Answer?(functions)

Looking at this problem, I need to find values of $p$ where $q = p \times [p]$ is an integer with $7 < q < 17$.

Let me denote $[p] = n$ (the greatest integer in $p$). This means $n \leq p < n+1$.

So $q = p \times n$, and I need:

• $7 < p \times n < 17$
• $q$ is an integer

Let me check different values of $n$:

For $n = 1$ (where $1 \leq p < 2$):

• Need: $7 < p < 17$
• But $p < 2$, so no solutions here.

For $n = 2$ (where $2 \leq p < 3$):

• Need: $7 < 2p < 17$, so $3.5 < p < 8.5$
• But $p < 3$, so this gives $2 \leq p < 3$
• Then $4 \leq 2p < 6$, which doesn't satisfy $q > 7$. No solutions.

For $n = 3$ (where $3 \leq p < 4$):

• Need: $7 < 3p < 17$, so $\frac{7}{3} < p < \frac{17}{3}$
• Combined with $3 \leq p < 4$: valid range is $3 \leq p < 4$
• This gives $9 \leq 3p < 12$
• For $q$ to be an integer with $7 < q < 17$: $q \in {9, 10, 11}$
• Solutions: $p = 3, \frac{10}{3}, \frac{11}{3}$ ✓

For $n = 4$ (where $4 \leq p < 5$):

• Need: $7 < 4p < 17$, so $\frac{7}{4} < p < \frac{17}{4}$
• That's $1.75 < p < 4.25$
• Combined with $4 \leq p < 5$: valid range is $4 \leq p < 4.25$
• This gives $16 \leq 4p < 17$
• For $q$ to be an integer: $q = 16$
• Solution: $p = 4$ ✓

For $n \geq 5$:

• We'd have $q = pn \geq 5n \geq 25 > 17$. No solutions.

Answer to Question 46:

The positive real values of $p$ are: $3, \frac{10}{3}, \frac{11}{3}, 4$

Number of values: 4

Answer to Question 47:

Product of all possible values: $3 \times \frac{10}{3} \times \frac{11}{3} \times 4 = \frac{3 \times 10 \times 11 \times 4}{9} = \frac{1320}{9} = \frac{440}{3}$

Product: $\dfrac{440}{3}$ (or $146\frac{2}{3}$ or approximately $146.67$)