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# mathematician solving problems with coffee

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A mathematician works for t hours per day and solves p problems per hour, where  and  are positive integers. One day, the mathematician drinks some coffee and discovers that he can now solve 4p+5 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?

May 13, 2024

#1
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Similar to the one you posted before, so I will only give partial work:

Comparison between before and after drinking coffee gives

\((t - 3)(4p + 5) = 2tp\\ 4tp - 12p + 5t - 15 = 2tp\\ 2tp - 12p + 5t - 15 =0 \\ 2p(t - 6) + 5(t - 6) = -15\\ (2p + 5)(6 - t) = 15\)

Note that since p is a positive integer, \(p \geq 1\). That means \(2p + 5 \geq 7\), and 2p + 5 is also a divisor of 15.

What is the only divisor of 15 that is greater than 7?

Please continue on your own from this point. If you are stuck, look at my answer to your previous similar problem. I gave full solution there.

May 13, 2024
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Thank you, using what you told me I arrived at p and t values of 5, and found that the problems the mathematician solved the day he drank coffee was \(\boxed{50}.\)

eramsby1010  May 13, 2024