Find the number of Zeros at the end of the expression -
`(3^123-3^122-3^121)(2^121-2^120-2^119)`

Options

A. a) 1
B. b) 0
C. c) 119
D. d) 120

Answer

To find the number of zeros at the end of the expression \((3^{123} - 3^{122} - 3^{121})(2^{121} - 2^{120} - 2^{119})\), we need to determine how many times the factors of 10 can be formed in the expression. Since \(10 = 2 \times 5\), we need to find the number of pairs of factors of 2 and 5 in the final expression.

### Step-by-Step Solution:

1. **Factor out the common terms in each part of the expression**:
- For the first part, \(3^{123} - 3^{122} - 3^{121}\):
- The lowest power of \(3\) is \(3^{121}\). Factor this out:
\[ 3^{121}(3^2 - 3^1 - 3^0) = 3^{121}(9 - 3 - 1) = 3^{121}(5) \]

- For the second part, \(2^{121} - 2^{120} - 2^{119}\):
- The lowest power of \(2\) is \(2^{119}\). Factor this out:
\[ 2^{119}(2^2 - 2^1 - 2^0) = 2^{119}(4 - 2 - 1) = 2^{119}(1) \]

2. **Combine the factored expressions**:
- Now, we can rewrite the entire expression:
\[ (3^{121} \cdot 5)(2^{119} \cdot 1) = 3^{121} \cdot 2^{119} \cdot 5 \]

3. **Identify the factors of 2 and 5**:
- From the expression \(3^{121} \cdot 2^{119} \cdot 5\):
- The number of factors of \(2\) is \(119\).
- The number of factors of \(5\) is \(1\).

4. **Calculate the number of pairs of \(2\) and \(5\)**:
- The number of zeros at the end of the expression is determined by the limiting factor, which is the number of pairs of \(2\) and \(5\):
\[ \text{Number of zeros} = \min(\text{number of factors of } 2, \text{number of factors of } 5) = \min(119, 1) = 1 \]

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