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Sunny00731

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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

Bunuel

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27 Apr 2025, 09:30

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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

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.