+86 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 + 2 problems per hour. In fact, he only works for t - 11 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?
+1768 Let's solve this step-by-step to find the number of problems the mathematician solves the day he drinks coffee.
Normal Day:
Hours worked (t) = t
Problems solved per hour (p) = p
Total problems solved in a normal day (normal_problems) = t * p
Coffee Day:
Hours worked (t_coffee) = t - 11 (since he works for fewer hours)
Problems solved per hour (p_coffee) = 4p + 2 (due to the coffee boost)
Total problems solved on the coffee day (coffee_problems) = (t - 11) * (4p + 2)
Twice the Problems: We are given that the coffee day problems are double the normal day problems. So, we can write the equation: coffee_problems = 2 * normal_problems
Solving the Equation: Substitute the expressions for each term: (t - 11) * (4p + 2) = 2 * (t * p)
Expand the equation: 4pt - 44p + 2t - 22 = 2tp
Combine terms with t and p: 2pt - 44p + 2t - 2tp = 22 pt - 44p + 2t = 22
Notice a Pattern: We can see that all the terms on the left-hand side of the equation have a common factor of t. Let's rewrite it: t(p - 44 + 2) = 22
Isolating t: Divide both sides by (p - 44 + 2): t = 22 / (p - 42)
Integer Requirement: Since t and p are positive integers, the value of (p - 42) must be a factor of 22 (which are 1, 2, 11, and 22). This means p must be 43, 44, 53, or 64 (because adding 42 to each value results in a factor of 22).
Checking Each Case: We can try plugging in each possible value of p (43, 44, 53, or 64) into the equation for t. However, for this scenario to work, t must also be an integer. Let's test:
p = 43: t = 22 / (43 - 42) = 22 (which is valid).
p = 44 (or any value greater than 43): The expression for t becomes negative, which is not possible.
Therefore, the only case that works is when p = 43 and t = 22.
Coffee Day Problems: Now that we know t = 22 and p = 43, we can find the number of problems solved on the coffee day (coffee_problems):
coffee_problems = (t - 11) * (4p + 2) coffee_problems = (22 - 11) * (4 * 43 + 2) coffee_problems = 11 * 174 coffee_problems = 1914
So, the mathematician solves 1914 problems on the day he drinks coffee. That's a lot of problems!
+9673 Comparison between before and after drinking coffee gives:
\((t - 11)(4p + 2) = 2tp\\ 4tp - 44p + 2t - 22 = 2tp\\ 2tp + 2t - 44p - 22 = 0\\ 2t(p + 1) - 44(p + 1) = -22\\ (22 - t)(p + 1) = 11\)
How is that possible? The only possibilities are \(\begin{cases} 22-t = 1\\ p+1=11 \end{cases} \text{ or } \begin{cases} 22-t=11\\ p+1=1 \end{cases}\). But p + 1 can't be 1 since p would be 0, which is not positive.
So, the solution is (t, p) = (21, 10), which means the mathematician solved 4p + 2 = 42 problems that day.
