M31-17

Question

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

Bunuel · Accepted Answer

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 all 5-digit integers will have an odd sum of their digits and the other half will have an even sum.
 
Let's consider the first ten 5-digit integers: 10,000 through 10,009. Half of them have an even sum of digits and the other half have an odd sum.
 
We can then move to the next ten 5-digit integers: 10,010 through 10,019. Again, half of them have an even sum and half have an odd sum.
 
This pattern continues for all 5-digit integers. 
 
Since there are  9*10^4  5-digit integers, half of them, which is  45*10^3 , will have an odd sum of their digits.

Answer: D

srivelivala · Answer

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?

shasadou · Answer

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.

Linhbiz · Answer

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

PerseveranceWins · Answer

I think this is a  high-quality  question and I agree with explanation.

Senthil7 · Answer

I think this is a  high-quality  question and I agree with explanation.

Ditstat · Answer

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.

CPGguyMBA2018 · Answer

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%.

smkashyap · Answer

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).

mbadude2017 · Answer

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  -> 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

KB04 · Answer

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.

theeni03 · Answer

I think this is a  high-quality  question and I agree with explanation.

RK007 · Answer

I think this is a  high-quality  question and I agree with explanation.

coreyander · Answer

Using   COUNTING Method

For sum to be ODD, we have 3 possibilities

1) All 5 digits odd ---- OOOOO
2) 3 Odd, 2 Even --- OOOEE
3) 1 Odd, 4 Even ---- OEEEE

Lets Count

1) All Odd   ----------> OOOOO = 5*5*5*5*5 = 3125

2) 3 Odd, 2 Even 
 I) First digit Odd: O---OOEE = 5*5*5*5*5 = 3125....... (1)
 But, last 4 digits can be arranged in different orders, OEOE, OEEO etc... Number of such items will 4!/(2!2!)=6. 
 Thus, we multiply (1) with 6 = 3125*6

II) First digit Even: E---OOOE = 4*5*5*5*5 = 2500.....(2)....  (First digit can't be 0, thus number of even terms for 1st digit is 4) 
 But, last 4 digits can be arranged in different orders, OOOE, OEOO etc... Number of such items will 4!/(3!)=4.
 Thus, we multiply (2) with 4 = 2500*4

3) 1 Odd, 4 even 
 I) First digit Even: E---EEEO = 4*5*5*5*5 = 2500.....(3)... (First digit can't be 0, thus number of even terms for 1st digit is 4) 
 But, last 4 digits can be arranged in different orders. Number of such items will 4!/(3!)=4.
 Thus, 2500*4
 
 II) First digit Odd: O---EEEE= 5*5*5*5*5 = 3125

Add all possibilities = 3125*8 + 2500*8 = 25000 + 20000 = 45000 = 45*10^3.... ANS (D)

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Blair15 · Answer

Is this true for all 4 digit, 3 digit ,etc numbers as well?

Posted from my mobile device

Bunuel · Answer

Yes, this is true for all integers, regardless of the number of digits. For example, consider two-digit integers from 10 to 99. The first number, 10, has an odd sum of its digits (1+0=1), then 11 has an even sum (1+1=2), and it alternates for 10 numbers until 19. The same happens from 20 to 29, from 30 to 39, and finally, from 90 to 99. This pattern of alternating between odd and even sums of digits in every set of 10 numbers is consistent for all n-digit integers.

So, for any range of n-digit integers, we can deduce that half of them will have an odd sum of digits and the other half will have an even sum of digits.

Shivang29 · Answer

I did not quite understand the question. not sure if its just me but the wording doesn't make any sense.

Bunuel · Answer

The wording of the question is clear. I’d recommend re-reading the discussion above carefully—you might find the clarification you need. Also, there’s another discussion on the same question (https://gmatclub.com/forum/how-many-pos ... 98547.html), which might be helpful. In the future, it would be great if you could phrase your questions more specifically and with more details.

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