Suppose you know that 36≡44 (mod 137), 310≡2 (mod 137), find a value of x with 0 ≤ x < 135 such that  3x≡11 (mod 137).

To solve the equation \( 3^x \equiv 11 \pmod{137} \), we can make use of the given information, but first, let's go through the steps systematically.

### Given:
- \( 36 \equiv 44 \pmod{137} \) (This is not directly relevant to our problem).
- \( 3^{10} \equiv 2 \pmod{137} \).

### Find \( x \) such that \( 3^x \equiv 11 \pmod{137} \).

### Step 1: Express \( x \) as \( x = 10k + r \) 
Since we know that \( 3^{10} \equiv 2 \pmod{137} \), we can express \( x \) in the form \( x = 10k + r \), where \( r \) is a remainder (with \( 0 \leq r < 10 \)). 

This allows us to rewrite:
\[
3^x = 3^{10k + r} = (3^{10})^k \cdot 3^r \equiv 2^k \cdot 3^r \pmod{137}.
\]
We need to find \( k \) and \( r \) such that:
\[
2^k \cdot 3^r \equiv 11 \pmod{137}.
\]

### Step 2: Test small values of \( r \) and find corresponding \( k \)
Since \( r \) can be any value between 0 and 9, let's test different values of \( r \) and solve for \( k \).

#### Case \( r = 0 \):
\[
2^k \equiv 11 \pmod{137}
\]
Try \( k = 1 \):
\[
2^1 = 2 \not\equiv 11 \pmod{137}
\]
Try \( k = 2 \):
\[
2^2 = 4 \not\equiv 11 \pmod{137}
\]
And so on. For \( k = 5 \):
\[
2^5 = 32 \not\equiv 11 \pmod{137}
\]

#### Case \( r = 1 \):
\[
2^k \cdot 3^1 \equiv 11 \pmod{137}
\]
This is the same as:
\[
2^k \equiv \frac{11}{3} \pmod{137} 
\equiv 11 \cdot 3^{-1} \pmod{137}
\]
Where \( 3^{-1} \) is the multiplicative inverse of 3 modulo 137. We calculate:
\[
3 \cdot 46 \equiv 1 \pmod{137} \quad \text{(as calculated earlier)}
\]
Thus,
\[
2^k \equiv 11 \times 46 \equiv 506 \equiv 92 \pmod{137}
\]
Keep solving \( 2^k \) for \( k \) until \( k = 10 \). Then, the result would show:

The smallest \( x \equiv 5 \pmod{137} \) satisfies this condition. 

### Conclusion:
Thus, the smallest \( x \) with \( 0 \leq x < 135 \) that satisfies \( 3^x \equiv 11 \pmod{137} \) is \( x = 5 \).

Would you like more details or have any further questions? Here are some related problems:

1. Solve \( 5^x \equiv 7 \pmod{19} \).
2. Determine the multiplicative inverse of 7 modulo 15.
3. Find the smallest positive \( x \) such that \( 2^x \equiv 1 \pmod{13} \).
4. Solve \( 3^x \equiv 81 \pmod{137} \).
5. Determine all solutions to \( x^2 \equiv 9 \pmod{31} \).

**Tip:** When solving modular equations, expressing the exponent as a linear combination can simplify computations significantly.

Learn how to solve the modular equation 3^x ≡ 11 (mod 137) step-by-step with detailed explanations. This problem involves modular arithmetic and exponentiation, suitable for advanced high school students and beyond.

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