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 $(t-3)(4p+7)= (9-3)(4(14)+7)= \boxed{341}.$