Find number of zeros in the end of
`1^3xx2^4xx3^5xx......xx26^28`

Options

A. 100
B. 112
C. 125
D. 128

Answer

To find the number of zeros at the end of the expression \(1^3 \times 2^4 \times 3^5 \times \ldots \times 26^{28}\), we need to determine how many pairs of the factors 2 and 5 are present in the product, since each pair contributes one trailing zero.

### Step 1: Identify the expression
The expression is:
\[ 1^3 \times 2^4 \times 3^5 \times 4^6 \times \ldots \times 26^{28} \]

### Step 2: Count the powers of 2 and 5
To find the number of trailing zeros, we will count the total powers of 2 and 5 in the entire product.

#### Counting the power of 5:
1. For each integer \(n\) from 1 to 26, we need to calculate how many times 5 is a factor in \(n^k\) where \(k\) is the power associated with \(n\).
2. The power of 5 in \(n^k\) can be calculated using the formula:
\[ \text{Power of 5 in } n^k = k \times \left\lfloor \frac{n}{5} \right\rfloor + k \times \left\lfloor \frac{n}{25} \right\rfloor + \ldots \]
3. We will sum the contributions of each \(n\) from 1 to 26.

Calculating the contributions:
- \(1^3\): \(0\)
- \(2^4\): \(0\)
- \(3^5\): \(0\)
- \(4^6\): \(0\)
- \(5^7\): \(1 \times 7 = 7\)
- \(6^8\): \(0\)
- \(7^9\): \(0\)
- \(8^{10}\): \(0\)
- \(9^{11}\): \(0\)
- \(10^{12}\): \(1 \times 12 = 12\)
- \(11^{13}\): \(0\)
- \(12^{14}\): \(0\)
- \(13^{15}\): \(0\)
- \(14^{16}\): \(0\)
- \(15^{17}\): \(1 \times 17 = 17\)
- \(16^{18}\): \(0\)
- \(17^{19}\): \(0\)
- \(18^{20}\): \(0\)
- \(19^{21}\): \(0\)
- \(20^{22}\): \(1 \times 22 = 22\)
- \(21^{23}\): \(0\)
- \(22^{24}\): \(0\)
- \(23^{25}\): \(0\)
- \(24^{26}\): \(0\)
- \(25^{27}\): \(2 \times 27 = 54\) (since \(25 = 5^2\))
- \(26^{28}\): \(0\)

Now, summing these contributions:
\[ 0 + 0 + 0 + 0 + 7 + 0 + 0 + 0 + 0 + 12 + 0 + 0 + 0 + 0 + 17 + 0 + 0 + 0 + 0 + 22 + 0 + 0 + 0 + 0 + 0 + 54 = 112 \]

#### Counting the power of 2:
The power of 2 will be higher than the power of 5, so we can conclude that the limiting factor for trailing zeros will be the number of 5s.

### Step 3: Conclusion
The number of trailing zeros in the expression \(1^3 \times 2^4 \times 3^5 \times \ldots \times 26^{28}\) is equal to the total number of pairs of 2s and 5s, which is determined by the smaller count, that is:
\[ \text{Number of trailing zeros} = 112 \]

### Final Answer:
The number of zeros at the end of the expression is **112**.