© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Divisibility Criteria
A number is divisible by 15 if and only if it is divisible by both 3 and 5.
• Divisible by 5 ⇒ The units digit must be 5 or 0. Here, from the digits {1,2,3,5}, the units digit must be 5.
• Divisible by 3 ⇒ The sum of all digits must be a multiple of 3.
Step 2: Fix the Last Digit
Since the number must end in 5, let the 4-digit number be written as (x, y, z, 5). The digits x, y, and z can be chosen from {1, 2, 3, 5} with repetition allowed.
Step 3: Apply the Divisibility by 3 Condition
For (x, y, z, 5) to be divisible by 3, we need:
x + y + z + 5 \equiv 0 \pmod{3}.
Since 5 ≡ 2 (mod 3), the condition becomes:
x + y + z + 2 \equiv 0 \pmod{3} \implies x + y + z \equiv 1 \pmod{3}.
We look for triplets (x, y, z) from {1, 2, 3, 5} whose sum is congruent to 1 modulo 3.
Step 4: Find All Valid Triplets
The valid triplets (x, y, z) that satisfy x + y + z \equiv 1 \pmod{3} are given in the solution as:
(1, 1, 5) → sum = 7, which is 1 mod 3
(1, 1, 2) → sum = 4, which is 1 mod 3
(2, 2, 3) → sum = 7, which is 1 mod 3
(2, 3, 5) → sum = 10, which is 1 mod 3
(3, 3, 1) → sum = 7, which is 1 mod 3
(5, 5, 3) → sum = 13, which is 1 mod 3
Step 5: Count the Permutations for Each Triplet
Since each triplet (x, y, z) can be arranged in 3 positions, we calculate the permutations as follows:
(1, 1, 5): There are 3 items with two 1’s identical. Number of distinct permutations = \frac{3!}{2!} = 3.
(1, 1, 2): Number of distinct permutations = \frac{3!}{2!} = 3.
(2, 2, 3): Number of distinct permutations = \frac{3!}{2!} = 3.
(2, 3, 5): All digits different. Number of distinct permutations = 3! = 6.
(3, 3, 1): Number of distinct permutations = \frac{3!}{2!} = 3.
(5, 5, 3): Number of distinct permutations = \frac{3!}{2!} = 3.
Step 6: Calculate the Total Number of 4-Digit Numbers
By summing all the permutations, we obtain:
3 + 3 + 3 + 6 + 3 + 3 = 21.
Final Answer
Therefore, the total number of 4-digit numbers (with repetition allowed) formed using the digits {1, 2, 3, 5} that are divisible by 15 is
21.
