Question

A five-digit number is formed using digits 1, 3, 5, 7, and 9 without repetition. What is the sum of all such possible numbers?

• A. 6666600
• B. 6666660
• C. 6666666
• D. None

Hint:

For sum of all permutations, multiply frequency per place × digit sum × place value sum.

Answer 1

The Correct Option is A

Given digits: \( \{1, 3, 5, 7, 9\} \) — all odd, and each used once.
Total 5-digit numbers = \( 5! = 120 \)
Each digit appears equally in each position (units, tens, hundreds, etc.).
So, each digit appears in each position \( \frac{5!}{5} = 24 \) times. Sum contributed by each digit: For digit \( d \), its total contribution is: \[ 24 \times d \times (10^0 + 10^1 + 10^2 + 10^3 + 10^4) = 24 \times d \times 11111 \] Sum of digits: \( 1 + 3 + 5 + 7 + 9 = 25 \) Total sum: \[ \text{Sum} = 24 \times 11111 \times 25 = (24 \times 25) \times 11111 = 600 \times 11111 = \boxed{6666600} \]