0
$\begingroup$

If a calendar has 427 days in the year and 8 days a week and the first day of their current year, which is 1027 falls on the second day of their week. What day of the week will the first day of the year 1050 fall?

I really do not understand this so could someone help me figure out and understand how to solve it please?

  • 0
    @user, I applaud you for filling in for SNS.2013-06-12

2 Answers 2

2

The question involves the time interval of $1050-1027=23$ years, which is $427\cdot 23 $ days. Which is some number of full weeks, plus some incomplete week (remainder), from 0 to 7 days. To answer the question, we don't need the number of full weeks, only the remainder. André Nicolas already gave two ways to find the remainder, but here's one more: $427\cdot 23 = 427\cdot 24 - 427 = 427\cdot 24 - 432 + 5$ which has remainder $5$ because both $24$ and $432$ are divisible by $8$. Advancing $5$ days of the week brings up the $7$th day.

1

427=3(Mod8); 23=7(Mod8); 427.23=3,7(Mod8)=21(Mod8)=5(Mod8). Add 5 days to the 1st. day of year 1027 , the first day of year 1050 will fall on day 7th.'