To find the number with the highest number of divisors, we need to consider the prime factorization of each number.
The prime factorization of each number from 1 to 15 is as follows:
1: $1$
2: $2$
3: $3$
4: $2^2$
5: $5$
6: $2 \cdot 3$
7: $7$
8: $2^3$
9: $3^2$
10: $2 \cdot 5$
11: $11$
12: $2^2 \cdot 3$
13: $13$
14: $2 \cdot 7$
15: $3 \cdot 5$
To find the number of divisors, we can use the formula that the number of divisors is equal to $(a_1 + 1)(a_2 + 1)\cdots(a_n + 1)$,
where $a_1, a_2, \dots, a_n$ are the exponents of the prime factorization.
Using this formula, we find that the number with the highest number of divisors is $\boxed{12}$,
which has 6 divisors: 1, 2, 3, 4, 6, and 12.
The answer is: 12
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Among the numbers 1, 2, 3, ..., 15, which number has the highest number of divisors?

### Answer:
