Question

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 $(a - b)^3$.

Answer 1

0

Step-by-step explanation:

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 = 0

Answer 2

0 or -3375

Step-by-step explanation:

The question is somewhat ambiguous because we can't be sure whether B will be positive integers which are multiples of both 3 and 2 or 3 and 2 separately.

For the former:

There a 4 multiples of 6 less than 30: 6, 12, 18, 24.

Positive integers (whole numbers) less than 30: 1-29 ; multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27 and of those, only 6, 12, 18 and 24 are multiples of both 2 and 3.

Therefore we have (4-4)^3 = 0

For the latter:

multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27 ; Multiples of 2: 2 4 6 8 10 12 14 16 18 20 22 24 26 28 (remove common multiples)

Therefore (4-19)^3 = -3375