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HINT: Suppose that the digits are $abc$.

• How many choices are there for $ab$?


• How many choices are there for $c$ if $a+b$ is even? How many choices are there for $c$ if $a+b$ is odd? Does it make any difference whether $a+b$ is odd or even?

A simple way of seeing that it is exactly half, is to realize that we have a bijection between 3 digit numbers by sending $N$ to $1099- N$. Note that since the sum of these numbers is odd, hence the pair of numbers have different parity.

Put explicitly, we are pairing up ${100, 999}, {101, 998} ldots, {544, 545 } $.

Ways to get an even sum:

$2$ odd $1$ even.

$3$ even.

First case:

First see that there are three posiblities for the place the even number takes: (the first number cant be 0 because then it would not be a 3 digit number. Therefore there are $(4*5*5)+2(5*5*5)$ ways

second case: all even. there are $4*5*5$ ways to do it therefore the total is $100+250+100=450$ numbers

Exactly half of them.

If you allow leading zeroes, there are $1000$ three-digit numbers, if you are strict, there are $900$ three-digit numbers, so the final answer is $500$ or $450$, depending on your definition.
Note that among any two consecutive three-digit numbers tha tdiffer only in their last digit, one has even and one has odd digit sum.

HINT If the final digit of $n$ (units digit) is not $9$ then $n$ and $n+1$ have one odd digit sum and one even sum. Can you see how to pair numbers up so that one is odd and one is even, and every three-digit number is included in the pairing?

HINT:

So, the number digits with odd value is even

If we take no digits with odd values, we shall have $4cdot 5cdot5=100$ combinations ,

If we take two digits with odd values,

we have following combinations $(O,O,E),(O,E,O),(E,O,O)$

Now, for the first digit in even case, it can assume $4$ values namely, ${2,4,6,8}$

and in odd case, it can assume, $5$ values namely, ${1,3,5,7,9}$

Since the even number = sum 3 even or sum 2 even, 1 odd so we have 4 case:

1. The number is $, overline{abc}, $ where $a,b,cin { 0,2,4,6,8}$. The number $a$ has $4$ choices, the number $b$ has $4$ choices, the number $c$ has $3$ choices. Therefore, we have $, 4*4*3=48,$ numbers.


2. The number is $, overline{abc}, $ where $ain { 2,4,6,8}$ and $b,c in {1,3,5,7,9 }$. The number $a$ has $4$ choices, the number $b$ has $5$ choices, the number $c$ has $4$ choices. Therefore, we have $, 4*5*4=80,$ numbers.


3. The number is $, overline{abc}, $ where $a,cin { 0,2,4,6,8}$ and $bin {1,3,5,7,9}$. The number $a$ has $4$ choices, the number $c$ has $4$ choices, the number $b$ has $5$ choices. Therefore, we have $, 4*4*5=80,$ numbers.


4. The number is $, overline{abc}, $ where $cin { 0,2,4,6,8}$ and $a, bin {1,3,5,7,9}$. The number $a$ has $5$ choices, the number $b$ has $4$ choices, the number $c$ has $5$ choices. Therefore, we have $, 5*4*5=100,$ numbers.
   The conclution, we have $, 48+80+80+100=308$ numbers.