Modular Arithmetic Calendars

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?

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I know that the answer is the 7th day of the week but I figured it a different way , I just need to understand the modular arithmetic way. – SNS Feb 27 '12 at 22:36
Note that $(427)(23)\equiv 5 \pmod 8$. So we have advanced by $5$ days of the week. – André Nicolas Feb 27 '12 at 22:45
Okay, but how do you solve (427)(23)=5 (mod 8)? – SNS Feb 27 '12 at 22:57
You want to find the remainder when $(427)(23)$ is divided by $8$. The remainder of $424$ is $0$, so for $427$ it is $3$. The remainder of $16$ is $0$, so the remainder of $23$ is $7$. Multiply $3$ by $7$. You get $21$, which has remainder $5$. Or else, with a calculator, multiply $427$ by $23$. I get $9821$. To find the remainder on division by $8$, divide by $8$ on the calculator. I get $1227.625$. Subtract $1227$ on the calculator, then multiply by $8$. We get $5$. – André Nicolas Feb 27 '12 at 23:05
Now that you get it, SNS, you can write it up as an answer and then, if no one finds a mistake in it, you can accept it after a while. – Gerry Myerson Feb 27 '12 at 23:29

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.