The Problem
A student came to her professor’s office, wanting to know her score on the final examination. The professor did not have the graded exams with her. But she looked up the student’s score in her gradebook, and asked the student if she wanted the number outright, or she wanted a puzzle to figure it out. The student opted for the puzzle.
The professor said, “Your score left a remainder of 1 when divided by 2, 3, 4, 5, and 6; and a remainder of zero when divided by 11.” As the student started walking away, the professor said, “Oh, just as a reminder, the exam was out of 150 points.” What was the student’s score?
As a bonus, can you find the next number that is larger than the student’s score and that satisfies all the divisibility conditions? (So it may not satisfy being less than 150.)
A Solution
To begin with, we seek an odd (the number leaves a remainder of 1 when divided by 2) multiple of 11 that is smaller than 150, that ends in "1" (the number leaves a remainder of one when divided by 5). So the following numbers are candidates: 11, 121. But 11 does not leave a remainder of 1 when divided by 3. On the other hand,
121=2(60)+1; 121=3(40)+1; 121=4(30)+1; 121=5(24)+1; 121=6(2)+1; and, of course, 121=11(11). So, the student scored 121 out of 150.
The bonus question was "What is the next number satisfying all the divisibility conditions?" The correct answer is 781, which can be arrived at by brute force (you are looking for multiples of 11 that end in "1," so there are not too many to check before you get to 781.
I enjoyed the observation made by one group of solvers: Numbers with the desired divisibility properties will occur after intervals of 660, since 660 is the least common multiple of 2, 3, 4, 5 and 6. So the sequence of numbers will be 121, 781, 1441, 2101, 2761, ... .
Brian Blakely
Patrick Reilly (problem AND bonus)
Samuel Marek
Harry Mushenheim (retired faculty member)
Miranda O'Brenski
Ben Rice (retired faculty member) and John Rice (alumnus)
Yujun Sun
Yitian Wang
Thanks to all who participated.