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07 Aug 2010, 06:10
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How many different 7 digit members are their sum of whose digits is even ?
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07 Aug 2010, 14:57
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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.
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.
07 Aug 2010, 15:46
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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.
* 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.
. Thanks Bunuel/IanStewart/crack700 for the explanation. 
