Question

If all the possible 3-digit numbers are formed using the digits 1, 3, 5, 7, 9 without repeating any digit, then the number of such 3-digit numbers which are divisible by 3 is:

• A. 6
• B. 12
• C. 18
• D. 24 \bigskip

Hint:

To find how many numbers are divisible by 3, check if the sum of their digits is divisible by 3.

Answer 1

The Correct Option is D

Step 1: The available digits are 1, 3, 5, 7, and 9. According to the divisibility rule for 3, a number is divisible by 3 if the sum of its digits is divisible by 3. 

Step 2: The total number of 3-digit numbers that can be formed from the digits 1, 3, 5, 7, and 9 without repetition is: \[ 5 \times 4 \times 3 = 60. \] Now, we need to determine how many of these 3-digit numbers are divisible by 3. 

 Step 3: To check divisibility by 3, we need to ensure the sum of the digits is divisible by 3. Let's evaluate the possible sums of digits formed from the set {1, 3, 5, 7, 9}: 1 + 3 + 5 = 9 (divisible by 3) 
1 + 3 + 7 = 11 (not divisible by 3) 
1 + 3 + 9 = 13 (not divisible by 3) 
1 + 5 + 7 = 13 (not divisible by 3) 
1 + 5 + 9 = 15 (divisible by 3) 
1 + 7 + 9 = 17 (not divisible by 3)
3 + 5 + 7 = 15 (divisible by 3) 
3 + 5 + 9 = 17 (not divisible by 3)
3 + 7 + 9 = 19 (not divisible by 3) 
5 + 7 + 9 = 21 (divisible by 3)
 Step 4: The valid combinations of digits that result in a sum divisible by 3 are: {1, 3, 5} 
{1, 5, 9} 
{3, 5, 7} 
{5, 7, 9}
Step 5: For each valid combination, the number of possible 3-digit numbers is the number of ways to arrange 3 distinct digits, which is \( 3! = 6 \). 

Step 6: Since there are 4 valid combinations and each combination can be arranged in 6 ways, the total number of 3-digit numbers divisible by 3 is: \[ 4 \times 6 = 24. \] Thus, the number of 3-digit numbers divisible by 3 is \( \boxed{24} \).