Friday the 13th Parts I, II and III in 2009: Feb, Mar and Nov. Is this unusual?

Update: See detailed info on Fri 13th at bottom. Also, several revisions have been made.

Update on Contest:
All teams' Answer Forms have been submitted and scored. The highest score was a perfect 11 out of 11. There were several other outstanding scores as well. More details to follow...

If you suffer from triskaidekaphobia, uh, well, maybe you can be Rip Van Winkle starting tonight, Thu Mar 12th, and wake up on August 14th next year. Next year will be much less scary with only one Friday the 13th!
Note: Triskaidekaphobia only refers to fear of the number 13. See the Wikipedia article for the very long word denoting fear of Friday the 13th!

Can this be a learning experience for our students?
Here are some questions to ask:

Feb 13th 2009 happens to fall on a Friday.

Q: What day of the week will it be 28 days later? Explain.
Ans: Friday
Explanation: The days of the week repeat every 7 days and 28 is a multiple of 7.

Q: What day of the month will it be 28 days after Feb 13th? Explain.
Ans: It will be the 13th of the next month, in this case, March.

Explanation--
BASIC RULE:
If today is the nth day of a given month and there are k days in this month, then k days from today will always be the nth day of the following month!
Why? Well, we can write k = (k-n) + n, which can be interpreted to mean that k days from today will be the nth day of the following month.

Ok, an example would help here:
Let's say today is the 5th day of some month and there are 31 days in that month. We can write 31 = (31-5) + 5 = 26 + 5. the "26" brings us to the end of the month and then we add 5 more days which brings us to the 5th of the next month. Make sense?

Putting this all together, 28 days from Friday Feb 13th will be Friday March 13th. The key was that 28 is BOTH the number of days in the month AND a multiple of 7!
That wasn't hard, but now for the tougher question:

Q: Why is there going to be a third Friday the 13th in 2009, namely Nov 13th?

Explanation:
Mar-31 Cum. Total = 31 (not div by 7)
Apr-30 Cum. Total = 61 (not div by 7)
May-31 Cum. Total = 92 (not div by 7)
Jun-30 Cum. Total = 122 (not div by 7)
Jul-31 Cum. Total = 153 (not div by 7)
Aug-31 Cum. Total = 184 (not div by 7)
Sep-30 Cum. Total = 214 (not div by 7)
Oct-31 Cum. Total = 245 (Div by 7!!)

Thus, 245 days after Fri Mar 13th will not only be the 13th of Nov but it will also land on a Friday!
Why? Because, 245 is the cumulative total of the number of days in the 8 months starting in Mar but it is also a multiple of 7. No simple formula here, just grinding it out.

Suggested Additional Questions:
(1) Is 3 the maximum number of Friday the 13ths in any calendar year?
Ans: Yes!
(2) When is the next calendar year in which this 3-peat (three Friday 13ths in one calendar year) will occur?
Ans: See below!
(3) Is 1 the minimum number, i.e, could there be a calendar year in which there are no Fri the 13ths?
Ans: See below!
(4) In which months could a 3-peat occur in a non-leap year? A leap year?
Ans: See below!
(5) Related to (4): 31+29+31 = 91: What does this tell you about Friday 13ths in a leap year?
(6) Other questions: From the students!


The following is from the excellent Wikipedia article on Friday the 13th...

The following months have a Friday the 13th: Month Years Dominical Letter January 2006, 2012, 2017, 2023 A, AG February 2004, 2009, 2015, 2026 D, DC March 2009, 2015, 2020, 2026 D, ED April 2001, 2007, 2012, 2018 G, AG May 2005, 2011, 2016, 2022 B, CB June 2003, 2008, 2014, 2025 E, FE July 2001, 2007, 2012, 2018 G, AG August 2004, 2010, 2021, 2027 C, DC September 2002, 2013, 2019, 2024 F, GF October 2006, 2017, 2023, 2028 A, BA November 2009, 2015, 2020, 2026 D, ED December 2002, 2013, 2019, 2024 F, GF

