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 determine how many times 10 can be factored from the product. Since \(10 = 2 \times 5\), we will find the number of factors of 2 and 5 in the expression.

### Step 1: Simplifying the first part \(2^{123} - 2^{122} - 2^{121}\)

1. **Factor out the common term**:
\[ 2^{121}(2^2 - 2^1 - 1) \]
This simplifies to:
\[ 2^{121}(4 - 2 - 1) = 2^{121}(1) \]
So, the first part simplifies to \(2^{121}\).

### Step 2: Simplifying the second part \(3^{234} - 3^{233} - 3^{232}\)

2. **Factor out the common term**:
\[ 3^{232}(3^2 - 3^1 - 1) \]
This simplifies to:
\[ 3^{232}(9 - 3 - 1) = 3^{232}(5) \]
So, the second part simplifies to \(3^{232} \times 5\).

### Step 3: Combine the results

3. **Combine the simplified parts**:
\[ (2^{121}) \times (3^{232} \times 5) = 2^{121} \times 3^{232} \times 5 \]

### Step 4: Count the factors of 2 and 5

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

### Step 5: Determine the number of zeros

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

Thus, the number of zeros at the end of the expression is **1**.

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