Question

Find the sum of all 3-digit natural numbers which contain at least one odd digit and at least one even digit.

Answer 1

Let $X$ denote the set of all 3-digit natural numbers
Let $O$ have only odd digits and $E$ have only even digits
Then $X$ $(O\cup E)$ is the set of all 3-digit natural numbers having atleast one odd digit and atleast one even digit. The desired sum is therefore

${\sum }_{x\in X}x={\sum }_{y\in Y}y-{\sum }_{z\in Z}z$
${\sum }_{x\in X}x={\sum }_{j=1}^{999}j-{\sum }_{k=1}^{99}k$
$=\frac{999\times 1000}{2}-\frac{99\times 100}{2}$
$=50\times 9891=494550$
Consider $O=(1,3,5,7,9)$. Suppose the digit in unit's place is $1$. We can fill the digit in ten's place in $5$ ways and the digit in hundred's place in $5$ ways. Thus there are $25$ numbers in the set $O$ each of which has $1$ in its unit's place. Similarly, there are $25$ numbers whose digit in unit's place is $3$; $25$ having its digit in unit's place as $5$; $25$ with $7$ and $25$ with $9$.
Thus the sum of the digits in unit's place of all the numbers in $O$ is
$25(1+3+5+7+9)=25\times 25=625$
A similar argument shows that the sum of digits in ten's place of all the numbers in $O$ is $625$ and that in hundred's place is also $625$. thus the sum of all the numbers in $O$ is
$625({10}^2+10+1)=625\times 111=69375$
Consider $E=(0,2,4,6,8)$.
The digit in hundred's place is never $0$. Suppose the digit in unit's place is $0$. There are $4\times 5=20$ such numbers. Similarly, $20$ numbers each having digits $2,4,6,8$ in their unit's place. Thus the sum of the digits in unit's place of all the numbers in $E$ is
$20(0+2+4+6+8)=20\times 20=400$
A similar reasoning shows that the sum of the digits in ten's place of all the numbers in $E$ is $400$, but the sum of the digits in hundred's place of all the numbers in $E$ is $25\times 25=500$. Thus the sum of all the numbers in $E$ is
$500\times {10}^2+400\times 10+400=54400$.
The required sum is
$494550-69375-54400=370775$