The following years have Fridays the 13th in these months: Year Months Dominical Letter 2001 April, July G 2002 September, December F 2003 June E 2004 February, August DC 2005 May B 2006 January, October A 2007 April, July G 2008 June FE 2009 February, March, November D 2010 August C 2011 May B 2012 January, April, July AG 2013 September, December F 2014 June E 2015 February, March, November D 2016 May CB 2017 January, October A 2018 April, July G 2019 September, December F 2020 March, November ED 2021 August C 2022 May B 2023 January, October A 2024 September, December GF 2025 June E 2026 February, March, November D 2027 August C 2028 October BA

This sequence, here given for 2001–2028, repeats every 28 years from 1901 to 2099. The months with a Friday the 13th are determined by the Dominical letter (G, F, GF, etc.) of the year. Any month that begins on a Sunday will contain a Friday the 13th, and there is at least one Friday the 13th in every calendar year.

The longest period that can occur without a Friday the 13th is fourteen months, either from July to September the following year (e.g. in 2001/2002 and 2012/13), or from August to October in a leap year (e.g. in 2027/28).

Patterns for non leap-years: First month occurring Second month Third month January October February March November April July May June August September December

Patterns for leap years: First month occurring Second month Third month January April July February August March November May June September December October

Remainders and Number Theory Challenges for Middle School and Beyond

Edit: #4 below has been corrected. I am indebted to one of mathmom's astute students for catching my error!

Number theory is part of many states' standards but usually only at a basic level (factors, multiples, primes, composites, gcf, lcm). Below you will find a problem for your students to work on (preferably with partner). It is not an introductory problem using remainders so they would have needed to do preliminary work beforehand.

Here are some suggestions for developing the foundation for today's challenge problem:

(1 ) List the first 5 positive integers which leave a remainder of 1 when divided by 2? Describe, in general, such positive integers.

(2) List the first 5 positive integers which leave a remainder of 3 when divided by 13? If you subtract 3 from each of these, what do you notice? Explain!

(3) List the first 5 positive integers which leave a remainder of 12 when divided by 13. If you subtract 12 from each of these, what do you notice? If, instead you ADD 1 to each of the 5 positive integers, what do you notice? Explain!

(4) What is the least positive integer N, greater than 1, which leaves a remainder of 1 when divided by 2, 3, 4 or 5? [Ans: 61]
Note: The word 'or' may be confusing or inaccurate here. Modify as needed!

Now for today's challenge (allow use of calculator):

What is the least positive integer which satisfies ALL of the following:
leaves a remainder of 1 when divided by 2
leaves a remainder of 2 when divided by 3
leaves a remainder of 3 when divided by 4
leaves a remainder of 4 when divided by 5
leaves a remainder of 5 when divided by 6
leaves a remainder of 6 when divided by 7
leaves a remainder of 7 when divided by 8
leaves a remainder of 8 when divided by 9.

Notes/Comments
This challenge looks harder than it is. Variations of these often appear on math contests for middle school and beyond. Simpler versions like example (4) above have appeared on the SATs.

Of course, modular arithmetic and congruences would make this problem trivial but that is non-standard and requires more time to develop.

I will not yet post the answer or possible solution...

A 'Simple' Traversal through a Number Grid -- Patterns, Functions, Algebra Investigation Part I

Here is an activity for Prealgebra and Algebra students. This introductory activity is not meant to be a conundrum for our crack problem-solvers out there, but the extensions below may prove more challenging.



Target Audience: Grades 6-9 (Prealgebra through Algebra 1)

Major Standards/Objectives:
(1) Representing numerical relationships and patterns algebraically
(2) Recognizing, interpreting and developing function notation
(3) Applying remainder concepts

A 2-column number matrix (grid) is shown above and assumed to continue indefinitely. We will be visiting (traversing) the numbers in the grid starting in the upper left corner with 1. Following the arrows we see that the tour proceeds right, then down, followed by left, then down and repeats.

First, some examples of the function notation we will be using to describe this traversal:
T(1) = 1 denotes that the 1st cell visited contains the number 1.
T(4) = 3 denotes that the 4th cell visited contains the number 3.
Similarly, T(6) = 6.

