Number of 4-digit numbers (the repetition of digits is allowed) which are made using the digits 1, 2, 3 and 5, and are divisible by 15, is equal to

For a 4-digit number to be divisible by 15, it must satisfy both of the following conditions simultaneously:

1. Divisibility by 5:

The number must end in 5 (since 0 is not available).

2. Divisibility by 3:

The sum of all four digits must be divisible by 3.

Hence, fix the last digit as 5 and choose the remaining three digits such that the total sum is divisible by 3.

Possible valid selections are:

Group 1: Sets with repeated 1s

$$\{1,1,2\}$$

Sum of digits:

$$1+1+2+5 = 9$$

Since 9 is divisible by 3, the number is divisible by 15.

Number of arrangements of $$1,1,2$$:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(1,1,2), (1,2,1), (2,1,1)$$

Total numbers: $$3$$

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$$\{1,1,5\}$$

Sum of digits:

$$1+1+5+5 = 12$$

Since 12 is divisible by 3, the number is divisible by 15.

Number of arrangements:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(1,1,5), (1,5,1), (5,1,1)$$

Total numbers: $$3$$

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Group 2: Sets with repeated 2s or 5s

$$\{2,2,3\}$$

Sum of digits:

$$2+2+3+5 = 12$$

Valid since 12 is divisible by 3.

Number of arrangements:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(2,2,3), (2,3,2), (3,2,2)$$

Total numbers: $$3$$

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$$\{5,5,3\}$$

Sum of digits:

$$5+5+3+5 = 18$$

Valid since 18 is divisible by 3.

Number of arrangements:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(5,5,3), (5,3,5), (3,5,5)$$

Total numbers: $$3$$

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Group 3: Sets with repeated 3s

$$\{3,3,1\}$$

Sum of digits:

$$3+3+1+5 = 12$$

Valid since 12 is divisible by 3.

Number of arrangements:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(3,3,1), (3,1,3), (1,3,3)$$

Total numbers: $$3$$

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Group 4: Sets with distinct digits

$$\{2,3,5\}$$

Sum of digits:

$$2+3+5+5 = 15$$

Valid since 15 is divisible by 3.

Since all digits are distinct, the number of arrangements is:

$$3! = 6$$

Possible arrangements:

$$(2,3,5), (2,5,3), (3,2,5), (3,5,2), (5,2,3), (5,3,2)$$

Total numbers: $$6$$

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Total number of 4-digit numbers divisible by 15:

$$3+3+3+3+3+6 = 21$$

Hence, the required number of 4-digit numbers is

$$\boxed{21}$$