For an integer $n>3$ denote by $n?$ the product of all primes less than $n$. Solve the equation $n?=2n+16$.

[i]V. Senderov [/i]
1. **Consider the case when \( n \) is composite:**
   - Let \( n \) be a composite number and \( p \) be a prime such that \( p \mid n \).
   - Since \( p \mid n \), it follows that \( p \mid n? \). Therefore, \( p \mid n? - 2n = 16 \).
   - The prime factors of 16 are 2. Hence, \( p = 2 \).
   - This implies \( n \) must be a power of 2, say \( n = 2^a \) where \( a > 1 \).
   - Substituting \( n = 2^a \) into the equation \( n? = 2n + 16 \):
     \[
     n? = 2^a? = 2 \cdot 2^a + 16 = 2^{a+1} + 16
     \]
   - For \( a > 1 \), \( 2^a? \) includes the product of all primes less than \( 2^a \). Since \( 2^a \) is a power of 2, the product \( 2^a? \) will include at least the primes 2 and 3.
   - Therefore, \( 2^a? \) is even, and \( 2^{a+1} + 16 \) is also even. However, \( 2^a? \) must be divisible by 3 for \( a > 1 \), making \( 2^a? \) not equal to \( 2^{a+1} + 16 \) since \( 2^{a+1} + 16 \) is not divisible by 3.
   - Hence, there is no solution when \( n \) is composite.

2. **Consider the case when \( n \) is prime:**
   - Let \( n \) be a prime number.
   - There exists a prime \( p \) such that \( \frac{n-1}{2} \leq p \leq n-1 \).
   - If \( \frac{n-1}{2} \leq 3 \), then \( n \leq 7 \). The possible values for \( n \) are 3, 5, and 7.
     - For \( n = 3 \):
       \[
       n? = 2 = 2 \cdot 3 + 16 = 22 \quad \text{(not a solution)}
       \]
     - For \( n = 5 \):
       \[
       n? = 2 \cdot 3 = 6 = 2 \cdot 5 + 16 = 26 \quad \text{(not a solution)}
       \]
     - For \( n = 7 \):
       \[
       n? = 2 \cdot 3 \cdot 5 = 30 = 2 \cdot 7 + 16 = 30 \quad \text{(solution)}
       \]
   - If \( \frac{n-1}{2} > 3 \), then \( n > 7 \).
     - For \( n > 7 \), we need to check if \( 2n + 16 \geq 2 \cdot 3 \cdot \frac{n-1}{2} \):
       \[
       2n + 16 \geq 3(n-1)
       \]
       \[
       2n + 16 \geq 3n - 3
       \]
       \[
       19 \geq n
       \]
     - Therefore, \( n \leq 19 \). The possible prime values for \( n \) are 11, 13, 17, and 19.
       - For \( n = 11 \):
         \[
         n? = 2 \cdot 3 \cdot 5 \cdot 7 = 210 \quad \text{and} \quad 2 \cdot 11 + 16 = 38 \quad \text{(not a solution)}
         \]
       - For \( n = 13 \):
         \[
         n? = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310 \quad \text{and} \quad 2 \cdot 13 + 16 = 42 \quad \text{(not a solution)}
         \]
       - For \( n = 17 \):
         \[
         n? = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 = 30030 \quad \text{and} \quad 2 \cdot 17 + 16 = 50 \quad \text{(not a solution)}
         \]
       - For \( n = 19 \):
         \[
         n? = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 = 510510 \quad \text{and} \quad 2 \cdot 19 + 16 = 54 \quad \text{(not a solution)}
         \]

Therefore, the only solution is \( n = 7 \).

The final answer is \( \boxed{ n = 7 } \)