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Archive for the ‘Problem of the Month’ Category

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 @
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



The problem, once again:

For any polygon, a “diagonal” is defined as a line segment that runs from one vertex  to another, running  through the polygon’s interior. Find a formula that determines the number of diagonals in any convex polygon with n sides. Once you have the formula, use it to figure out the number of diagonals in a convex polygon with 1,000 sides (don’t try this by hand! — that’s why algebra was invented).

The winning answer was provided by Chris Mark. The formula, for a convex polygon with n sides, is this:  Number of Diagonals =  [n(n– 3)]/2. For n = 1000, the number of diagonals = 498,500.

The reasoning behind the formula. A polygon has as many vertices as sides. So a polygon with n sides also has n vertices. Now, consider any vertex of the n-gon. From that vertex the number of diagonals that can be drawn is (n – 3). That is because we cannot draw a diagonal to 3 vertices:  the vertex chosen, and the two adjacent vertices. So the expression (n – 3) = the number of diagonals that can be drawn from any vertex. We multiply (n – 3) by n to obtain the total number of diagonals that can be drawn from all n vertices. But if we simply multiply n by (n – 3), we’d be counting each diagonal twice. To eliminate that problem, we divide the product, [ n(n – 3)], by 2, and that provides the correct formula:

Diagonals  =  [n(n – 3)]/2

Applying this formula to a convex polygon with 1,000 sides, we see that the number of diagonals =  498,500.

In addition to providing the answer, Chris pointed out that the problem need not be restricted to regular polygons, as it was when posted. This formula works for all convex polygons, regular or not.

Thanks, Chris. And thanks to everyone who submitted answers.

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.


The problem:

It’s your friend’s 73rd birthday. You’ve put together a surprise party and baked a special coconut meringue cake. But at the last minute you realize that — golly gee! — you forgot to get candles.

Rummaging through your drawers with just five minutes before your friend’s scheduled arrival, you find that you do have 14 candles. And being a brilliant mathematician, you realize that you can represent the number 73 with these 14 candles, using every candle. How do you do it?

As a hint, here’s a model showing how to do a problem like this, if you are celebrating someone’s 44th birthday, when you have just 13 candles. Notice that each dot on the top row is one candle.


Note that you may use icing to create the symbols: +, –, x, ÷, and you may also put in exponents, using candles to show the value of the exponent.

Have fun!

Send answers to:

Make the Subject line: POTM

Please include your full name, where you live, and if you don’t mind, describe your connection to math and math education (for example: teacher, tutor, math enthusiast, etc.).

The first person to send in a correct answer receives a $20 gift certificate toward the purchase of any Singing Turtle Press products. I’ll fill the winner in on the details by email.

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ANSWER: October Problem of the Month

The problem, once again, was this:

Here’s the OCTOBER POTM —


A) Each of the variables x, y, and z, stands for a unique natural number from 1 thru 9 inclusive.

B) The following, therefore, represents a number in the millions place: 3,x2y,1z3

C) This number: 3,x2y,1z3 is divisible by 9.

Question: What are three other 7-digit numbers containing all of these same digits: 3, x, 2, y, 1, z, and 3 that are also divisible by 9?

The correct answer was provided by Kevin Pickard, 48, a computer programmer and homeschooling parent of “two very smart girls,” ages 7 and 10. Kevin lives in Markham, ON, Canada.
Here is Kevin’s answer, in his own words:
Given that 3,x2y,1z3 is divisible by 9, then all of its digits
will add up to 9 eventually by repeated sums (eg. if x=4, y=2, z=3
then 3 + 4 + 2 + 2 + 1 + 3 + 3 = 18 and 1 + 8 = 9). So any arrangement
of these digits will still add up to 9 by repeated sums. These new
arrangements will therefore also be divisible by 9. So 3 other 7-digit
numbers that are also divisible by 9 would be as follows.
Plugging in the example values of x,y & z from above confirms
the result.
Josh’s note: In this problem, ANY new combination of the 7 given digits will produce a number divisible by 9. So in a sense this is a “trick problem.” The way the problem was written made some people think that only certain combinations of the digits would result in a number divisible by 9. The problem shows that the rule for divisibility by 9 is very flexible. When applying this divisibility rule, you need not take into account the order of the digits whatsoever. All that matters is that the sum of the digits is divisible by 9.

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Here’s the problem —

Al’s Discount Clothing store is trying to entice customers with a

special “20-40” Percent Sale.


Here’s how it works. For all sale items, the Al’s

first takes off 20%. Then the store takes an

additional 40% off the sale price.




a)  How much is the discount, under the terms

of this sale, for a bundle of clothes with retail

value of $250?


b)  How much would the discount be if Al’s had

just run a straight 60% discount sale?


c)  Why do you think that Al’s is running the

sale in this way?


Send replies to:

Make the subject:  POTM