To find the number of zeros at the end of the product \(3^3 \times 4^4 \times 5^5 \times \ldots \times 49^{49}\), we need to determine how many times the product can be divided by 10. Since \(10 = 2 \times 5\), we will count the number of factors of 2 and 5 in the product. The number of zeros at the end will be determined by the limiting factor, which is the smaller count of 2s or 5s.
### Step-by-Step Solution:
1. **Identify the Product**:
The product is given as \(3^3 \times 4^4 \times 5^5 \times \ldots \times 49^{49}\). This means we are multiplying each integer \(n\) raised to the power of \(n\) from 3 to 49.
2. **Count the Factors of 5**:
To find the number of factors of 5 in this product, we need to consider each term \(n^n\) where \(n\) is a multiple of 5. The multiples of 5 between 3 and 49 are: 5, 10, 15, 20, 25, 30, 35, 40, 45.
- For \(5^5\): contributes 5 factors of 5.
- For \(10^{10}\): contributes \(10/5 = 2\) factors of 5.
- For \(15^{15}\): contributes \(15/5 = 3\) factors of 5.
- For \(20^{20}\): contributes \(20/5 = 4\) factors of 5.
- For \(25^{25}\): contributes \(25/5 + 25/25 = 5 + 1 = 6\) factors of 5.
- For \(30^{30}\): contributes \(30/5 = 6\) factors of 5.
- For \(35^{35}\): contributes \(35/5 = 7\) factors of 5.
- For \(40^{40}\): contributes \(40/5 = 8\) factors of 5.
- For \(45^{45}\): contributes \(45/5 = 9\) factors of 5.
Now we sum these contributions:
\[
5 + 2 + 3 + 4 + 6 + 6 + 7 + 8 + 9 = 50
\]
3. **Count the Factors of 2**:
Next, we count the number of factors of 2 in the product. The multiples of 2 between 3 and 49 are: 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48.
- For \(4^4\): contributes \(4/2 + 4/4 = 2 + 1 = 3\) factors of 2.
- For \(6^6\): contributes \(6/2 + 6/6 = 3 + 1 = 4\) factors of 2.
- For \(8^8\): contributes \(8/2 + 8/4 + 8/8 = 4 + 2 + 1 = 7\) factors of 2.
- For \(10^{10}\): contributes \(10/2 + 10/10 = 5 + 1 = 6\) factors of 2.
- For \(12^{12}\): contributes \(12/2 + 12/4 + 12/12 = 6 + 3 + 1 = 10\) factors of 2.
- For \(14^{14}\): contributes \(14/2 + 14/14 = 7 + 1 = 8\) factors of 2.
- For \(16^{16}\): contributes \(16/2 + 16/4 + 16/8 + 16/16 = 8 + 4 + 2 + 1 = 15\) factors of 2.
- For \(18^{18}\): contributes \(18/2 + 18/18 = 9 + 1 = 10\) factors of 2.
- For \(20^{20}\): contributes \(20/2 + 20/4 + 20/10 = 10 + 5 + 2 = 17\) factors of 2.
- For \(22^{22}\): contributes \(22/2 + 22/22 = 11 + 1 = 12\) factors of 2.
- For \(24^{24}\): contributes \(24/2 + 24/4 + 24/8 + 24/24 = 12 + 6 + 3 + 1 = 22\) factors of 2.
- For \(26^{26}\): contributes \(26/2 + 26/26 = 13 + 1 = 14\) factors of 2.
- For \(28^{28}\): contributes \(28/2 + 28/4 + 28/7 + 28/28 = 14 + 7 + 4 + 1 = 26\) factors of 2.
- For \(30^{30}\): contributes \(30/2 + 30/6 + 30/15 = 15 + 5 + 2 = 22\) factors of 2.
- For \(32^{32}\): contributes \(32/2 + 32/4 + 32/8 + 32/16 + 32/32 = 16 + 8 + 4 + 2 + 1 = 31\) factors of 2.
- For \(34^{34}\): contributes \(34/2 + 34/34 = 17 + 1 = 18\) factors of 2.
- For \(36^{36}\): contributes \(36/2 + 36/4 + 36/9 + 36/36 = 18 + 9 + 4 + 1 = 32\) factors of 2.
- For \(38^{38}\): contributes \(38/2 + 38/38 = 19 + 1 = 20\) factors of 2.
- For \(40^{40}\): contributes \(40/2 + 40/4 + 40/8 + 40/10 + 40/20 = 20 + 10 + 5 + 4 + 2 = 41\) factors of 2.
- For \(42^{42}\): contributes \(42/2 + 42/42 = 21 + 1 = 22\) factors of 2.
- For \(44^{44}\): contributes \(44/2 + 44/4 + 44/11 + 44/44 = 22 + 11 + 4 + 1 = 38\) factors of 2.
- For \(46^{46}\): contributes \(46/2 + 46/46 = 23 + 1 = 24\) factors of 2.
- For \(48^{48}\): contributes \(48/2 + 48/4 + 48/8 + 48/12 + 48/16 + 48/24 + 48/48 = 24 + 12 + 6 + 4 + 3 + 2 + 1 = 52\) factors of 2.
Now we sum these contributions:
\[
3 + 4 + 7 + 6 + 10 + 8 + 15 + 10 + 17 + 12 + 22 + 14 + 26 + 22 + 31 + 18 + 32 + 20 + 41 + 22 + 38 + 24 + 52 = 370
\]
4. **Determine the Number of Zeros**:
The number of zeros at the end of the product is determined by the smaller of the two counts:
\[
\text{Number of zeros} = \min(\text{count of 2s}, \text{count of 5s}) = \min(370, 50) = 50
\]
### Final Answer:
The number of zeros at the end of the product \(3^3 \times 4^4 \times 5^5 \times \ldots \times 49^{49}\) is **50**.
