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Odd digits only

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Consider every $3$ -digit number whose digits are all odd, such as $357$ or $911$ . What is the sum of all these numbers?

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Hint

Match each number with the one whose digits are $10$ minus its digits.

Solution

Each digit has $5$ choices, so there are $5 \times 5 \times 5 = 125$ such numbers.
Replacing every digit $d$ by $10 - d$ gives a number of the same kind, its partner, for example $357$ and $753$ ; only $555$ is its own partner.
Each number and its partner add up to $1110$ , since $357 + 753 = 1110$ .
Adding each number to its partner, over all $125$ numbers, gives $125 \times 1110$ and counts every number $2$ times.
So the sum is $125 \times \frac{1110}{2} = 69375$ .

Answer: $69375$