How many factors of the number `16!` Don't have their unit digit 5?

Options

A. (A) 252
B. (B) 7201
C. (C) 5124
D. (D) 5040

Answer

To find how many factors of the number \(16!\) do not have their unit digit as 5, we can follow these steps:

### Step 1: Calculate the prime factorization of \(16!\)

First, we need to find the prime factorization of \(16!\). The prime factors of \(16!\) can be determined by finding the highest powers of all prime numbers less than or equal to 16.

- **For 2:**
\[ \text{Power of } 2 = \left\lfloor \frac{16}{2} \right\rfloor + \left\lfloor \frac{16}{4} \right\rfloor + \left\lfloor \frac{16}{8} \right\rfloor + \left\lfloor \frac{16}{16} \right\rfloor = 8 + 4 + 2 + 1 = 15 \]

- **For 3:**
\[ \text{Power of } 3 = \left\lfloor \frac{16}{3} \right\rfloor + \left\lfloor \frac{16}{9} \right\rfloor = 5 + 1 = 6 \]

- **For 5:**
\[ \text{Power of } 5 = \left\lfloor \frac{16}{5} \right\rfloor = 3 \]

- **For 7:**
\[ \text{Power of } 7 = \left\lfloor \frac{16}{7} \right\rfloor = 2 \]

- **For 11:**
\[ \text{Power of } 11 = \left\lfloor \frac{16}{11} \right\rfloor = 1 \]

- **For 13:**
\[ \text{Power of } 13 = \left\lfloor \frac{16}{13} \right\rfloor = 1 \]

Thus, the prime factorization of \(16!\) is:
\[ 16! = 2^{15} \times 3^6 \times 5^3 \times 7^2 \times 11^1 \times 13^1 \]

### Step 2: Determine the total number of factors of \(16!\)

The total number of factors of a number can be calculated using the formula:
\[ (\text{power of } p_1 + 1)(\text{power of } p_2 + 1)(\text{power of } p_3 + 1) \ldots \]

For \(16!\):
\[ \text{Total factors} = (15 + 1)(6 + 1)(3 + 1)(2 + 1)(1 + 1)(1 + 1) = 16 \times 7 \times 4 \times 3 \times 2 \times 2 \]

Calculating this:
\[ 16 \times 7 = 112 \]
\[ 112 \times 4 = 448 \]
\[ 448 \times 3 = 1344 \]
\[ 1344 \times 2 = 2688 \]
\[ 2688 \times 2 = 5376 \]

So, \(16!\) has a total of \(5376\) factors.

### Step 3: Count factors that have a unit digit of 5

A factor has a unit digit of 5 if it includes at least one factor of 5 and does not include any factors of 2 (since \(2 \times 5 = 10\) which ends with 0).

- **Possible powers of 5:** Since the maximum power of 5 in \(16!\) is 3, the possible powers of 5 are \(1, 2, \text{ or } 3\) (3 choices).

- **Possible powers of 2:** The only valid power of 2 is \(0\) (1 choice).

- **Possible powers of 3:** The powers of 3 can be \(0, 1, 2, 3, 4, 5, 6\) (7 choices).

- **Possible powers of 7:** The powers of 7 can be \(0, 1, 2\) (3 choices).

- **Possible powers of 11:** The powers of 11 can be \(0, 1\) (2 choices).

- **Possible powers of 13:** The powers of 13 can be \(0, 1\) (2 choices).

Now, we multiply the number of choices:
\[ \text{Factors with unit digit 5} = 3 \times 1 \times 7 \times 3 \times 2 \times 2 = 252 \]

### Step 4: Calculate factors that do not have a unit digit of 5

To find the number of factors that do not have a unit digit of 5, we subtract the number of factors with a unit digit of 5 from the total number of factors:
\[ \text{Factors without unit digit 5} = \text{Total factors} - \text{Factors with unit digit 5} \]
\[ = 5376 - 252 = 5124 \]

### Final Answer

Thus, the number of factors of \(16!\) that do not have their unit digit as 5 is \(5124\).

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