Question

An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is

• A. \( 18 \times 7! \)
• B. \( 24 \times 7! \)
• C. \( 36 \times 7! \)
• D. \( 72 \times 7! \)

Hint:

A number is divisible by 9 if the sum of its digits is divisible by 9. Use this to determine valid combinations.

Answer 1

The Correct Option is C

Step 1: Divisibility Rule for 9: A number is divisible by 9 if the sum of its digits is divisible by 9. Step 2: Digits used: From 0 to 9 (i.e., 10 digits total), choose 8 **distinct** digits such that their sum is divisible by 9. Total sum of digits 0 through 9: \[ 0 + 1 + 2 + \ldots + 9 = \frac{9 \times 10}{2} = 45 \] We are to select 8 digits out of these 10 such that their sum is divisible by 9. So, we need to **remove 2 digits** whose sum is divisible by 9. Step 3: Count valid digit sets: We find all pairs \( (a, b) \) such that \( a + b = 9, 18, \) or \( 27 \) (since total = 45, removing 9, 18, or 27 keeps the remaining sum divisible by 9): - \( a + b = 9 \): possible pairs = (0,9), (1,8), (2,7), (3,6), (4,5) 5 pairs - \( a + b = 18 \): (9, 9) invalid (can't repeat), and other valid unique pairs: (8,10) invalid, (7,11) invalid etc. So only valid pairs that actually exist within 0-9 = (9,9) [but repeating not allowed] 0 pairs - \( a + b = 27 \): check combinations within 0–9 whose sum is 27 and distinct (9, 8, 7, ..., 0): only (9, 8, 1), (9, 7, 2), etc. would be 3 digits. So only 5 valid 2-digit removal sets where sum = 9 So, total such pairs: **5** Step 4: For each of those 5 valid digit sets: Number of 8-digit numbers = permutations of those 8 digits But since first digit can't be 0, subtract cases where 0 is leading. Total ways for each set: \[ 8! - 7! = 40320 - 5040 = 35280 \] Total ways over 5 sets: \[ 5 \times (8! - 7!) = 5 \times 35280 = 176400 \] Factorized form: \[ = 36 \times 7! \] % Tip