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# How many 5-digit positive integers exist the sum of whose

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Senior Manager
Joined: 14 Jul 2006
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How many 5-digit positive integers exist the sum of whose [#permalink]

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05 Aug 2006, 06:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many 5-digit positive integers exist the sum of whose digits are odd?

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Senior Manager
Joined: 22 May 2006
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Re: Comb problem [#permalink]

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05 Aug 2006, 12:23
apollo168 wrote:
How many 5-digit positive integers exist the sum of whose digits are odd?

Possible combinations
1.ooooo
2.eeooo
3.eeeeo

1. 5^5
2. [5!/(2!*3!)]*5^5 - 5^4 (-5^4 to remove numbers start with 0)
3. [5!/4!]*5^5 - 5^4 (-5^4 to remove numbers start with 0)

Hence, 5^5 + 49*5^4 + 24*5^4
= 5^4(5+49+24) = 78*5^4 = 48750.
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Senior Manager
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05 Aug 2006, 18:33
Care to explain more?
I used a simpler method to get a different answer.
There are 3 out of possible 6 combinations of odd and even digits, that result in an odd sum.

There are 99999-10000 = 89999 5 digit numbers totally.
So , the number of odd sum 5 digit nos are 89999/2 or 45000

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Senior Manager
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05 Aug 2006, 22:36
ArvGMAT wrote:
Care to explain more?
I used a simpler method to get a different answer.
There are 3 out of possible 6 combinations of odd and even digits, that result in an odd sum.

There are 99999-10000 = 89999 5 digit numbers totally.
So , the number of odd sum 5 digit nos are 89999/2 or 45000

Sorry for the late respond. I had to go out and grab somethin to eat.

Good point.. you are absolutely right! Thanks!

I made silly mistake
Possible combinations
1.ooooo
2.eeooo
3.eeeeo

1. 5^5
2. [5!/(2!*3!)]*5^5 - [4]*5^4 ( to remove numbers start with 0)
3. [5!/4!]*5^5 - [4]*5^4 ( to remove numbers start with 0)

Hence, 5^5 + (50-4)*5^4 + (25-4)*5^4
= 5^4(5+46+21) = 78*5^4 = 45000

As you said.. there are equal numbers of those integers..
Easiest it can be = (99999-10000 + 1)/2 = 45000
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CEO
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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05 Aug 2006, 23:07
Total 5 digit numbers = 99999-9999 = 90000

Half of these will have odd sum of digits and half will have even sum of digits.
Answer = 90000/2 = 45000.
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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Senior Manager
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Location: united states

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06 Aug 2006, 08:25
freethinking, could you exlain how you did the following:

2. [5!/(2!*3!)]*5^5 - [4]*5^4 ( to remove numbers start with 0)
3. [5!/4!]*5^5 - [4]*5^4 ( to remove numbers start with 0)

number of ways

1)ODD1, ODD, ODD, ODD, ODD

2)ODD1, ODD, ODD, EVEN, EVEN

3)ODD1, EVEN, EVEN, EVEN, EVEN

1) can be accomplished in 5^5 ways

n1 = (5)^5

2) ODD1 can be chosen in 5 ways. Then we have to choose 2 ODD numbers and two EVEN numbers. They all can be choosen in 5 ways each and they can be placed in 4! ways.

n2 = (4!)*(5)^5 = 51000.

I know i commmitted an error here. I haven't subtracted the repititive combinations here. Could somebody help?

3) same thing can be done with this case.

3) OD
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06 Aug 2006, 08:25
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# How many 5-digit positive integers exist the sum of whose

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