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M25-20

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M25-20 [#permalink]

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Re M25-20 [#permalink]

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New post 16 Sep 2014, 01:23
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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: \(40+32 = 72\).


Answer: C
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Re: M25-20 [#permalink]

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New post 28 Feb 2015, 12:23
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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: \(40+32 = 72\).


Answer: C


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.

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Re: M25-20 [#permalink]

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New post 01 Mar 2015, 04:30
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: \(40+32 = 72\).


Answer: C


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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M25-20 [#permalink]

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New post 01 Mar 2015, 18:54
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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: \(40+32 = 72\).


Answer: C



Amazingly elegant method.
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Re: M25-20 [#permalink]

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New post 25 May 2015, 07:26
Thanks for your explanation!
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!]

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Re: M25-20 [#permalink]

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New post 25 May 2015, 07:31
josegf1987 wrote:
Thanks for your explanation!
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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Collection of Questions:
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Re M25-20 [#permalink]

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New post 04 Nov 2015, 16:03
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 ?

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Re M25-20 [#permalink]

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New post 29 Jun 2016, 12:01
I don't agree with the explanation. Why can't 9 be an option for the first digit?

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Re: M25-20 [#permalink]

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New post 30 Jun 2016, 08:01

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Re: M25-20 [#permalink]

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New post 11 Nov 2016, 14:42
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

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Re: M25-20 [#permalink]

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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.

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Re: M25-20 [#permalink]

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New post 01 Apr 2017, 10:18
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M25-20 [#permalink]

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New post 02 Apr 2017, 05:19
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?

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Re: M25-20 [#permalink]

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New post 02 Apr 2017, 05:28
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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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Kudos [?]: 118722 [0], given: 12006

Re: M25-20   [#permalink] 02 Apr 2017, 05:28
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