A mathematician works for \(t\) hours per day and solves 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: p = 7 and t = 8. Before drinking coffee, he worked for t = 8 hours per day and solved p=7 problems per hour=7*8=56 problems per day.
On the day he drank his coffee , he worked for (8 - 4) = 4 hours and solved (3*7 +7) =28 problems per hour for a total of: 28 * 4 = 112 problems on that day.
