A number is divisible by 5 and 27. Find all the numbers which divide the given number.

Answer

To solve the problem of finding all the numbers that divide a number which is divisible by 5 and 27, we can follow these steps:

### Step 1: Find the prime factorization of 5 and 27.
- The prime factorization of 5 is simply \(5\) (since 5 is a prime number).
- The prime factorization of 27 is \(3^3\) (since \(27 = 3 \times 3 \times 3\)).

### Step 2: Find the Least Common Multiple (LCM) of 5 and 27.
- To find the LCM, we take the highest power of each prime factor from the factorizations.
- The LCM will be \(5^1\) and \(3^3\).
- Therefore, the LCM is:
\[ LCM = 5^1 \times 3^3 = 5 \times 27 = 135 \]

### Step 3: Find all the divisors of 135.
- To find the divisors, we can list the factors of 135.
- The prime factorization of 135 is \(3^3 \times 5^1\).
- The formula for finding the number of divisors is \((e_1 + 1)(e_2 + 1)...\) where \(e_1, e_2, ...\) are the powers of the prime factors.
- Here, \(e_1 = 3\) (for 3) and \(e_2 = 1\) (for 5).
- Thus, the number of divisors is \((3 + 1)(1 + 1) = 4 \times 2 = 8\).

### Step 4: List all the divisors of 135.
- To find the actual divisors, we can calculate:
- \(1\)
- \(3\)
- \(5\)
- \(9\) (which is \(3^2\))
- \(15\) (which is \(3 \times 5\))
- \(27\) (which is \(3^3\))
- \(45\) (which is \(3^2 \times 5\))
- \(135\) (which is \(3^3 \times 5\))

So, the divisors of 135 are: **1, 3, 5, 9, 15, 27, 45, 135**.

### Final Answer:
The numbers that divide the given number (which is divisible by 5 and 27) are: **1, 3, 5, 9, 15, 27, 45, 135**.

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