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# M25-20

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Math Expert
Joined: 02 Sep 2009
Posts: 51035

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16 Sep 2014, 00:23
1
18
00:00

Difficulty:

95% (hard)

Question Stats:

52% (01:33) correct 48% (01:42) wrong based on 485 sessions

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How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 60
C. 72
D. 81
E. 104

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Math Expert
Joined: 02 Sep 2009
Posts: 51035

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16 Sep 2014, 00:23
8
8
Official Solution:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 60
C. 72
D. 81
E. 104

In the range 800 - 900:

1 choice for the first digit: 8

5 choices for the third digit: 1, 3, 5, 7, 9

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Total for this range: $$1*5*8 = 40$$.

In the range 900 - 999:

1 choice for the first digit: 9

4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's the first digit)

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Total for this range: $$1*4*8 = 32$$.

Total: $$40+32 = 72$$.

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Joined: 29 Mar 2014
Posts: 14

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28 Feb 2015, 11:23
1
Bunuel wrote:
Official Solution:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 60
C. 72
D. 81
E. 104

In the range 800 - 900:

1 choice for the first digit: 8

5 choices for the third digit: 1, 3, 5, 7, 9

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Total for this range: $$1*5*8 = 40$$.

In the range 900 - 999:

1 choice for the first digit: 9

4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's the first digit)

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Total for this range: $$1*4*8 = 32$$.

Total: $$40+32 = 72$$.

Hi Bunuel,

I just wanted to clarify this question, as i didn't understand it properly. If the question is asking how many three digit odd integers then why are we including the numbers from 801 - 899...shouldn't these numbers not be counted because these numbers have an 8 in them? (even number) . I thought the question is asking the possibilities of 3 digit odd integers that are all different..example 913 - all odd and the digits are different. I hope my question makes sense, i'm just a bit confused. Thank you.
Math Expert
Joined: 02 Sep 2009
Posts: 51035

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01 Mar 2015, 03:30
pinkcupcake wrote:
Bunuel wrote:
Official Solution:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 60
C. 72
D. 81
E. 104

In the range 800 - 900:

1 choice for the first digit: 8

5 choices for the third digit: 1, 3, 5, 7, 9

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Total for this range: $$1*5*8 = 40$$.

In the range 900 - 999:

1 choice for the first digit: 9

4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's the first digit)

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Total for this range: $$1*4*8 = 32$$.

Total: $$40+32 = 72$$.

Hi Bunuel,

I just wanted to clarify this question, as i didn't understand it properly. If the question is asking how many three digit odd integers then why are we including the numbers from 801 - 899...shouldn't these numbers not be counted because these numbers have an 8 in them? (even number) . I thought the question is asking the possibilities of 3 digit odd integers that are all different..example 913 - all odd and the digits are different. I hope my question makes sense, i'm just a bit confused. Thank you.

801, 803, 805, 807, ... are all odd integers.
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Joined: 27 Jun 2014
Posts: 68
Location: New Zealand
Concentration: Strategy, General Management
GMAT 1: 710 Q43 V45
GRE 1: Q161 V163

GRE 2: Q159 V166
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01 Mar 2015, 17:54
1
Bunuel wrote:
Official Solution:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 60
C. 72
D. 81
E. 104

In the range 800 - 900:

1 choice for the first digit: 8

5 choices for the third digit: 1, 3, 5, 7, 9

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Total for this range: $$1*5*8 = 40$$.

In the range 900 - 999:

1 choice for the first digit: 9

4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's the first digit)

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Total for this range: $$1*4*8 = 32$$.

Total: $$40+32 = 72$$.

Amazingly elegant method.
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"Hardwork is the easiest way to success." - Aviram

One more shot at the GMAT...aiming for a more balanced score.

Intern
Joined: 20 Apr 2015
Posts: 4

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25 May 2015, 06:26
Is there an alternative method to solve this question? trying to think how could this problem be solve by applying the combinatorics formula with repeating elements [N!][/A!B!]
Math Expert
Joined: 02 Sep 2009
Posts: 51035

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25 May 2015, 06:31
josegf1987 wrote:
Is there an alternative method to solve this question? trying to think how could this problem be solve by applying the combinatorics formula with repeating elements [N!][/A!B!]

Check other ways of solving HERE.

