# Alternate solutions to an olympiad problem

I recently wrote the following question for a math olympiad:

Newton, one of the founding father's of Calculus passed away on the 31st of March $$1727$$. Lagrange, another one of the father's of Calculus was born on the $$25$$th of January $$1736$$. Given that: Sunday $$=0$$, Monday $$=1$$.... Saturday $$=6$$ and assuming that these dates follow the Gregorian calendar, determine the sum of the two days.

$$A)4$$ $$B)5$$ $$C)3$$ $$D)7$$ $$E)25$$

My Solution:

Firstly let us note that any date before the $$28$$th of February on a non leap year is the same date next year. Since there are $$365$$ days this implies the following congruence holds: $$365 \equiv 1$$ (mod $$7$$). The remainder of $$1$$ means that there is an increase of $$1$$ day in the week. So for example, if $$1$$st January is a Monday, it becomes Tuesday the following year.Using the previous logic, similar working shows that for a leap year, we $$+2$$ days.(e.g. "Monday" becomes "Wednesday".) Therefore: Therefore if $$31/1/1732$$ was a Monday(which it actually was) then $$31/3/1735$$ is a Thursday. Following this logic and continuing it holds that $$25/1/1736$$ was a Wednesday. Since Wednesday $$= 3$$ and Thursday $$= 4$$, the sum is $$7$$, therefore the answer is $$D$$.

• "determine what is the sum of the two days they were born." You gave Newton's death date rather than his birth date. Going to need to either change that or give Newton's age at death (84 yrs 27 days). May 12 at 14:58
• @eyeballfrog Yes, my mistake. Corrected now. May 12 at 15:38
• Many issues: Why are you now asking for the sum of the days they died when you're only giving Lagrange's birth date (your "correction" didn't fix anything)? Why are you using the date 31/3/1735 in your calculation when it's nowhere in the problem statement? Why doesn't the problem tell us that 31/1/1732 is a Monday (without that info, this seems to at best be a knowledge test on the Gregorian calendar)? May 14 at 4:41
• In the end, if you translate the two dates given in the problem to numbers as given in the problem, their sum is 4, not 7. Oddly enough, if you retroactively fit the Julian calendar (whose leap year rule is simpler) to today's date and consider the two dates given in the problem, their sum would be 7, but this would be all kinds of wrong. May 14 at 4:45
• @BrianMoehring Sorry, I was having a very stressful day yesterday so I wasn't thinking properly. Let me fix the problem. May 14 at 4:52