Which is the smallest positive odd number that has as many divisors as 360?
Let's determine the number of divisors of 360 from its prime factorization according to the known relationship.

If \( n = p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdots \cdots p_{k}^{\alpha_{k}} \), then the number of divisors of \( n \) is

\[
d(n) = (\alpha_{1}+1)(\alpha_{2}+1) \ldots (\alpha_{k}+1).
\]

For \( 360 = 2^{3} \cdot 3^{2} \cdot 5 \), we have \( d(360) = (3+1)(2+1)(1+1) = 24 \).

Now, let's find the smallest odd number \( N \) that has 24 divisors. Since the number we are looking for is odd, only odd prime numbers can appear in its prime factorization. To make \( N \) as small as possible, the smallest odd prime numbers should be used; let's start with 3.

\( N = 3^{23} \): this is a very large number, and we can definitely find a smaller one.

\( N = 3^{\alpha_{1}} \cdot 5^{\alpha_{2}}, (\alpha_{1}+1)(\alpha_{2}+1) = 24 \). Factoring 24 into two factors:

\[
24 = 2 \cdot 12 = 3 \cdot 8 = 4 \cdot 6.
\]

For a base greater than 1, the exponential function is strictly increasing, and if we increase the base, the value of the power increases for the same exponent. Therefore, to get the smallest number, the exponent of 3 should be larger than the exponent of 5.

From the two-factor factorizations of 24: \( N = 3^{11} \cdot 5 \) or \( 3^{7} \cdot 5^{2} \) or \( 3^{5} \cdot 5^{3} \).

Let's compare these products:

\[
3^{11} \cdot 5 = 3^{7} \cdot 3^{4} \cdot 5 = 3^{7} \cdot 3^{2} \cdot 3^{2} \cdot 5 > 3^{7} \cdot 3^{2} \cdot 5^{2} > 3^{7} \cdot 5^{2}
\]

and

\[
3^{7} \cdot 5^{2} = 3^{5} \cdot 3^{2} \cdot 5^{2} > 3^{5} \cdot 5^{3}
\]

Thus, the smallest is \( N = 3^{5} \cdot 5^{3} = 30375 \).

Now, let's take the next prime number, 7, and examine the three-factor factorizations of 24.

\( 24 = 2 \cdot 2 \cdot 6 = 2 \cdot 3 \cdot 4 \), in this case, the smallest \( N = 3^{5} \cdot 5 \cdot 7 \) or \( 3^{3} \cdot 5^{2} \cdot 7 \).

Similarly to the previous case, it can be seen that

\[
3^{3} \cdot 5^{2} \cdot 7 = 4725 < 3^{5} \cdot 5 \cdot 7
\]

If we factor 24 into four factors (which is only possible in one way, since \( 24 = 2^{3} \cdot 3 \)) and take the next prime number, 11, we get \( N = 3^{2} \cdot 5 \cdot 7 \cdot 11 = 3465 \).

Since 24 cannot be factored into more than four primes, this is the smallest number that satisfies the conditions of the problem.

Tünde Sipos (Heves, Eötvös J. Gymnasium, 12th grade)