A mathematician works for $t$ hours per day and solves $p$ problems per hour, where $t$ and $p$ are positive integers. One day, the mathematician drinks some coffee and discovers that he can now solve $4p+7$ problems per hour. In fact, he only works for $t-3$ hours that day, but he still solves twice as many problems as he would in a normal day. How many problems does he solve the day he drinks coffee?

Question

A mathematician works for $t$ hours per day and solves $p$ problems per hour, where $t$ and $p$ are positive integers. One day, the mathematician drinks some coffee and discovers that he can now solve $4p+7$ problems per hour. In fact, he only works for $t-3$ hours that day, but he still solves twice as many problems as he would in a normal day. How many problems does he solve the day he drinks coffee?

Answer 1

On a normal day, he solves $tp$ problems. The day he drinks coffee, he only works for $t-3$ hours, so he solves $(t-3)(4p+7)$ problems. We have the equation $$2(tp)= (t-3)(4p+7).$$ Expanding both sides of the equation gives \begin{align*}

2tp&= (4p+7)(t-3) \\
2tp&= 4pt-12p+7t-21 \\
0&= 2pt-12p+7t-21.
\end{align*}Taking out $p$ and $t$ gives \begin{align*}
0&= p(2t-12)+7t-21 \\
0&= (2t-12)p+7t-21.
\end{align*}Since $p$ and $t$ are positive integers, we can conclude that $$2t-12=7.$$Therefore, we have $t=\tfrac{7+12}{2}=9.$ Since we know $t=9,$ we can substitute to find $$p(9)= 3p = 7(9)-21 = 42.$$Dividing by $3,$ we find that $p=14,$ so our final answer is $(t-3)(4p+7)= (9-3)(4(14)+7)= \boxed{341}.$

Answer 2

Let's solve this step-by-step.

Step 1: Let's assume the mathematician normally works for $t$ hours per day and solves $p$ problems per hour. Therefore, the mathematician solves a total of $t \times p$ problems in a normal day.

Step 2: On the day the mathematician drinks coffee, he discovers that he can solve $4p+7$ problems per hour. However, he only works for $t-3$ hours that day.

Step 3: We know that on the day he drinks coffee, he solves twice as many problems as he would in a normal day. Therefore, we can set up the following equation:

$(t-3) \cdot (4p+7) = 2 \cdot (t \cdot p)$

Step 4: Let's simplify the equation:

$4tp - 12p + 7t - 21 = 2tp$

Step 5: We can rearrange the equation:

$2tp - 4tp - 7t + 12p - 21 = 0$

Step 6: Combine like terms:

$-2tp + 12p - 7t - 21 = 0$

Step 7: Factor out the common terms:

$-(2t - 12p + 7) = 0$

Step 8: Set each factor equal to zero:

$2t - 12p + 7 = 0$

Step 9: Solve for $p$:

$p = \frac{2t + 7}{12}$

Step 10: Since $p$ is a positive integer, we need to find values of $t$ that satisfy the equation. Let's try different values of $t$ to find an integer value of $p$:

For $t = 5$,
$p = \frac{2(5) + 7}{12} = \frac{17}{12} \rightarrow$ Not an integer

For $t = 6$,
$p = \frac{2(6) + 7}{12} = \frac{19}{12} \rightarrow$ Not an integer

For $t = 7$,
$p = \frac{2(7) + 7}{12} = \frac{21}{12} \rightarrow$ Not an integer

For $t = 8$,
$p = \frac{2(8) + 7}{12} = \frac{23}{12} \rightarrow$ Not an integer

For $t = 9$,
$p = \frac{2(9) + 7}{12} = \frac{25}{12} \rightarrow$ Not an integer

For $t = 10$,
$p = \frac{2(10) + 7}{12} = \frac{27}{12} \rightarrow$ Not an integer

For $t = 11$,
$p = \frac{2(11) + 7}{12} = \frac{29}{12} \rightarrow$ Not an integer

For $t = 12$,
$p = \frac{2(12) + 7}{12} = \frac{31}{12} \rightarrow$ Not an integer

For $t = 13$,
$p = \frac{2(13) + 7}{12} = \frac{33}{12} \rightarrow$ Not an integer

For $t = 14$,
$p = \frac{2(14) + 7}{12} = \frac{35}{12} \rightarrow$ Not an integer

Step 11: From the calculations above, we can see that there is no integer value of $t$ that satisfies the equation. Therefore, there is no solution to this problem. The mathematician cannot solve twice as many problems as he would in a normal day after drinking coffee.