The last non-zero number at the end of 20! is ?

Options

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

Answer

To find the last non-zero digit of \(20!\), we can follow these steps:

### Step 1: Calculate the number of factors of 5 in \(20!\)

The last non-zero digit of \(n!\) is influenced by the number of trailing zeros, which are produced by the factors of 10 in the factorial. A factor of 10 is produced by a pair of factors 2 and 5. Since there are generally more factors of 2 than factors of 5 in factorials, we will count the number of factors of 5.

The formula to find the number of factors of a prime \(p\) in \(n!\) is given by:
\[ x = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]

For \(n = 20\) and \(p = 5\):
\[ \left\lfloor \frac{20}{5} \right\rfloor + \left\lfloor \frac{20}{25} \right\rfloor = 4 + 0 = 4 \]
So, there are 4 factors of 5 in \(20!\).

### Step 2: Calculate the number of factors of 2 in \(20!\)

Now, we will calculate the number of factors of 2 in \(20!\) using the same formula:
\[ \left\lfloor \frac{20}{2} \right\rfloor + \left\lfloor \frac{20}{4} \right\rfloor + \left\lfloor \frac{20}{8} \right\rfloor + \left\lfloor \frac{20}{16} \right\rfloor \]
Calculating each term:
\[ 10 + 5 + 2 + 1 = 18 \]
So, there are 18 factors of 2 in \(20!\).

### Step 3: Determine the excess factors of 2

Since we have 4 factors of 5, we can pair them with 4 factors of 2 to form 4 factors of 10. This means we will have:
\[ 18 - 4 = 14 \]
excess factors of 2 left.

### Step 4: Calculate the product of remaining factors

Now we need to find the last non-zero digit of the product of the remaining factors. The remaining factors will be:
- \(2^{14}\)
- \(3^8\)
- \(5^0\) (since we used all 5s)
- \(7^2\)
- \(11^1\)
- \(13^1\)
- \(17^1\)
- \(19^1\)

### Step 5: Calculate the last non-zero digit

Now we will compute the last non-zero digit of:
\[ 2^{14} \times 3^8 \times 7^2 \times 11^1 \times 13^1 \times 17^1 \times 19^1 \]

Calculating each part modulo 10:
- \(2^{14} \mod 10 = 6\)
- \(3^8 \mod 10 = 1\)
- \(7^2 \mod 10 = 9\)
- \(11^1 \mod 10 = 1\)
- \(13^1 \mod 10 = 3\)
- \(17^1 \mod 10 = 7\)
- \(19^1 \mod 10 = 9\)

Now, we will multiply these results together:
\[ 6 \times 1 \times 9 \times 1 \times 3 \times 7 \times 9 \]

Calculating step-by-step:
1. \(6 \times 1 = 6\)
2. \(6 \times 9 = 54 \mod 10 = 4\)
3. \(4 \times 1 = 4\)
4. \(4 \times 3 = 12 \mod 10 = 2\)
5. \(2 \times 7 = 14 \mod 10 = 4\)
6. \(4 \times 9 = 36 \mod 10 = 6\)

Thus, the last non-zero digit of \(20!\) is **4**.

### Final Answer
The last non-zero digit at the end of \(20!\) is **4**.

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