Kiss those Math Headaches GOODBYE!

Archive for the ‘Problem of the Week’ 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.

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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

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.

Problem of the Week-10/1/2010


Problem of the Week – 10/1/2010

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.

Send your answers as comments to this blog post. You need not worry about incorrect answers being posted. I will post only those answers that are correct, and I will post the first five correct answers on Monday.

Feel free to share this problem with anyone who might like to try it.

Reflection of a triangle about the y axis

Reflection across the y-axis

Problem of the Week – Answer


Answer to the 9/24/2010 Problem of the Week

The problem —

At Gamesville High, students love their clubs. While 20% of the children in the Hex Club are also members of the Backgammon Club, 80% of children in the Backgammon Club are also members of the Hex Club. The Backgammon Club has 35 children. The question:  how many children are in the Hex Club?

Solution:  Here is the solution, provided by the only person who got it right (name below).

20%=0.2, 80%= 0.8.
Let X = the number of students in the Hex Club
Let Y = the number of students in the Backgammon Club
So:
0.2X=0.8 Y because the 20% and 80% are the same children, the same number of people. It just looks different because the percentages show a relationship between the total numbers in each club.

Speaking of total numbers, the problem tells us how many are in the Backgammon Club: 35. So:
Y=35

We now have two equations. We can substitute the value of Y from the second for the Y in the first and solve for X.
0.2X=0.8 (35)
0.2X=28
X=140
There are 140 students in the Hex Club

Sharron Herring

Sharron has answered my problems many times in the past. So thanks for sharing that solution, Sharon.

Next problem will be posted this Friday, Oct. 1.