What is the smallest positive integer $n$ such that $2n$ is a perfect square and $3n$ is a perfect cube?
If $2n$ is a perfect square, then $n$ must be divisible by 2. Now if $3n$ is a perfect cube and $n$ is divisible by 2, then $n$ must be divisible by $3^2=9$ and by $2^3=8$. Therefore, the smallest positive integer $n$ such that $2n$ is a perfect square and $3n$ is a perfect cube is $9\times8=\boxed{72}$.
72
Level 3
Number Theory
Problem: What is the smallest positive integer $n$ such that $2n$ is a perfect square and $3n$ is a perfect cube?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the smallest positive integer \( n \) such that \( 2n \) is a perfect square and \( 3n \) is a perfect cube, we start by expressing \( n \) in terms of its prime factorization. Let \( n = 2^a \cdot 3^b \cdot p_1^{c_1} \cdots p_k^{c_k} \), where \( p_i \) are other prime factors.

For \( 2n \) to be a perfect square, the exponent of each prime factor in its prime factorization must be even. Thus, we have:
- For \( 2n = 2^{a+1} \cdot 3^b \cdot p_1^{c_1} \cdots p_k^{c_k} \), \( a+1 \) must be even, implying \( a \) is odd.
- \( b \) must be even.
- Each \( c_i \) must be even.

For \( 3n \) to be a perfect cube, the exponent of each prime factor must be divisible by 3. Thus, we have:
- For \( 3n = 2^a \cdot 3^{b+1} \cdot p_1^{c_1} \cdots p_k^{c_k} \), \( a \) must be divisible by 3.
- \( b+1 \) must be divisible by 3, implying \( b \equiv 2 \mod 3 \).
- Each \( c_i \) must be divisible by 3.

Now, we analyze the conditions for \( a \) and \( b \):
1. \( a \) is odd and divisible by 3. The smallest odd integer that satisfies this is \( a = 3 \).
2. \( b \) is even and \( b \equiv 2 \mod 3 \). The smallest even integer satisfying this condition is \( b = 2 \).

Next, we consider the other prime factors \( p_i \). Since there are no restrictions on them, we can set \( c_i = 0 \) for all \( i \).

Thus, we have:
- \( n = 2^3 \cdot 3^2 = 8 \cdot 9 = 72 \).

Finally, we verify:
- \( 2n = 2 \cdot 72 = 144 \), which is \( 12^2 \), a perfect square.
- \( 3n = 3 \cdot 72 = 216 \), which is \( 6^3 \), a perfect cube.

Therefore, the smallest positive integer \( n \) that satisfies both conditions is \( 72 \).<|END_OF_LATENT|> The final answer is $72$.