+84 A mathematician works for \(t\) hours per day and solves \(p\) problems per hour, where \(t\) and \(p\) are positive integers and \(1 . One day, the mathematician drinks some coffee and discovers that he can now solve \(3p + 7\) problems per hour. In fact, he only works for \(t-4\) 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?
a=1; b=1;c=(3*a+7)*(b-4); if(c==2*b*a, goto4, goto5);printc, a, b; a++;if(a<100, goto2, 0);a=1;b++;if(b<100, goto2, discard=0;
OUTPUT: t = 5 and p = 1. Before drinking coffee, he worked for t = 5 hours and solved p=1 problem per hour=5*1 =5 problems. On the day he drank his coffee , he worked for(t - 4) =(5 - 4) = 1 hour and solved (3p + 7) per hour =(3*1 +7) =10 problems.
+26393 A mathematician works for \(t\) hours per day and solves \(p\) problems per hour, where \(t\) and \(p\) are positive integers and \(1 < p < 20\) .
One day, the mathematician drinks some coffee and discovers that he can now solve \(3p+7\) problems per hour.
In fact, he only works for \(t-4\) 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?
Formula: \((3p+7)(t-4) = 2tp\)
\(\begin{array}{|rcll|} \hline (3p+7)(t-4) &=& 2tp \\ 3pt-12p+7t-28 &=& 2tp \\ pt-12p+7t-28 &=& 0 \\ t(p+7)-12p-28 &=& 0 \\ t(p+7)&=& 12p+28 \\ t(p+7)&=& 4(3p+7) \\ t &=& 4\left(\dfrac{3p+7}{p+7}\right) \\ t &=& 4\left(\dfrac{3p+21-14}{p+7}\right) \\ t &=& 4\left(\dfrac{3(p+7)-14}{p+7}\right) \\ \mathbf{t} &=& \mathbf{4\left(3-\dfrac{14}{p+7}\right)} \\ \hline \end{array} \)
\(t\) and \(p\) are positive integers
\(\begin{array}{|rcll|} \hline \dfrac{14}{p+7} &=& 0 \qquad t > 0 \checkmark \\ \dfrac{14}{p+7} &=& 1 \qquad t > 0 \checkmark \\ \dfrac{14}{p+7} &=& 2 \qquad t > 0 \checkmark \\ \dfrac{14}{p+7} &=& 3 \qquad t = 0 \\ \dfrac{14}{p+7} &>& 3 \qquad t < 0 \\ \hline \end{array} \)
\(\mathbf{p=\ ?}\)
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{14}{p+7}} &=& \mathbf{0} \\ 14 &=& 0(p+7) \\ 14 &=& 0 \qquad | \text{ does not yield any value of p.} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{14}{p+7}} &=& \mathbf{1} \\ 14 &=& 1(p+7) \\ 14 &=& p+7 \\ \mathbf{p} &=& \mathbf{7}\ \checkmark \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{14}{p+7}} &=& \mathbf{2} \\ 14 &=& 2(p+7) \\ 14 &=& 2p+14 \\ 0 &=& 2p \\ p &=& 0\qquad 1 < p < 20,\ \text{ no solution, p > 1!} \\ \hline \end{array}\)
\(\mathbf{t=\ ?}\)
\(\begin{array}{|rcll|} \hline t &=& 4\left(\dfrac{3p+7}{p+7}\right) \quad | \quad \mathbf{p=7} \\ t &=& 4\left(\dfrac{3*7+7}{7+7}\right) \\ t &=& 4\left(\dfrac{4*7}{2*7}\right) \\ t &=& 4\left(\dfrac{4 }{2 }\right) \\ t &=& 4*2 \\ \mathbf{t} &=& \mathbf{8} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline 2tp &=& 2*8*7 \\ \mathbf{2tp} &=& \mathbf{112} \\ \hline \end{array}\)
The mathematician solved 112 problems the day he drank coffee.
