Sokal da tries to find out the largest positive integer n such that if n transforms to base-7, then it looks like twice of base-10. $156$ is such a number because $(156)_{10}$ = $(312)_7$ and 312 = 2$\times$156. Find out Sokal da's number.
1. **Understanding the Problem:**
   We need to find the largest positive integer \( n \) such that when \( n \) is transformed to base-7, it looks like twice of \( n \) in base-10. Mathematically, we need to find \( n \) such that:
   \[
   (2n)_{10} = (n)_{7}
   \]

2. **Expressing the Condition Mathematically:**
   Let \( n \) be the number in base-10. When \( n \) is converted to base-7, it should be equal to \( 2n \) in base-10. This can be written as:
   \[
   n = a_k \cdot 7^k + a_{k-1} \cdot 7^{k-1} + \cdots + a_1 \cdot 7 + a_0
   \]
   where \( a_i \) are the digits in base-7 representation of \( n \).

3. **Setting Up the Equation:**
   We need:
   \[
   2n = a_k \cdot 7^k + a_{k-1} \cdot 7^{k-1} + \cdots + a_1 \cdot 7 + a_0
   \]
   This implies:
   \[
   2n = n_7
   \]
   where \( n_7 \) is the base-7 representation of \( n \).

4. **Finding the Largest \( n \):**
   We need to find the largest \( n \) such that the above condition holds. Let's test some values:
   - For \( n = 156 \):
     \[
     2 \times 156 = 312
     \]
     Converting \( 312 \) to base-7:
     \[
     312_{10} = 3 \cdot 7^2 + 1 \cdot 7^1 + 2 \cdot 7^0 = 3 \cdot 49 + 1 \cdot 7 + 2 = 147 + 7 + 2 = 156
     \]
     This satisfies the condition.

   - For \( n = 315 \):
     \[
     2 \times 315 = 630
     \]
     Converting \( 630 \) to base-7:
     \[
     630_{10} = 1 \cdot 7^3 + 2 \cdot 7^2 + 6 \cdot 7^1 + 0 \cdot 7^0 = 1 \cdot 343 + 2 \cdot 49 + 6 \cdot 7 + 0 = 343 + 98 + 42 = 483
     \]
     This does not satisfy the condition.

5. **Verification:**
   The largest \( n \) that satisfies the condition is indeed \( 156 \).

The final answer is \( \boxed{ 156 } \).