): 8 options
7 7 _ (same): 4 options
7 _ _ (same): 4 options
8 8 _ : 5 options
8 _ _ : 5 options
9 9 _ (where the blank is NOT 9): 4 options
9 _ 9 (same): 8 options
9 9 9 (same): 1 option
9 _ _ (same): 4 options

So we have 135 - 44, or 91 possible numbers.
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