How many different 7 digit members are their sum of whose : GMAT Problem Solving (PS)
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# How many different 7 digit members are their sum of whose

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How many different 7 digit members are their sum of whose [#permalink]

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07 Aug 2010, 05:10
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How many different 7 digit members are their sum of whose digits is even ?
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Re: How to solve this question [#permalink]

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07 Aug 2010, 09:13
My attempt:

7 digit number is of the form abcdefg. In this the sum of the digits would end up as a even number only if the following is true.

a) There number of even number among these numbers -- a, b, c, d, e, f and g should be either 1 or 3 or 5 or 7.

Also the number of even and odd single digit numbers are 5. (1,3,5,7,9 and 0, 2, 4, 6, 8)

To summarize, let us take the even digit to be e and odd digit to be o.

Hence the number abcdefg could be either

ooooooe OR ooooeee OR ooeeeee OR eeeeeee

Number of ways to choose a even or odd digit is 5.

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Re: How to solve this question [#permalink]

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07 Aug 2010, 12:58
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zareentaj wrote:
How many different 7 digit members are their sum of whose digits is even ?

Here is how i approached it.

Sum of the 7 digits is even if the number has 1 or 3 or 5 or 7 even digits
There are 5 even digits available 0,2,4,6,8 and 5 odd digits available 1,3,5,7,9. However the first digit in the 7 digit number should not be 0. Otherwise it will become a 6 digit number.

7E 0O = $$5^7-5^6$$
There are 5 even numbers available. So each of 7 the digits can be filled with an even number in $$5^7$$ ways, but this also contains the numbers having 0 in the first digit location.
If first digit has a 0 , rest 6 digits can be filled with an even number in $$5^6$$ ways

5E 2O = $$C^7_5*5^5*5^2-C^6_4*5^4*5^2$$

We can first pick the 5 spots having even numbers in $$C^7_5$$ ways ,fill them each with even number in $$5^5$$ ways and fill the remaining 2 digits with odd numbers in $$5^2$$ ways. Again we need to subtract the numbers having 0 in the first digit location
If first digit has a 0 , we can pick the 4 spots out of the remaining 6 having even numbers in $$C^6_4$$ ways ,fill them each with even number in $$5^4$$ ways and fill the remaining 2 digits with odd numbers in $$5^2$$ ways.

3E 4O = $$C^7_3*5^3*5^4-C^6_2*5^2*5^4$$

We can first pick the 3 spots having even numbers in $$C^7_3$$ ways ,fill them each with even number in $$5^3$$ ways and fill the remaining 4 digits with odd numbers in $$5^4$$ ways. Again we need to subtract the numbers having 0 in the first digit location
If first digit has a 0 , we can pick the 2 spots out of the remaining 6 having even numbers in $$C^6_2$$ ways ,fill them each with even number in 5^2 ways and fill the remaining 4 digits with odd numbers in $$5^4$$ ways.

1E 6O = $$C^7_1*5^1*5^6-5^6$$

We can first pick the 1 spot having an even number in $$C^7_1$$ ways ,fill it with an even number in $$5^1$$ ways and fill the remaining 6 digits with odd numbers in $$5^6$$ ways. Again we need to subtract the numbers having 0 in the first digit location
If first digit has a 0 , we can fill the remaining 6 digits with odd numbers in $$5^6$$ ways.

Total no of 7 digit numbers whose sum of digits is even=
$$5^7-5^6$$+$$C^7_5*5^7-C^6_4*5^6$$+ $$C^7_3*5^7-C^6_2*5^6$$+$$C^7_1*5^7-5^6$$
=$$4*5^6+90*5^6+160*5^6+34*5^6$$
=$$288*5^6$$

Zareentaj - What's the OA ?
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Re: How to solve this question [#permalink]

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07 Aug 2010, 13:57
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I guess the question is:

How many different 7-digit numbers are there the sum of whose digits are even?

From the first glance I'd say exactly half of the number will have the sum of their digits odd and half even, why should there be more even than odd?

Consider first ten 7-digit integers: 1,000,000-1,000,009 - half has even sum and half odd;
Next ten 7-digit integers: 1,000,010-1,000,019 - also half has even sum and half odd;
And so on.

There are $$9*10^6$$ 7-digit integers, half of it, $$45*10^5$$ will have odd sum of their digits, or as crack700 wrote $$288*5^6$$, ($$288*5^6=2^5*9*5^6=(9*5)*(2^5*5^5)=45*10^5$$).

