Let $a$ be the number of positive multiples of $6$ that are less than $30$. Let $b$ be the number of positive integers that are less than $30$, and a multiple of $3$ and a multiple of $2$. Compute $(a - b)^3$.
Let $x$ be a multiple of $6$. Then $x = 6 \cdot n$ for some integer $n$. So $x = 2 \cdot (3n)$ and $x = 3 \cdot (2n)$. This means that $x$ is a multiple of $3$ and $x$ is a multiple of $2$. So multiples of $6$ must be multiples of $2$ and multiples of $3$.

Every number that is a multiple of both 2 and 3 must also be a multiple of the least common multiple of 2 and 3, which is 6.  Hence any number that is a multiple of $3$ and a multiple of $2$ is a multiple of $6$.

We have shown that the numbers that are multiples of $6$ and the numbers that are multiples of $2$ and multiples of $3$ are exactly the same numbers, since any multiple of $6$ is a multiple of $2$ and a multiple of $3$, and any number that is a multiple of $2$ and a multiple of $3$ is a multiple of $6$. So we must have $a = b$. A number minus itself is zero, so our final answer is   $$(a - b)^3 = 0^3 = \boxed{0}.$$
0
Level 3
Prealgebra
Problem: Let $a$ be the number of positive multiples of $6$ that are less than $30$. Let $b$ be the number of positive integers that are less than $30$, and a multiple of $3$ and a multiple of $2$. Compute $(a - b)^3$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the number of positive multiples of 6 that are less than 30, we identify the multiples of 6: 6, 12, 18, 24. The next multiple, 30, is not included since we are looking for numbers less than 30. Thus, there are 4 multiples of 6, so a = 4.

Next, we determine the number of positive integers less than 30 that are multiples of both 3 and 2. A number that is a multiple of both 3 and 2 is a multiple of their least common multiple, which is 6. Therefore, we need to count the multiples of 6 that are less than 30, which we already found to be 4. Thus, b = 4.

Now we compute (a - b). We have a = 4 and b = 4, so a - b = 4 - 4 = 0.

Finally, we calculate (a - b)^3. Since a - b = 0, we find that (0)^3 = 0.

The final answer is 0.<|END_OF_LATENT|> The final answer is $0$.