Problem 44 - Discrete Mathematics with Applications

Question

Each day of the week, beginning with Sunday, can be identified by a code \(x,\) where \(0 \leq x \leq 6 .\) January 1 of any year \(y\) can be determined using the following formula^{** } $$ x \equiv\left(y+\left\lfloor\frac{y-1}{4}\right\rfloor-\left\lfloor\frac{y-1}{100}\right\rfloor+\left\lfloor\frac{y-1}{400}\right\rfloor\right) \bmod 7 $$ Using this formula determine the first day in each year. $$3000$$

Step-by-Step Solution

Answer

The first day of the year 3000 would be a \(Wednesday\), based on the given formula and the computation of the day of the week code \(x = 3\).

Step 1: In order to solve the problem, we first need to identify the variable given. In this case, the year \(y = 3000\). #Step 2: Plug the values into the formula#

Now that we have the year \(y\), we need to plug this value into the formula given for the day of the week code \(x\), which is: $x \equiv\left(y+\left\lfloor\frac{y-1}{4}\right\rfloor-\left\lfloor\frac{y-1}{100}\right\rfloor+\left\lfloor\frac{y-1}{400}\right\rfloor\right) \bmod 7 $ Replace \(y\) with 3000: $x \equiv\left(3000+\left\lfloor\frac{3000-1}{4}\right\rfloor-\left\lfloor\frac{3000-1}{100}\right\rfloor+\left\lfloor\frac{3000-1}{400}\right\rfloor\right) \bmod 7$ #Step 3: Calculate all mathematical terms individually#

Step 2: Now, let's compute every mathematical term inside the floor functions: \(\left\lfloor\frac{2999}{4}\right\rfloor = 749\) (as we take the largest integer less than or equal to the result) \(\left\lfloor\frac{2999}{100}\right\rfloor = 29\) \(\left\lfloor\frac{2999}{400}\right\rfloor = 7\) #Step 4: Plug the computed values back into the formula#

Now that we have computed the values of the floor functions, we should plug these into the original expression: \(x \equiv\left(3000 + 749 - 29 + 7\right) \bmod 7\) #Step 5: Perform the calculations and apply modulo 7#

Step 3: Now, let's compute the value of \(x\): \(x \equiv (3727) \bmod 7\) To find \(x\), we divide 3727 by 7 and take the remainder: \(3727 \div 7 = 532\) (with a remainder of 3) Therefore, \(x \equiv 3\). #Step 6: Interpret the result as a day of the week#

Now, we need to interpret the value of \(x\) as a day of the week. From the information provided in the problem, we can infer the following: 0: Sunday 1: Monday 2: Tuesday 3: Wednesday 4: Thursday 5: Friday 6: Saturday As \(x = 3\), the first day of the year 3000 would be Wednesday.

Key Concepts

Modulo Arithmetic

Modulo Arithmetic is like a mathematical game of finding remainders. Think back to your elementary school days when you were learning division. Let's say we divide 17 by 5. We know 5 fits into 17 three times because 5 times 3 equals 15. But there's something left over, a remainder. In this case, the remainder is 2, which is simply 17 minus 15. This remainder is what modulo arithmetic focuses on. This is expressed mathematically as:

• 17 mod 5 equals 2.

Modulo arithmetic is crucial in many areas of mathematics, including finding the day of the week for a given date. In our calendar calculations, the result of taking a number modulo 7 gives us a number between 0 and 6, which corresponds to the days of the week from Sunday to Saturday.

Floor Function

The floor function is an operation in mathematics that takes a real number and rounds it down to the nearest integer less than or equal to the number. It's like chopping off decimal points and keeping the whole part. For example, if you take the number 5.7, the floor of 5.7 is 5. This function is immensely useful, especially in calculations involving division.

In the formula used for determining the day of the week for January 1 of a given year, the floor function helps us calculate whole numbers even when the division of the year by another number (like 4, 100, or 400) doesn't result in a clean integer. For instance:

• Floor of (2999/4) = 749 tells us how many complete sevens fit inside 2999 when it's been grouped by fours.
• Likewise, floor of (2999/100) = 29 and floor of (2999/400) = 7 are similarly calculated.

Thus, the floor function is an ingenious way to deal with non-whole numbers in equations.

Calendar Calculation

Calendar calculations often induce a puzzled expression but break down into simple steps when explained. Our primary task is to determine the day of the week for a specific date, using only the year (in this case, the year 3000). The formula provided utilizes the important operations of modulo arithmetic and floor functions.

Here's a recap of the steps using the year 3000:

• Calculate each term involving a floor function separately, like floor(2999/4), floor(2999/100), and floor(2999/400).
• Plug these values into the equation: 3000 + 749 - 29 + 7.
• Compute the total and apply modulo 7 to find the day as a number.

Ultimately, the formula tells us that January 1, 3000 corresponds to the code 3, which stands for Wednesday. The method simplifies complex calendar queries using logical steps and basic arithmetic to yield reliable results.

Day of the Week

Determining the day of the week for any particular date is a fascinating puzzle that ties back into modulo arithmetic. Each day is assigned a code between 0 and 6 to make calculations simpler.

• 0 corresponds to Sunday.
• 1 corresponds to Monday.
• 2 corresponds to Tuesday.
• 3 stands for Wednesday.
• 4 equals Thursday.
• 5 relates to Friday.
• 6 matches Saturday.

When you solve the equation for a given year and find your modulo result as 3, you know it corresponds to Wednesday. This coding system allows us to unambiguously determine the weekday just by knowing the corresponding numerical result, making it a manageable task even when dealing with large numbers like the year 3000.