Hope it helps.
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Manager
Joined: 28 Dec 2013
Posts: 68

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04 Nov 2015, 15:03
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I don't get explanation for the second digit part can you please list out all possibilities ?
Intern
Joined: 25 Mar 2013
Posts: 7
Location: India
GMAT 1: 700 Q50 V39
WE: Information Technology (Consulting)

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29 Jun 2016, 11:01
I don't agree with the explanation. Why can't 9 be an option for the first digit?
Math Expert
Joined: 02 Sep 2009
Posts: 51035

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30 Jun 2016, 07:01
Vishal Kumar wrote:
I don't agree with the explanation. Why can't 9 be an option for the first digit?

Please re-read the solution above. We do consider 9 as the hundreds digit there.
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Joined: 23 Apr 2016
Posts: 22
Location: Finland
GPA: 3.65

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11 Nov 2016, 13:42
3 digit integers such that all three digits are different between 800-899 :
100th place is reserved for 8,
10th place can be taken by any of the remaining 9 digits (0,1,2,3,4,5,6,7,9)
Unit's pace can be taken by any of the remaining 8 digits
so 9*8 = 72 such numbers
Similary there are 72 such numbers between 900-999
Total number 72 + 72 =144
Between 800-999 (both inclusive) there are 200 numbers, so there should be equal number of odd and even numbers:
hence total odd numbers = 144/2 = 72
Intern
Joined: 16 Jan 2017
Posts: 6

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01 Apr 2017, 09:12
I dont get the explanation for the 2nd digit.
800-899
1st digit: 1 choice (8)
3rd digit: 5 choices (1,3,5,7,9)
2nd digit: 10 digits - 1 digit - 5 digits (0,2,4,6)

This way, you cannot get 811. If I use your solution above, it does not exclude 811, 822, 833, 844 etc.
Math Expert
Joined: 02 Sep 2009
Posts: 51035

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01 Apr 2017, 09:18
Kontaxis wrote:
I dont get the explanation for the 2nd digit.
800-899
1st digit: 1 choice (8)
3rd digit: 5 choices (1,3,5,7,9)
2nd digit: 10 digits - 1 digit - 5 digits (0,2,4,6)

This way, you cannot get 811. If I use your solution above, it does not exclude 811, 822, 833, 844 etc.

The first step when solving a question ALWAYS must be reading CAREFULLY:
How many odd three-digit integers greater than 800 are there such that all their digits are different?
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Joined: 16 Jan 2017
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02 Apr 2017, 04:19
Bunuel wrote:
The first step when solving a question ALWAYS must be reading CAREFULLY:
How many odd three-digit integers greater than 800 are there such that all their digits are different?

Hi Bunuel,

I understand the question stem to explicitly exclude cases where some of the 3 digits are the same, i.e. numbers such as 811, 822, 833 should be excluded. That is why I have less choices for the 2nd digit in my solution as you have. That way I ensure that none of the 3 digits are the same. Could you please elaborate on this?
Math Expert
Joined: 02 Sep 2009
Posts: 51035

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02 Apr 2017, 04:28
Kontaxis wrote:
Bunuel wrote:
The first step when solving a question ALWAYS must be reading CAREFULLY:
How many odd three-digit integers greater than 800 are there such that all their digits are different?

Hi Bunuel,

I understand the question stem to explicitly exclude cases where some of the 3 digits are the same, i.e. numbers such as 811, 822, 833 should be excluded. That is why I have less choices for the 2nd digit in my solution as you have. That way I ensure that none of the 3 digits are the same. Could you please elaborate on this?

So, you say that it should be - In the range 800 - 900: 1*5*4? That's not right. For the 2nd digit you can use any digit apart from those used for the first and the thrid one, so the number of options is 10 - 1 - 1 = 8.

If still not convinced check other ways of solving HERE.
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Joined: 19 Jun 2017
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08 Sep 2017, 14:01
Think of the digits as the menu choices for a three-course meal: Appetizers (hundreds digit), Main (tens digit), and Desserts (units digit)

Appetizer menu has 2 choices: {8,9} -- eat "9"
Dessert menu has 4 choices: {1,3,5,7} i.e. again, "9" is already eaten

Total three-course meal possibilities = 2*9*4 = 72
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Location: Pakistan
GMAT 1: 770 Q48 V50
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25 Jun 2018, 12:13
It took me a little time to understand the question. Like, I had to read it thrice to formulate a concept in my head but once that was done, it was an easy ride.
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Re: M25-20 &nbs [#permalink] 25 Jun 2018, 12:13
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# M25-20

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