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Sunny00731
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M31-17 27 Apr 2025, 07:33
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I don’t quite agree with the solution. Since the question is about integers so technically there are a total of 2X(9X10^4) hence the number of even integers will be 9X10^4
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Bunuel
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M31-17 27 Apr 2025, 09:30
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Sunny00731

Bunuel
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

I don’t quite agree with the solution. Since the question is about integers so technically there are a total of 2X(9X10^4) hence the number of even integers will be 9X10^4
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The total number of 5-digit integers is fixed: from 10000 to 99999, which gives exactly 9*10^4 numbers. There is no reason to multiply by 2.
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wArrior902
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M31-17 22 Jun 2025, 01:24
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I like the solution - it’s helpful.
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adarsh2209
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M31-17 26 Aug 2025, 12:03
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I like the solution - it’s helpful.
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saepeducimus
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M31-17 08 Dec 2025, 04:33
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I like the solution - it’s helpful.
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M31-17 09 May 2026, 03:51
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I like the solution - it’s helpful.
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cdb
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M31-17 13 Jul 2026, 07:15
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Ditstat
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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You did not consider the full cases for option 2 & 3, 3 odd and 2 even will have to bifurcated in two cases: 1. When first digit is odd, then remaining condition: 6*5^5 2. When first digit is even, then remaining cases : 8*5^4.

For case three also: two cases: 1. When first digit is odd, then remaining condition: 5^5 2. When first digit is even, then remaining cases : 16*5^4.

This way total cases will be if combine 45*10^3.
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cdb
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M31-17 13 Jul 2026, 07:19
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Sunny00731
I don’t quite agree with the solution. Since the question is about integers so technically there are a total of 2X(9X10^4) hence the number of even integers will be 9X10^4
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Try this: It took 10 min btw :)

Case1: All five odd number digit: OOOOO
Total cases: 5^5

Case 2: 3 odd and 2 even will have to bifurcated in two cases:
1. When first digit is odd, then remaining condition: 6*5^5
2. When first digit is even, then remaining cases : 8*5^4.

Case:3 1 Odd and 4 even
1. When first digit is odd, then remaining condition: 5^5
2. When first digit is even, then remaining cases : 16*5^4.

This way total cases will be if combine 45*10^3.
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