Find the number of Zeros at the end of the given expression -
`(8^123 - 8^122 - 8^121) (3^223 - 3^222 - 3^221)`

Options

A. a) 1
B. b) 2
C. c) 0
D. d) 3

Answer

To find the number of zeros at the end of the expression \((8^{123} - 8^{122} - 8^{121}) (3^{223} - 3^{222} - 3^{221})\), we need to determine how many times the factors of 10 can be formed from the expression. Since \(10 = 2 \times 5\), we need to find the minimum of the number of factors of 2 and 5 in the expression.

### Step 1: Simplifying the first part \((8^{123} - 8^{122} - 8^{121})\)

1. Factor out the smallest power of 8:
\[ 8^{121}(8^2 - 8^1 - 1) \]
This simplifies to:
\[ 8^{121}(64 - 8 - 1) = 8^{121}(55) \]

### Step 2: Simplifying the second part \((3^{223} - 3^{222} - 3^{221})\)

1. Factor out the smallest power of 3:
\[ 3^{221}(3^2 - 3^1 - 1) \]
This simplifies to:
\[ 3^{221}(9 - 3 - 1) = 3^{221}(5) \]

### Step 3: Combining the results

Now we have:
\[ (8^{121} \cdot 55) \cdot (3^{221} \cdot 5) \]
This can be rewritten as:
\[ 8^{121} \cdot 3^{221} \cdot 55 \cdot 5 \]

### Step 4: Finding the prime factorization

1. **Prime factorization of \(8^{121}\)**:
\[ 8 = 2^3 \implies 8^{121} = (2^3)^{121} = 2^{363} \]

2. **Prime factorization of \(55\)**:
\[ 55 = 5 \cdot 11 \]

3. **Prime factorization of \(5\)**:
\[ 5 = 5^1 \]

### Step 5: Total prime factors

Combining these, we get:
\[ 2^{363} \cdot 3^{221} \cdot 5^{2} \cdot 11^{1} \]

### Step 6: Counting the factors of 2 and 5

- The number of factors of \(2\) is \(363\).
- The number of factors of \(5\) is \(2\).

### Step 7: Finding the number of zeros

The number of zeros at the end of the expression is determined by the limiting factor, which is the number of \(5\)s:
\[ \text{Number of zeros} = \min(363, 2) = 2 \]

Thus, the number of zeros at the end of the expression is **2**.

---