Kiss those Math Headaches GOODBYE!

Archive for the ‘Challenge Problem’ Category

Monday the 13th


Today is Monday, the 13th.

So what, right?

Well, maybe not so fast …

If you have a mathematical/logical bent of mind, you might find that interesting.

Friday the 13th is generally considered a bad luck day. So if that is the case, you might wonder if Monday the 13th would be the logical opposite to Friday the 13th, a good luck day. Afterall, Friday is the end of the workweek, and Monday is the beginning of the workweek.

So in that sense, can it be said that Monday and Friday are opposites? And what might that imply.

So here is the challenge. Compose a logical argument as to whether or not Monday the 13th should be considered a lucky day.

That is the challenge for Monday, the 13th of June 2011.

HINT:  You may want to include information about the “truth value” (truthiness, as Steven Colbert likes to say) of statements and their converses.

REWARD:  The first person who presents a compelling logical argument, one way or the other, wins a $10 gift certificate toward the purchase of any Singing Turtle Press products. All comments must be posted by 1 a.m. on Tuesday, the 14th of June, this year.

James Bond Math Challenge


Math in the movies … if there ever was a cool way to explore math, this has to be it. And if you missed my earlier posts on this, check them out here and here.

Math is Cool!

I was looking through the links to movies with math themes, and a question came up.

On the site showing the movies, the text says that there are “mathematical themes and patterns motivated by math” in the introduction scene for the James Bond movie, Casino Royale, this clip:

I’ve watched the clip a few times, and I have my own ideas as to mathematical themes and patterns.

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Movie Math: Wake Students Up with Silver Screen Riddles


Last days of the school year … kids getting “antsy.”

Harder and harder to keep their attention … so what’s a teacher to do?

Answer:  Let the media help us with the media generation.

In my May 16 post, I pointed you to a website that showed how math is used in major motion pictures.

Math is Cool!

In this post I’d like to focus on one such reference to math in the movies, and show how you can turn it into a fun “End-of-Year” lesson.

The clip of Die Hard below has a great scene in which the Bruce Willis character needs to solve a mathematical puzzle in less than five minutes to avoid getting blown up. It’s an exciting scene, and the math is interesting.

I suggest that you first have your class watch this clip.

After watching it, review the solution with your class.
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Answer to Fun Math Problem #2


ANSWER TO FUN MATH PROBLEM #2

The problem, once again, reads as follows. Before you go out to lunch, you glance at the clock above your desk. When you come back from lunch, you glance at the clock again, and you notice something strange: the minute and the hour hand have exchanged places from the positions they had just before you went to lunch.

The question is:  how long were you away?

I received several answers to this problem, but the first person who got it right was Adrian W. Langman, of Port Angeles, TX.

Here is Adrian’s answer, in his own words:

It’s probably true that the hour hand is near the 12 at the beginning and near the 1 at the end. So it’s about 5 minutes after noon at the beginning, and just a bit after 1 at the end.

(Or, it could be a tad before 11 in the morning at the beginning, and about 5 minutes before noon at the end, if the worker is an early riser.  But this problem is just the geometric mirror image of the one hypothesized above, so the duration of the lunch break will be exactly the same.)

Obviously the lunch break is about 55 minutes. But to find the exact length, let M be the number of minutes past noon at the beginning. I’ll use the obvious coordinate system – the origin at the center of the clock, the clock hands radial lines, the 12 at 0 degrees, and the 3 at 90 degrees.

At M minutes past noon, the minute hand is at M/60 x 360 degrees, i.e. 6M degrees, and the hour hand is at M/60 x 30 degrees (since it’s M/60 of the way from the 12 to the 1, which is at 30 degrees), i.e. M/2 degrees.

So at the end of lunch, since they’ve switched places, the hour hand is at 6M degrees and the minute hand is at M/2 degrees.

Since the minute hand is at M/2 degrees, it is 1/6(M/2) minutes past 1 o’clock, i.e. M/12 minutes past 1 o’clock.

Since the hour hand is at 6M degrees, it’s at 6M-30 degrees past the numeral 1, so it’s 2(6M-30) minutes after 1o’clcock.

Setting M/12 = 2(6M-30) and solving for M yields 720/143 (which is approximately 5.035).

So you left at 720/143 minutes after 12, and returned at 60/143 minutes after 1 o’clock.

So you were gone for 60 + 60/143 – 720/143 minutes,

i.e. 7920/143 minutes, i.e. 55 and 55/143 minutes, which reduces to 55 and 5/13 minutes.

Fun Math Problem #2


Here is the second in my series of “Fun Math Problems.”

Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit! I will post the answer to the problems two days later, after people have had time to respond.

To provide your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.
Please show how you worked the problem. Thanks. I will post the names of the first three people who get this right.

The Problem:  Before you go out to lunch, you glance at the clock above your desk. When you come back from lunch, you glance at the clock again, and you notice something strange. The minute and the hour hand have exchanged places from the positions they had just before you went to lunch.

The question is:  how long were you away?

Rusting clock face

Image by The Hidaway (Simon) via Flickr

Answer, Fun Math Problem


Answer to problem about the circular and square pegs and holes.

The “fit” for each situation is the following ratio:
(Area of Inner Figure) ÷ (Area of Outer Figure)

For the square peg in a round hole —
Call the radius of the circle r.
Then the diagonal of square “peg” = 2r
Notice that by slicing the square along its diagonal,
we get a 45-45-90 triangle, with the diagonal being
the hypotenuse and the sides being the two equal legs.
Using the proportions in a 45-45-90 triangle,
side of square peg = r times the square root of 2
Multiplying this side of the square by itself gives
us the area of the square, which comes out as:
2 times the radius squared

This being the case,
Area of square is: 2 times radius squared, and
Area of circle is: Pi times radius squared, and so …

Cancelling the value of the radius squared, we get:
Ratio of (Area of square) to (Area of circle) is:
2÷Pi = 0.6366

For the round peg in a square hole —
Call radius of the circle r.
And since the diameter of the circle is the same length as
the side of the square, the side of the square = 2r
Multiplying the side of the square by itself to get the
area of the square, we find that the area of the square
is given by: 4 times radius squared.

