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M31-17

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Official Solution:

How many positive 5-digit integers have the odd sum of their digits?

A. \(9*10^2\)
B. \(9*10^3\)
C. \(10^4\)
D. \(45*10^3\)
E. \(9*10^4\)


Exactly half of the number will have odd sum of their digits and half even.

Consider first ten 5-digit integers: 10,000 - 10,009 - half has even sum and half odd;

Next ten 5-digit integers: 10,010 - 10,019 - also half has even sum and half odd;

And so on.

Since there are \(9*10^4\) 5-digit integers, then half of it, so \(45∗10^3\) will have odd sum of their digits.


Answer: D
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Re: M31-17 [#permalink]

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New post 31 Aug 2015, 11:07
Why should exactly half of the numbers have odd sum of their digits and half even?? Is it due to the principle of symmetry? Can you please give me an example?

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srivelivala wrote:
Why should exactly half of the numbers have odd sum of their digits and half even?? Is it due to the principle of symmetry? Can you please give me an example?


there are 6 possible combinations of digits for a 5 digit number, of these 6 combinations only 3 would suffice our criteria, i.e. half:

1. all odd - 11111 (sum is odd)
2. all even - 22222 (sum is even)
3. 1 even 4 odd - 21111 (even)
4. 2 even 3 odd - 22111 (odd)
5. 3 even 2 odd - 11222 (even)
6. 4 even 1 odd - 22221 (odd)

To calculate the possible number of 5 digit integers we can use simple counting:

9 * 10 * 10 * 10 * 10 = 9 * 10^4 (Note a digit cannot start from 0 so fo the first place we take 9 digits out of 10, for the rest places the digits can repeat). Hence half of this is 45 * 10^3. Answer D.
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New post 25 May 2016, 18:39
It's a tricky question, thanks for the tip & answer.
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New post 28 Jul 2016, 10:40
I think this is a high-quality question and I agree with explanation.

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New post 24 Aug 2016, 01:53
I think this is a high-quality question and I agree with explanation.

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New post 30 Oct 2016, 05:00
Hi,
I came up with this method:

In order to sum up the digits to be odd, there are only 3 ways;
1) OOOOO (5^5)
2) EEOOO (4x5^4) as the first digit cannot be zero
3) EEEEO (4*5^4) as the first digit cannot be zero

And I got 13 x 5^4.

What did I do wrong??

Thanks in advance.

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Re: M31-17 [#permalink]

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New post 21 Jan 2017, 21:00
Ditstat wrote:
Hi,
I came up with this method:

In order to sum up the digits to be odd, there are only 3 ways;
1) OOOOO (5^5)
2) EEOOO (4x5^4) as the first digit cannot be zero
3) EEEEO (4*5^4) as the first digit cannot be zero

And I got 13 x 5^4.

What did I do wrong??

Thanks in advance.

I have also approached the problem in the same way. Can any expert clarify as to what am I missing here?

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Re: M31-17 [#permalink]

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New post 26 Jan 2017, 21:31
E+E = E
E+O = O
O+E = O
O+O=E

There are only 2 ways out of 4 possible ways digits of an integer can form odd. exactly 50%.

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Re: M31-17 [#permalink]

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New post 28 Feb 2017, 02:26
Ditstat wrote:
Hi,
I came up with this method:

In order to sum up the digits to be odd, there are only 3 ways;
1) OOOOO (5^5)
2) EEOOO (4x5^4) as the first digit cannot be zero
3) EEEEO (4*5^4) as the first digit cannot be zero

And I got 13 x 5^4.

What did I do wrong??

Thanks in advance.


I guess we would have to do something about the arrangement of EEOOO. 4*5^4 is the number of arrangements with position of Evens and odds fixed. Doesn't consider EOEOO, for example. Ideally if we knew Es and Os to be distinct, we would multiple by 5! On the other hand, if we knew Es and Os to be same, we would use the 'repetition' formula. But we have a mix of distinct and similar items.

Consider 22111 (EEOOO); we would have 21211 (EOEOO).
But if we consider 24111 (EEOOO), we would have 21411 and 41211 (EOEOO).

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niks18 wrote:
Ditstat wrote:
Hi,
I came up with this method:

In order to sum up the digits to be odd, there are only 3 ways;
1) OOOOO (5^5)
2) EEOOO (4x5^4) as the first digit cannot be zero
3) EEEEO (4*5^4) as the first digit cannot be zero

And I got 13 x 5^4.

What did I do wrong??

Thanks in advance.

I have also approached the problem in the same way. Can any expert clarify as to what am I missing here?


I don't claim to be an expert but I think this may help.

If you want an odd sum, you need an even number of even digits and odd number of odd digits.

Combos of 5 odds (e.g. OOOOO) = 5^5
Total: 5^5

Combos of 3 odds, 2 evens (e.g. EEOOO or OOOEE) = 5!/3!2! = 10
But, if the first number is even, there are only 4 possibilities for that digit (2, 4, 6, 8; 0 will give you a < a 5 digit integer so we can't use it). How many combos of 3 odds, 2 evens will give you an even in the front? E[EOOO] -> 1c1*4c3 = 4 combos with even in front, 10-4=6 combos with odd in front
Numbers with odd digit in front (6 of the 10 combos): 6*(5^5)
Numbers with even digit in front(4 of the 10 combos): 4*(4*5^4)
Total: 6*5^5+4*(4*5^4)

Combos of 1 odd, 4 evens (e.g. EEEEO): = 5!/4! = 5
Numbers with even digit in front (4 of the 5 combos): 4*(4*5^4)
Numbers with odd digit in front (1 of the 5 combos): 1*(5^5)

Sum them all up:
Total numbers with odd digit in front: 1*5^5+6*5^5+1*5^5 = 8*5^5 = 25,000
Total numbers with even digit in front: 4*(4*5^4)+4*(4*5^4)=8*4*5^4=20,000
Sum: 45,000 = 45*10^3 = D

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Re: M31-17 [#permalink]

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New post 09 Oct 2017, 03:36
Exactly half of the numbers will have odd sum of their digits.

Total numbers between 10,000 and 99,999 is (Largest integer-smallest integer + 1)

So, 99,999-10,000+1=90,000

Half of it will be 45,000 or 45x10^3

Answer is C.

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Re: M31-17   [#permalink] 09 Oct 2017, 03:36
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