How many positive 5-digit integers have the odd sum of their digits?

Question

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

askhere · Answer

We are looking at numbers between 10000 and 99999 both inclusive. 
There are 90000 numbers. 
Now for 
10000 : sum of digits is odd; 
10001 :sum of digits is even; 
10002 : sum of digits is odd ; so on and so forth. So every alternate number is such that the sum of digit is odd.
( Exception for the above statement :
When it is 10009 the sum is even and for 10010 again the sum is even; But if you look at 10019 :sum is odd; 10020 : sum is odd 
and this pattern continues so basically the number of odd sum of digits and even sum of digits are equal )

This means exactly half of the numbers will have odd sum of their digits. i.e 45000

Answer :  D

Ambarish

RudeyboyZ · Answer

Dear Ambarish,

How were you able to figure out the trend, could you explain this a little in detail...!

sorry for sounding a little stupid.

Warm Regards,

Rudraksh.

Jackal · Answer

My attempt:

Given set to work on is 10,000 to 99,999 which is 90,000 numbers.
First sum is 1 followed by 2 then 3 up to 45.
Half of them is odd and half even so 45,000 numbers have an odd digit sum.
I will go with like Ambarish for 45*10^3

askhere · Answer

Dear Rudraksh,

Its good if you remember in a sequence of consecutive integers : (where the number of "numbers in the sequence" is even. In this case 90000.)
"No: of integers with even sum of digits = No: of integers with odd sum of digits" 
This will give you the answer directly as 90000/2 = 45000.

I am happy to elaborate:
In GMAT it is very important that we figure out a pattern so that we can solve it easily. 
As I mentioned we were looking for numbers from 10000 to 99999 which gives you a odd sum of digits.

10000 : sum of digits = 1+0+0+0+0 = 1 which is odd
10001 : sum of digits = 1+0+0+0+1 = 2 which is even. As long as you add one to the  units  digit the sum of the digits alternates between odd and even.
The exception is when you look at the consecutive numbers where  tens  digit gets changed. (You can even check this for hundreds and thousands digits) 
For example: 
10009: sum of digits = 1+0+0+0+9 = 10 which is  even 
10010: sum of digits = 1+0+0+1+0 = 2 which is again  even . 
So you have two consecutive numbers which gives you sum as  even  and destroyed the pattern. But hang on, if you look at 
10019: sum of digits = 1+0+0+1+9 = 11 which is  odd 
10020: sum of digits = 1+0+0+2+0 = 3 which is again  odd 
So again you have two consecutive numbers which gives you sum as  odd  and compensated for the case in which you got two even sums in succession.
This keeps on repeating and one can see that exactly half the numbers from 10000 to 99999 will be even and the other half odd.

Hope you understood.

Ambarish

chetan2u · Answer

total five digits number= 9*10*10*10*10.. half will have sum as even and other half odd..
so odd sum=9*10*10*10*10/2=45*10^3.. D

RudeyboyZ · Answer

Dear Ambarish, Great explanation, the concept is much clear to me now. it just a little confusing when reading it in the beginning, haven't touched Mathematics for a while till like now. :-D

Thank you and regards, Rudraksh.

mejia401 · Answer

Alternative solution:
 9  choices for the ten thousands integer
 10  choices for the thousands integer
 10  choices for the hundreds integer
 10  choices for the tens integer
 5  choices for the units integer, to control whether the sum is odd or even.

9 * 5 * 10 * 10 * 10 = 45*10^{3}

IMO Answer D.

vitaliyGMAT · Answer

Different approach (although longer one). We can get odd sum of 5 numbers in 3 cases: Odd + Odd + Odd + Odd + Odd Odd + Odd + Odd + Even + Even Odd + Even + Even + Even + Even Case #1 Odd + Odd + Odd + Odd + Odd = 5^5 Case #2 a) Ten thousands place is even --> Even + Even + Odd + Odd + Odd = 4*5^4* \frac{4!}{3!} = 4^2*5^4 b) Ten thousands place is odd --> Odd + Even + Even + Odd + Odd = 5^5* \frac{4!}{2!*2!} = 6*5^5 Case #3 a) Ten thousands place is even --> Even + Even + Even + Even + Odd = 4*5^4*\frac{4!}{3!} = 4^2*5^4 b) Ten thousands place is odd --> Odd + Even + Even + Even + Even = 5^5 Cases are independent and we can add individual outcomes. Total # of integers = 5^5 + 4^2*5^5 + 6*5^5 + 4^2*5^4 + 5^5 = 5^4 (5 + 16 + 30 + 16 + 5) = 5^4*72 = 9*8*5^4 = 45*10^3 The fastest approach is when we actually convince ourselves that there are only two possible outcomes for the sum of the digits – it’s either odd or even – and then divide total # of outcomes ( 9*10*10*10*10 = 9*10^4 ) by 2 . But it’s good to confirm it by applying other approaches :-D

.

GMATinsight · Answer

For digit sum of a 5 digit number to be ODD

Case 1 : All digits of the number are Odd

Such numbers can be made in  5*5*5*5*5  because we have 5 odd digits to use at 5 different places to make a 5 digit number
Total cases = 5*5^4

Case 2 : 3 digits of the number are odd and 2 are even

- Starting from even digit --- 4C3* 4(can't be zero) *5*5*5*5 ----- 4C3 is used to select 3 places out of 4 to be filled by odd digits----- = 16*5^4
- Starting from odd digit ---4C2*5*5*5*5*5 ----- 4C2 is used to select 2 places out of 4 to be filled by even digits----- = 30*5^4
Total cases = 46*5^4

Case 3 : 1 digits of the number is odd and 4 are even

- Starting from even digit --- 4C1* 4(can't be zero) *5*5*5*5 ----- 4C1 is used to select 1 places out of 4 to be filled by odd digit----- = 16*5^4
- Starting from odd digit --- 5*5*5*5*5 = 5*5^4
Total cases = 21*5^4

Total Possible such numbers = (5+46+21)*5^4 = 72*5^4 = 45*10^3

Answer: Option D

Archit3110 · Answer

total such integers would be half of the total digits in the given set
so total integers ; 99999-10000+1 ; 90,000 /2 ; 45*10^3
IMO D

bumpbot · Answer

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How many positive 5-digit integers have the odd sum of their digits?

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