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Question Stats:
76% (02:01) correct 24% (02:21) wrong based on 45 sessions
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How many positive positive 5digit integers have the odd sum of their digits? A. \(9*10^2\) B. \(9*10^3\) C. \(10^4\) D. \(45*10^3\) E. \(9*10^4\)
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Official Solution:How many positive 5digit 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 5digit integers: 10,000  10,009  half has even sum and half odd; Next ten 5digit integers: 10,010  10,019  also half has even sum and half odd; And so on. Since there are \(9*10^4\) 5digit integers, then half of it, so \(45∗10^3\) will have odd sum of their digits. Answer: D
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31 Aug 2015, 12:07
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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25 May 2016, 19:39
It's a tricky question, thanks for the tip & answer.
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Re M3117
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28 Jul 2016, 11:40
I think this is a highquality question and I agree with explanation.



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24 Aug 2016, 02:53
I think this is a highquality question and I agree with explanation.



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Re: M3117
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30 Oct 2016, 06:00
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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26 Jan 2017, 22:31
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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28 Feb 2017, 03:26
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, 104=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: M3117
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09 Oct 2017, 04:36
Exactly half of the numbers will have odd sum of their digits.
Total numbers between 10,000 and 99,999 is (Largest integersmallest integer + 1)
So, 99,99910,000+1=90,000
Half of it will be 45,000 or 45x10^3
Answer is C.



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30 May 2018, 02:01
KB04 wrote: Exactly half of the numbers will have odd sum of their digits.
Total numbers between 10,000 and 99,999 is (Largest integersmallest integer + 1)
So, 99,99910,000+1=90,000
Half of it will be 45,000 or 45x10^3
Answer is C. Thanks the solution helps !!!
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24 Jul 2018, 04:32
KB04 wrote: Exactly half of the numbers will have odd sum of their digits.
Total numbers between 10,000 and 99,999 is (Largest integersmallest integer + 1)
So, 99,99910,000+1=90,000
Half of it will be 45,000 or 45x10^3
Answer is C. Shouldn't it be 9999910001?



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Re: M3117
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16 Sep 2018, 11:28
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. 1) OOOOO (five odds) 5^5 2) EEOOO (two evens three odds) 16*5^4 + 6*5^5 3) EEEEO (four evens and an odd) 5^5 + 16*5^4 So 16*5^4 + 6*5^5 + 5^5 + 16*5^4 + 5^5 = 32*5^4 + 8*5^5) = 5^4(32+40) = 5^4*72 = 5^4*2^3*9 = 5^3*2^3*45 = 10^3*45



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04 Oct 2018, 14:17
this is a very tricky question. Thank you.



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Got this wrong but here's how I got to the solution while reviewing the question. Take smaller two digit numbers 10  19 and add them up. Exactly half of them (10,12,14,16,18) add up to an odd number. Hence deduced that exactly half of the 5digit numbers will be odd when digits are added. 99999  10000 + 1 = 90,000/2 = 45,000 = 45 * 10^3



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03 Jul 2019, 00:16
I think this is a highquality question and I agree with explanation.










