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# Tricky question

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Intern
Joined: 25 Apr 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 6

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19 Jun 2012, 22:20
Hi,
I have got two questions that I would like to understand the difference in solutions.

1) How many even, three digit integers greater than 700 with distinct, non- zero digits are there
2) How many odd three-digit integers greater than 800 are there such that all their digits are different?

Thx
Intern
Joined: 07 Nov 2011
Posts: 40
Location: India
Concentration: Marketing, Other
GMAT Date: 10-26-2012
GPA: 2.5
WE: Operations (Consulting)
Followers: 0

Kudos [?]: 5 [0], given: 8

Re: Tricky question [#permalink]

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19 Jun 2012, 22:41
If the first digit is 7 then u cant use the no in tens place .. so u will be having 8 other different nos .. and in last place u will be having 7. So in total 1*8*7 = 56
This is for the numbers starting with 7 . Similar goes for nos starting with 8 & 9 .. So the total nos available will be equal to 56 *3 = 168
Joined: 29 Mar 2012
Posts: 324
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 30

Kudos [?]: 441 [0], given: 23

Re: Tricky question [#permalink]

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20 Jun 2012, 00:11
Hi,

1) You have to find the even (3 digit) numbers greater than 700, with non-zero digits as well as distict digits. So, available digits would be (1 to 9)
Even numbers starting with 7 = 1*7*4 = 28 (Hundredth digit is 7 - so, only 1 choice, unit digit can be (2, 4, 6, 8). Now tens digit will not be 7 & a digit chosen at units place - 7 possibilities)
Even numbers starting with 8 = 1*7*3 = 21
Even numbers starting with 9 = 1*7*4 = 28
Total numbers = 77

2) You have to find the odd (3digit) numbers greater than 800, all distict digits. Available digits would be (0 to 9)
Odd numbers starting with 8 = 1*8*5 = 40 (Hundredth digit is 8 - so, only 1 choice. Unit digit can be (1, 3, 5, 7, 9). Now tens digit will not be 8 & a digit chosen at units place - 8 possibilities)
Odd numbers starting with 9 = 1*8*4 = 40 (Hundredth digit is 9 - so, only 1 choice. Unit digit can be (1, 3, 5, 7). Now tens digit will not be 9 & a digit chosen at units place - 8 possibilities)
Total numbers = 32

Let me know, if you need any further help on this.

Regards,
Math Expert
Joined: 02 Sep 2009
Posts: 39042
Followers: 7751

Kudos [?]: 106498 [1] , given: 11626

Re: Tricky question [#permalink]

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20 Jun 2012, 01:20
1
KUDOS
Expert's post
Suren2012 wrote:
Hi,
I have got two questions that I would like to understand the difference in solutions.

1) How many even, three digit integers greater than 700 with distinct, non- zero digits are there
2) How many odd three-digit integers greater than 800 are there such that all their digits are different?

Thx

The first question is discussed here: how-many-even-3-digit-integers-greater-than-700-with-9358.html
The second question is discussed here: how-many-odd-three-digit-integers-greater-than-800-are-there-94655.html

Topic is locked.
_________________
Re: Tricky question   [#permalink] 20 Jun 2012, 01:20
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# Tricky question

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