Hope it's clear.
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Re: How to solve this question [#permalink]

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07 Aug 2010, 14:46
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I like Bunuel's approach, or you can think of things as follows. Let's start by ignoring the units digit:

* we have 9 choices for the first digit, and 10 choices for each remaining digit, so there are 9*10^5 choices for the first six digits.

* now the sum of the first six digits is either even or odd. If it's even, we need our last digit to be even, so we have 5 choices. If it's odd, we need our last digit to be odd, so we have 5 choices. So no matter how we choose our first six digits, we have 5 choices for our units digit, and the answer must then be 9 * 10^5 * 5 = 45* 10^5.
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Re: How to solve this question [#permalink]

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07 Aug 2010, 15:37
Bunuel/Ian
Can this approach be extended in general i.e., How many N digit integers are there whose sum is even/odd?
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Re: How to solve this question [#permalink]

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07 Aug 2010, 15:41
crack700 wrote:
zareentaj wrote:
How many different 7 digit members are their sum of whose digits is even ?

Here is how i approached it.

Sum of the 7 digits is even if the number has 1 or 3 or 5 or 7 even digits
There are 5 even digits available 0,2,4,6,8 and 5 odd digits available 1,3,5,7,9. However the first digit in the 7 digit number should not be 0. Otherwise it will become a 6 digit number.

7E 0O = $$5^7-5^6$$
There are 5 even numbers available. So each of 7 the digits can be filled with an even number in $$5^7$$ ways, but this also contains the numbers having 0 in the first digit location.
If first digit has a 0 , rest 6 digits can be filled with an even number in $$5^6$$ ways

5E 2O = $$C^7_5*5^5*5^2-C^6_4*5^4*5^2$$

We can first pick the 5 spots having even numbers in $$C^7_5$$ ways ,fill them each with even number in $$5^5$$ ways and fill the remaining 2 digits with odd numbers in $$5^2$$ ways. Again we need to subtract the numbers having 0 in the first digit location
If first digit has a 0 , we can pick the 4 spots out of the remaining 6 having even numbers in $$C^6_4$$ ways ,fill them each with even number in $$5^4$$ ways and fill the remaining 2 digits with odd numbers in $$5^2$$ ways.

3E 4O = $$C^7_3*5^3*5^4-C^6_2*5^2*5^4$$

We can first pick the 3 spots having even numbers in $$C^7_3$$ ways ,fill them each with even number in $$5^3$$ ways and fill the remaining 4 digits with odd numbers in $$5^4$$ ways. Again we need to subtract the numbers having 0 in the first digit location
If first digit has a 0 , we can pick the 2 spots out of the remaining 6 having even numbers in $$C^6_2$$ ways ,fill them each with even number in 5^2 ways and fill the remaining 4 digits with odd numbers in $$5^4$$ ways.

1E 6O = $$C^7_1*5^1*5^6-5^6$$

We can first pick the 1 spot having an even number in $$C^7_1$$ ways ,fill it with an even number in $$5^1$$ ways and fill the remaining 6 digits with odd numbers in $$5^6$$ ways. Again we need to subtract the numbers having 0 in the first digit location
If first digit has a 0 , we can fill the remaining 6 digits with odd numbers in $$5^6$$ ways.

Total no of 7 digit numbers whose sum of digits is even=
$$5^7-5^6$$+$$C^7_5*5^7-C^6_4*5^6$$+ $$C^7_3*5^7-C^6_2*5^6$$+$$C^7_1*5^7-5^6$$
=$$4*5^6+90*5^6+160*5^6+34*5^6$$
=$$288*5^6$$

Zareentaj - What's the OA ?

Kudos for the patience displayed. Although if taken this approach in GMAT I wonder if it can be done in 2 mins? Assume Bunuel or Ian's approach is what they would be looking for...
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Re: How to solve this question [#permalink]

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07 Aug 2010, 16:09
Wow! . Thanks Bunuel/IanStewart/crack700 for the explanation.
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Re: How to solve this question [#permalink]

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08 Aug 2010, 00:39
mainhoon wrote:
Bunuel/Ian
Can this approach be extended in general i.e., How many N digit integers are there whose sum is even/odd?

For $$n\geq{2}$$ - yes.
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Re: How many different 7 digit members are their sum of whose [#permalink]

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20 Aug 2014, 04:49
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Re: How many different 7 digit members are their sum of whose [#permalink]

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02 Feb 2016, 09:55
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Re: How many different 7 digit members are their sum of whose   [#permalink] 02 Feb 2016, 09:55
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