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Step-by-Step Solution
Step 1: Identify the Constraints
We need to form three-digit numbers from the digits 1, 2, 3, 4, 5 without repeating any digit. Each valid number must lie between 100 and 999 (effectively between 100 and 1000 as stated), and the number should be divisible by 3 or 5.
Step 2: Condition for Divisibility by 3
A number is divisible by 3 if the sum of its digits is divisible by 3. We must pick sets of three different digits (from 1, 2, 3, 4, 5) whose sum is a multiple of 3.
Possible digit triplets (whose sum is divisible by 3) using 1, 2, 3, 4, 5 are:
• (1, 2, 3)
• (2, 3, 4)
• (1, 3, 5)
• (3, 4, 5)
Each triplet can form 3! = 6 distinct 3-digit numbers because we can permute the three digits in 6 ways. Hence, total numbers divisible by 3 (from these digits) are:
4 \times 3! = 4 \times 6 = 24.
Step 3: Condition for Divisibility by 5
A number is divisible by 5 if its last digit (units place) is either 0 or 5. Here, we only have 5 as the valid last digit. Thus, we must pick the digit 5 for the units place and pick two other distinct digits from {1, 2, 3, 4} for the hundreds and tens places.
Possible digit pairs (for hundreds and tens places) from {1,2,3,4} are:
• (1, 2)
• (1, 3)
• (1, 4)
• (2, 3)
• (2, 4)
• (3, 4)
So, there are 6 such pairs, and each pair can be permuted in 2! = 2 ways for the hundreds and tens places. Therefore, total numbers divisible by 5 are:
6 \times 2! = 6 \times 2 = 12.
Step 4: Count the Numbers Divisible by Both 3 and 5
For a 3-digit number to be divisible by both 3 and 5, its digit sum must be divisible by 3 and its last digit must be 5. Combinations that satisfy both are:
• (1, 3, 5)
• (3, 4, 5)
For each valid set of digits, we fix the digit 5 in the units place. The other two digits can be arranged in 2! = 2 ways in the hundreds and tens places. Hence, total numbers divisible by both 3 and 5 are:
2 \times 2! = 2 \times 2 = 4.
Step 5: Apply the Inclusion-Exclusion Principle
According to the inclusion-exclusion principle, to find the total number of Valid three-digit numbers divisible by 3 or 5, we compute:
\text{Count}(\text{Divisible by 3 or 5}) = \text{Count}(\text{Divisible by 3}) + \text{Count}(\text{Divisible by 5}) - \text{Count}(\text{Divisible by both 3 and 5})
Substituting the values:
= 24 + 12 - 4 = 32.
Final Answer
The total number of 3-digit numbers (lying between 100 and 1000) that can be formed using the digits {1, 2, 3, 4, 5} without digit repetition and are divisible by 3 or 5 is
32.