This being the case,
Area of circle is: Pi times radius squared
Area of square is: 4 times radius squared, and so …

Ratio of (Area of circle) to (Area of square) is therefore:
Pi ÷ 4 = 0.7854

Of the two ratios, the ratio of the circular peg in a square hole
is greater than that of the square peg in a circular hole.

Therefore we can say that the circular peg in a square hole
provides a better fit than a square peg in a circular hole.

And that is the answer!

Problem of the Week – 10/18/2010


[Note:  I really am not getting a check from the New Mexico Tourism Department for this post, though I wouldn’t mind if they sent me one!]

Katja and Anthony are on a sightseeing trip in the western United States. Beginning where they land in Santa Fe, NM, they drive 80 miles east to see the historic wild west town of Las Vegas, NM. Then they travel 50 miles north to visit the Kiowa National Grasslands . Next they drive 140 miles west to visit Chaco Canyon National Historical Park. Finally they journey 130 miles south to visit El Malpais National Monument. When they reach El Malpais, how many miles are they from their starting point in Santa Fe?

Please explain how you found your answer, and send answers either as comments to this post, or as emails w/ subject POTW, sent to josh@SingingTurtle.com    I will not post your comments unless and until I determine that it is correct. And then, only on the day when I send out the answer on my blog.

 

Chaco's smaller kivas numbered around 100, eac...

Kivas in Chaco Canyon, New Mexico

 

Problem of the Week — Answer


Answer to the 10/1/2010 Problem of the Week

The problem:  Certain digits appear the same when reflected across horizontal lines or vertical lines. This week’s problem:  which two-digit numerals appear the same when reflected across a horizontal line? Which two-digit numerals appear the same when reflected across a vertical line? To answer, provide the list for the horizontal line and the list for the vertical line.

Solution, sent in by Jo Ehrlein, of Oklahoma City, OK:

Assuming you write the #1 with no serifs, then here are the single digits that are the same when reflected across a horizontal line:1, 3, 8, 0.  That means that the 2 digit numbers that are the same when reflected across a horizontal line are:  10, 11, 13, 18, 30, 31, 33, 38, 80, 81, 83, 88

2 digit numbers are only the same when reflected across a vertical line if both digits are the same AND the individual digits are the same when reflected across a vertical line. The single digits that meet that criteria are1, 8, 000 isn’t a valid 2 digit number.That means the 2 digit numbers that are the same when reflected across a vertical axis are 11 and 88.

Well done, Jo!

So the winner’s circle this week has one member:

Jo Ehrlein, Oklahoma City, OK

And, in Jo’s honor, here is our ceremonial picture of Oklahoma City, home of Ralph Ellison,author of Invisible Man,  if I recall correctly.

 

Oklahoma City

Oklahoma City, OK

 
Congratulations to everyone who worked on this problem. I had some detailed answers that were partially correct.

FYI:  Starting this coming week, I’m going to post the Problem of the Week on Monday for teachers who want to use it early in the week. Answers will be posted mid-week.

Challenge Problem – the ANSWER


Hi everyone,

Here is the answer to yesterday’s challenge problem, the probability problem about catching fish.

First, there are 6 ways to catch the three fish in three casts, getting exactly one trout, one carp and one bass.

You could get the fish in any of these six orders:

TCB / TBC / CTB / CBT / BTC / BCT

Next you find the probability of getting one of these possibilities. Let’s take the first one, TCB.

Keep in mind that after you catch the first fish, the number of fish left goes down by one, to 11; after catching the second fish, the number of fish goes down to 10.

The probability for the TCB possibility is calculated by multiplying:  6/12 x 4/11 x 2/10 = 2/55

When you think about the five other ways to catch the fish, you’ll see that the order of the numerators changes, but the denominators remain 12, 11, 10. So the probability for catching the three fish in any of the six ways is always the same:  2/55.

To get the probability for all six catches, just multiply the probability of one catch by 6:

6/1 x 2/55  =  12/55

And that is the answer: the probability of catching exactly one trout, one carp and one bass is 12/55, which works out to about 21.8%, meaning that this should happen a little more than 1/5 of the time.

I didn’t see anyone submit any answers, but feel free to send them in. Remember that I will post only the correct answers, so no one has to worry about seeing an incorrect answer posted.

Have a great day!

—  Josh

Interactive Challenge Problem — Send in responses


O.K., time to wake up the ol’ brain cells.

Here’s a little challenge problem. I’ll post the problem today, and then I’ll post the answer the next day. My thought is that this would be relevant for many kinds of people.

Teachers can use this as a fun class-opener. Homeschoolers can use it to start their math studies. And anyone who enjoys math can use it to sharpen math skills. So enjoy.

It would also be fun to see how many people get it right, so please send in your answers as comments. I will post only the correct answers.

The problem:

A lake contains exactly 12 fish: 6 trout, 4 carp, and 2 bass.
On any given cast, you catch exactly one fish, and no kind of fish is biting any more than any other kind. (i.e.:  Your odds of catching fish are governed by mathematical probability alone.)
What is the probability that in three casts you will catch exactly one trout, one carp and one bass?
To get credit, you must provide the correct answer and show how you solved the problem.