The number of zeros at the end of
`(2^(123) - 2^(122) - 2^(121)) xx (3^(234) - 3^(233) - 3^(232))` :

Options

A. 0
B. 1
C. 121
D. none of these

Answer

To find the number of zeros at the end of the expression \((2^{123} - 2^{122} - 2^{121}) \times (3^{234} - 3^{233} - 3^{232})\), we need to simplify the expression and determine how many times 10 is a factor in the result. Since \(10 = 2 \times 5\), we will look for the minimum of the number of factors of 2 and factors of 5 in the expression.

### Step-by-Step Solution:

1. **Factor out common terms in the first part:**
\[ 2^{123} - 2^{122} - 2^{121} = 2^{121}(2^2 - 2^1 - 1) \]
Here, we factor out \(2^{121}\).

2. **Simplify the expression inside the parentheses:**
\[ 2^2 - 2^1 - 1 = 4 - 2 - 1 = 1 \]
Thus, we have:
\[ 2^{123} - 2^{122} - 2^{121} = 2^{121} \times 1 = 2^{121} \]

3. **Factor out common terms in the second part:**
\[ 3^{234} - 3^{233} - 3^{232} = 3^{232}(3^2 - 3^1 - 1) \]
Here, we factor out \(3^{232}\).

4. **Simplify the expression inside the parentheses:**
\[ 3^2 - 3^1 - 1 = 9 - 3 - 1 = 5 \]
Thus, we have:
\[ 3^{234} - 3^{233} - 3^{232} = 3^{232} \times 5 \]

5. **Combine the results:**
Now, we can combine both parts:
\[ (2^{121}) \times (3^{232} \times 5) = 2^{121} \times 3^{232} \times 5 \]

6. **Count the factors of 2 and 5:**
- The number of factors of 2 is \(121\).
- The number of factors of 5 is \(1\) (from the \(5\) in the expression).

7. **Determine the number of zeros at the end:**
The number of zeros at the end of a number is determined by the minimum of the number of factors of 2 and 5:
\[ \text{Number of zeros} = \min(121, 1) = 1 \]

### Final Answer:
The number of zeros at the end of the expression is **1**.