Find the hundred's digit in the co-efficient of `x^(17)` in the expansion of `(1+x^(5)+x^(7))^(20)`.

Answer

To find the hundred's digit in the coefficient of \( x^{17} \) in the expansion of \( (1 + x^5 + x^7)^{20} \), we will use the multinomial theorem. Here’s a step-by-step solution:

### Step 1: Understand the multinomial expansion
The multinomial expansion of \( (a_1 + a_2 + a_3)^n \) is given by:
\[ \sum \frac{n!}{r_1! r_2! r_3!} a_1^{r_1} a_2^{r_2} a_3^{r_3} \]
where \( r_1 + r_2 + r_3 = n \).

### Step 2: Identify the terms in our case
In our case, we have:
- \( a_1 = 1 \)
- \( a_2 = x^5 \)
- \( a_3 = x^7 \)
- \( n = 20 \)

We want to find the coefficient of \( x^{17} \).

### Step 3: Set up the equation for powers
We need to find \( r_2 \) and \( r_3 \) such that:
\[ 5r_2 + 7r_3 = 17 \]
and also satisfy:
\[ r_1 + r_2 + r_3 = 20 \]

### Step 4: Solve the equations
From the second equation, we can express \( r_1 \) as:
\[ r_1 = 20 - r_2 - r_3 \]

Substituting \( r_1 \) into the first equation gives us:
\[ 5r_2 + 7r_3 = 17 \]

### Step 5: Use trial and error to find integer solutions
Let’s try different integer values for \( r_2 \) and \( r_3 \):

1. **If \( r_2 = 1 \)**:
\[ 5(1) + 7r_3 = 17 \implies 7r_3 = 12 \implies r_3 = \frac{12}{7} \text{ (not an integer)} \]

2. **If \( r_2 = 2 \)**:
\[ 5(2) + 7r_3 = 17 \implies 10 + 7r_3 = 17 \implies 7r_3 = 7 \implies r_3 = 1 \]
Now, substituting \( r_2 = 2 \) and \( r_3 = 1 \) into the second equation:
\[ r_1 = 20 - 2 - 1 = 17 \]

So, we have \( r_1 = 17, r_2 = 2, r_3 = 1 \).

### Step 6: Calculate the coefficient
Now we can find the coefficient using:
\[ \text{Coefficient} = \frac{20!}{r_1! r_2! r_3!} = \frac{20!}{17! \cdot 2! \cdot 1!} \]
Calculating this:
\[ = \frac{20 \times 19}{2 \times 1} = \frac{380}{2} = 190 \]

### Step 7: Find the hundred's digit
The coefficient is \( 190 \). The hundred's digit is \( 1 \).

### Final Answer
Thus, the hundred's digit in the coefficient of \( x^{17} \) in the expansion of \( (1 + x^5 + x^7)^{20} \) is **1**.

---