STUDENT/READER ACTIVITY/INVESTIGATION

(a) Determine T(1), T(5), T(9), T(13), T(17).
(b) 1, 5, 9, 13, 17, ... all leave a remainder of ___ when divided by 4. (Fill in the blank)
Therefore, these numbers can be represented algebraically as 4n + 1, n = 0,1,2,3,...
(c) Based on (a) and (b), it appears that T(4n+1) = _______, where n = 0,1,2,3...
(d) Determine T(2), T(6), T(10), T(14)
(e) 2,6,10,14,... all leave a remainder of ___ when divided by 4. Therefore, these numbers can be represented algebraically as ______, n = _________ (Fill in blanks)
(f) Based on (d) and (e), it appears that T( _____ ) = _____, n = __________.

Note: The instructor may choose to start n from zero or one throughout this activity. I will vary it depending on our needs. It is important for students to see how restrictions (domain of a variable) is critical for an accurate description and that more than one set of restrictions is possible (provided they are equivalent).

Since T(3) = 4 and T(4) = 3, we cannot say that T(n) = n for all n. The numbers 3 and 4 leave remainders of 3 and 0 respectively when divided by 4. We will need a different rule for these kinds of numbers. Let's collect some more data:

(g) By extending the table, determine T(7) and T(8); T(11) and T(12); T(15) and T(16)
(h) Without extending the table, make a conjecture about the values of T(35) and T(36).
(i) Numbers such as 4,8,12,16,... can be represented algebraically as ____, n= 1,2,3,...
(j) Numbers such as 3,7,11,15,... can be represented algebraically as ____, n = 1,2,3,...

Note: Again, the instructor may not like varying the restrictions here. Adjust as needed.

(h) Ok, so you're an expert now. Well, prove it:
T(100) = ______; T(153) = _____; T(999) = ______
Show or explain your method.

EXTENSIONS

Surely, a 3-column number grid or even a 5-column number grid can't be that much more difficult to solve using the same kind of traversal (move to the right until you come to the end, go down, move left until you come to the end, move down, lather, rinse, repeat...). ENJOY!

Ok, for our experts: Try an n x n grid!

DISCLAIMER: As with all of the investigations I publish, these are essentially original creations and therefore have not been proofread or edited by others. You are the 'others!'. You may not only find errors but alternate and perhaps superior ways to present these ideas.
Also, please adhere to the Guidelines for Attribution in the sidebar.

'Left-Overs' before the Super Bowl: Crazy Eights, Squares, Remainders and Algebra

Ok, so most normal people are not thinking about the significance of the digit '8' in 2008 the day before the Super Bowl. Sorry, but in this post there will be no predictions about the score, no 'over-unders', no boxes, no betting at all. You do have to admit that this is a great time for lovers of mathematics. People are actually interested in mathematical odds and chances of all kinds of weird number combinations occurring in the score on Sunday night. However, this post will focus instead on the number 8, the units' digit in 2008. The Super Bowl comments above will no doubt soon become outdated but the mathematics below will live on! Who knows, maybe the number 8 will turn out to have special significance on Feb 3, 2008? Remember, I said that here before the game!!

2008 is a special number for so many reasons, being divisible by 4 of course: Leap Year, Prez Election year, Summer Olympics and much more. In fact, 2008 is divisible not only by 4 but also by 8 itself. In the good ol' days, some students were even taught the divisibility rules for 2, 4 and 8:
Divisible by 2: If the 'last' digit is divisible by 2 (of course!)
Divisible by 4: If the number formed by the last TWO digits is divisible by 4
Divisible by 8: If the number formed by the last three digits is divisible by 8.

Let's demonstrate this for 2008:
2008 us divisible by 2 because 8 is divisible by 2
2008 is divisible by 4 because '08' is divisible by 4
2008 is divisible by 8 because '008' is divisible by 8

A little weird with those zeros and not particularly interesting, right? Anyone care to guess a rule for divisibility by 16? Interesting, but none of this is the issue for today....

