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

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.

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

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

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

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.

Thank you and regards,

Rudraksh.

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.

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 .

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

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

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