Consider this word problem:
If the first day of the year is a Monday, what is the 260th day? Answer: Monday
Why does this equation work to calculate what day of the week it is:
260 = (7w + 1) [w stands for week]
How does it work and how would you even develop this equation to get the answer?
I understand that if you divide 260 by 7 you get 37.1 . Meaning all the days fit into weeks excluding one day; the first day of the week.. Monday.
But what if the first day of the year is Wednesday. What would you do then ?
Thanks!!
$\endgroup$ 12 Answers
$\begingroup$Use modular arithmetic. Let's say the first day of the year is day $x$, which takes a value from $0$ to $6$, with and $0 \equiv \text{ Monday}$, so that $x$ is the number of days after the start of the first week that the transition to the new year occurs.
Now, to find what day of the week the $y$th day of the year will be, note that $y-1$ days have passed between the $1$st an $y$th day, so add on an extra $y-1$ days to whatever the first day was ($x$). $$(y-1)+x \mod 7$$ where $... \mod 7$ is the remainder left over when you take away $7$ the maximum number of times from $...$ (that still leaves a positive integer).
In the first case, $x= \text{ Monday }=0$ and $y=260$, so $$(260-1)+0 \mod 7 \equiv 259-7-7-7-... \mod 7 $$$$259-7(w) \mod 7$$$$\equiv 0 \mod 7$$ Which is Monday
If Wednesday is the first day, then let $x=2$, so the $260$th day is $$(260-1)+2 \mod 7 \equiv 2 \mod 7$$
Edit: the answer with less modular arithmetic.
If the $1$st day is Monday, then the $8$th, $15$th, $22$nd... days will also be Monday. These numbers have the general form $7w+1$.
If the $1$st day is Monday, then the $2$nd, $9$th, $16$th, $23$rd... days will be Tuesday. These numbers have the general form $7w+2$.
The $3$rd, $10$th, $17$th, $24$th... days will be Wednesday. These numbers have the general form $7w+3$.
In general, why do Mondays have the form $7w+1$? Day $1$ is a Monday, and any day a multiple of $7$ up from that will also be a Monday (or, will $\equiv 1 \mod 7$). So you can add $7$ to $1$ as many times as you want without affecting the resulting number's Mondayness.
If the first day is a Wednesday, all days that year will be shifted down $2$ from the Monday situation. Thus Monday has the form $(7w+1)-2=7w-1=7w+6$, Tuesday $7w$, W. $7w+1$, Th. $7w+2$, F. $7w+3$, Sa. $7w+4$, and Sunday $7w+5$.
$\endgroup$ 6 $\begingroup$Consider the following table:
Monday-1, 8, 15, 22, ... Tuesday-2, 9, 16, 23, ... Wednesday-3, 10, 17, 24, ... Thursday-4, 11, 18, 25, ... Friday-5, 12, 19, 26, ... Saturday-6, 13, 20, 27, ... Sunday-7, 14, 21, 28, ...Since there exist only seven days in a week, and they always happen in the same sequence, it follows that Mondays only happen on the (1+(7n))th of the year, Tuesdays on the (2+(7n))th day, ..., Sundays on the (7+(7n))th day of the year, where n is a natural number. Thus, all Mondays fall into the equivalence class [1], Tuesdays into the equivalence class [2], ..., Sundays into the equivalence class [7]. Thus, for the yth day of the year there exists some number (y-x) which is divisible by 7, where x is the number within the brackets and belongs to {1, 2, 3, 4, 5, 6, 7}. For example, for the 39th day of the year (39-4)=35 which is divisible by 7. So, the 39th day falls into the [4] equivalence class. Consequently, the 39th day of the year is a Thursday. Similarly, 39=((7*5)+4).
If you divide 260 by 7, you get 37 and (1/7)th (NOT 37.1... which means 37.1000... and NO approximation is NOT appropriate here). Another way of looking at this comes as to try and divide the nth day by 7 and see what you get for the remainder. You either have a remainder of 0, (1/7), (2/7), (3/7), (4/7), (5/7), or (6/7). If you have a remainder of 0, then that day is a Sunday, since 0 is in the [7] equivalence class. If you have a remainder of (x/7), then that day corresponds to the [x] equivalence class. In your case you have a remainder of (1/7), so the day corresponds to the [1] equivalence class, which is the equivalence class for Mondays.
If the first day of the year is a Wednesday, then you can write the following table:
Wednesday-1, 8, 15, ... Thursday-2, 9, 16, ... Friday-3, 10, 17, ... Saturday-4, 11, 18, ... Sunday-5, 12, 19, ... Monday-6, 13, 20, ... Tuesday-7, 14, 21, ...Thus, Wednesday now belong to the [1] class, Thursdays to the [2] class, ..., and Tuesdays to the [7] (or [0]) class. So, if you want to know the 260th day of the year, you can divide by 7 and check the remainder. You get a remainder of (1/7) placing the 260th day in the [1] equivalence class. Thus, if the first day of the year is "day x", where "day x" belongs to {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sundary}, then the 260th day of the year is also "day x".
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