Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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?

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.
_________________

It's a tricky question, thanks for the tip & answer.
_________________

[4.33] In the end, what would you gain from everlasting remembrance? Absolutely nothing. So what is left worth living for? This alone: justice in thought, goodness in action, speech that cannot deceive, and a disposition glad of whatever comes, welcoming it as necessary, as familiar, as flowing from the same source and fountain as yourself. (Marcus Aurelius)

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

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?

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).

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