To solve the problem, we need to find the number of terms in the expansion of \( (1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5)^{20} \) and then calculate \( O + T \) based on the digits of \( 3^{\lambda} \).
### Step 1: Identify the polynomial
The polynomial given is:
\[
1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5
\]
This can be recognized as the expansion of \( (1 + x)^5 \) using the binomial theorem.
### Step 2: Rewrite the polynomial
We can rewrite the polynomial as:
\[
(1 + x)^5
\]
Thus, the expression becomes:
\[
(1 + x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5
\]
### Step 3: Raise to the power of 20
Now, we need to raise this polynomial to the power of 20:
\[
[(1 + x)^5]^{20} = (1 + x)^{100}
\]
### Step 4: Determine the number of terms
The number of terms in the expansion of \( (1 + x)^{100} \) is given by the formula \( n + 1 \), where \( n \) is the exponent. Here, \( n = 100 \):
\[
\text{Number of terms} = 100 + 1 = 101
\]
Thus, \( \lambda = 101 \).
### Step 5: Calculate \( 3^{\lambda} \)
Now we need to compute \( 3^{\lambda} = 3^{101} \).
### Step 6: Find the units and tens digits of \( 3^{101} \)
To find the units and tens digits of \( 3^{101} \), we can use the properties of powers of 3.
1. **Units digit**: The units digits of powers of 3 cycle every 4 terms:
- \( 3^1 = 3 \) (units digit = 3)
- \( 3^2 = 9 \) (units digit = 9)
- \( 3^3 = 27 \) (units digit = 7)
- \( 3^4 = 81 \) (units digit = 1)
- The cycle is \( 3, 9, 7, 1 \).
Since \( 101 \mod 4 = 1 \), the units digit of \( 3^{101} \) is **3**.
2. **Tens digit**: To find the tens digit, we can calculate \( 3^{101} \mod 100 \) using the Chinese Remainder Theorem or directly calculate the last two digits of \( 3^{101} \).
- We find \( 3^{101} \mod 100 \):
- Using Euler's theorem, since \( \phi(100) = 40 \), we have \( 3^{40} \equiv 1 \mod 100 \).
- Thus, \( 3^{101} = 3^{40 \cdot 2 + 21} \equiv (3^{40})^2 \cdot 3^{21} \equiv 1^2 \cdot 3^{21} \mod 100 \).
- Now we need to compute \( 3^{21} \mod 100 \):
- \( 3^1 = 3 \)
- \( 3^2 = 9 \)
- \( 3^3 = 27 \)
- \( 3^4 = 81 \)
- \( 3^5 = 243 \equiv 43 \mod 100 \)
- \( 3^6 = 3 \cdot 43 = 129 \equiv 29 \mod 100 \)
- \( 3^7 = 3 \cdot 29 = 87 \)
- \( 3^8 = 3 \cdot 87 = 261 \equiv 61 \mod 100 \)
- \( 3^9 = 3 \cdot 61 = 183 \equiv 83 \mod 100 \)
- \( 3^{10} = 3 \cdot 83 = 249 \equiv 49 \mod 100 \)
- Continuing this process, we find \( 3^{21} \equiv 3 \mod 100 \).
Thus, the last two digits of \( 3^{101} \) are 03, meaning:
- Units digit \( O = 3 \)
- Tens digit \( T = 0 \)
### Step 7: Calculate \( O + T \)
Finally, we compute:
\[
O + T = 3 + 0 = 3
\]
### Final Answer
Thus, the value of \( O + T \) is:
\[
\boxed{3}
\]