BACKGROUND FOR PROBLEM/INVESTIGATION/ACTIVITY
Today, we are are interested in the squares of numbers and their remainders when divided by 8. Notice that 4² is divisible by 8 but 6² is not. So we cannot say that the square of any even number is divisible by 8. What about the squares of odd numbers when divided by 8?
1² leaves a remainder of 1 when divided by 8
3² leaves a remainder of 1 when divided by 8
5² leaves a remainder of 1 when divided by 8
7² leaves a remainder of 1 when divided by 8

What is going on here? That's for your crack investigative team to decipher.

TARGET AUDIENCE: Our readers of course; Middle schoolers through algebra

PROBLEM/INVESTIGATION FOR READERS/STUDENTS
1. Discover, state and prove a general rule for the remainder when the square of an even number is divided by 8.
2. Discover, state and prove a general rule for the remainder when the square of an odd number is divided by 8.

Comments:
(1) These are well-known relationships and not very difficult questions. Just something to extend thinking about divisibility, remainders and the use of algebra to deduce and prove generalizations. Prealgebra students may be able to explain their findings without algebra!
(2) 'Discovering' or stating the rule for question (2) is transparent from the examples above. Instructors may prefer 'data-gathering' and making a table first. That is, have students develop a table for the squares of the first 10 positive integers and their remainders when divided by 8. Proving the result for the squares of odd integers is more challenging, even algebraically. Most will see the remainder when dividing by 4, but 8 is slightly trickier.
(3) Those who are more comfortable with congruences and modular arithmetic can approach these questions another way.

What is the Largest 3-digit Multiple of 7? - A Middle School Activity on Remainders, Multiples and the Division Algorithm

Please don't forget to give proper attribution when using this activity in the classroom (see sidebar).

STUDENT/READER CHALLENGE #1:
What is the largest 3-digit multiple of 7?

Would you expect most students to reach for the calculator and, without much thinking, test 999,998,997, etc., until they reach 994?

This is an opportunity to review the Division Algorithm, the conceptual meaning of remainder and how important this integer can be when solving problems involving multiples, factors, and various number theory questions (not to mention those repeating pattern problems so popular on standardized tests).

From the calculator, students obtain a result like 1000/7 = 142.8571429. The last digit is rounded which disguises the repeating decimal but some will know that. You ask for the remainder from this division problem and you may get an answer like 6/7 or blank stares or some decimal. Because we live in a calculator environment, students may have already acquired ways to obtain the remainder from the calculator display. Older students with their sophisticated graphing calculators may have a function that returns the remainder (like mod) or some application they downloaded for this purpose.

Here's one way students find remainders on their calculator:

Ignore the decimal, that is, look at the greatest integer value of the quotient, namely 142. Then, the remainder can be obtained from: 6 = 1000 - 142⋅7. This is just another form of how students should check their division:
142⋅7 + 6 = 1000

Students often think about this procedurally, not conceptually. In fact, a thorough understanding of remainders and the division algorithm would make the questions in this post fairly simple.

In general, suppose B is divided by A, producing an integer 'quotient' Q and a remainder R, where R is an integer satisfying A > R ≥ 0. Then the Division Algorithm states:
QA + R = B or R = B - QA

As an aside, some students are taught or discover another calculator approach for finding the remainder:
Subtract (discard) the integer part of the decimal, leaving 0.8571429, then multiply this result by the original divisor 7. Like magic, the display reads 6. Students should realize that the calculator internally stores more places than it displays! It may be worth it to demonstrate why this method works or have students investigate this, so I'll make it part of today's challenge:

STUDENT/READER CHALLENGE #2:
Explain why the above procedure works for finding remainders. You may want to employ an algebraic derivation, using A, B, Q and R as above.

Students should now use remainder concepts to solve the original problem and the following:

STUDENT/READER CHALLENGE #3
What is the largest 6-digit multiple of 7? Explain your method carefully.

Note: If time permits (like right before a vacation), you may want to introduce modulo arithmetic (congruences, etc.) to solve the remainder problem in a more compact form:
10 ≡ 3 (mod 7) → 10⁶ ≡ 3⁶ ≡ (3²)³ ≡ 2³ ≡ 1 (mod 7), using the fact that 9 is congruent to 2 modulo 7. Thus, the remainder is 1 and the result follows. Of course, there's a great amount of overhead in developing this much number theory, but if you're looking a holiday challenge...