Find the last non-zero digit of 96!.

Options

A. 2
B. 4
C. 6
D. 8

Answer

To find the last non-zero digit of \( 96! \), we can follow these steps:

### Step 1: Express \( n \) in the form \( 5a + b \)
We need to express \( 96 \) in the form \( 5a + b \).
- Here, \( a = 19 \) and \( b = 1 \) since \( 96 = 5 \times 19 + 1 \).

### Step 2: Calculate \( 2^a \)
Next, we calculate \( 2^a \) where \( a = 19 \).
- \( 2^{19} \)

### Step 3: Find \( a! \) and \( b! \)
Now we need to calculate \( a! \) and \( b! \):
- \( a! = 19! \)
- \( b! = 1! = 1 \)

### Step 4: Combine the results
We combine the results to find the last non-zero digit using the formula:
\[ \text{Last non-zero digit} = 2^{19} \times 19! \times 1! \]
This simplifies to:
\[ 2^{19} \times 19! \]

### Step 5: Reduce \( 2^{19} \) modulo 10
To find the last non-zero digit, we need to reduce \( 2^{19} \) modulo 10. The powers of 2 cycle every 4:
- \( 2^1 = 2 \)
- \( 2^2 = 4 \)
- \( 2^3 = 8 \)
- \( 2^4 = 6 \)
- Then it repeats: \( 2^5 = 2 \), \( 2^6 = 4 \), etc.

Since \( 19 \mod 4 = 3 \), we have:
\[ 2^{19} \equiv 8 \mod 10 \]

### Step 6: Calculate \( 19! \) modulo 10
Now we need to compute \( 19! \) modulo 10, but we must ignore the factors of 10 (which come from pairs of 2 and 5).

- Count the number of 2's and 5's in \( 19! \):
- Number of 5's: \( \left\lfloor \frac{19}{5} \right\rfloor = 3 \)
- Number of 2's: \( \left\lfloor \frac{19}{2} \right\rfloor + \left\lfloor \frac{19}{4} \right\rfloor + \left\lfloor \frac{19}{8} \right\rfloor + \left\lfloor \frac{19}{16} \right\rfloor = 9 + 4 + 2 + 1 = 16 \)

Since we have more 2's than 5's, we can ignore 3 pairs of (2, 5) which gives us \( 10^3 \) (or 1000), and we are left with \( 19 - 3 = 16 \) factors of 2.

### Step 7: Calculate the last non-zero digit of \( 19! \)
We can calculate \( 19! \) ignoring the factors of 10:
\[ 19! = 1 \times 2 \times 3 \times 4 \times 6 \times 7 \times 8 \times 9 \times 11 \times 12 \times 13 \times 14 \times 16 \times 17 \times 18 \times 19 \]
Calculating this modulo 10, we can multiply and reduce at each step to find the last non-zero digit.

### Step 8: Combine results
Finally, we combine \( 2^{19} \) and \( 19! \) to find the last non-zero digit of \( 96! \).

### Final Result
After performing all calculations, the last non-zero digit of \( 96! \) is found to be **6**.

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