Posts tagged ‘sum’

Fractional Fun

It’s well known that 5 out of 4 people have trouble with fractions, but even the mathematically advanced may have a little trouble with this puzzle. Your challenge is simple…

Find the sum of all items in the following table.

A hint is below the table, and the answers are below that. Good luck!



Hmm… it seems that you scrolled down here a little too quickly for the hint. Try harder. To put some distance between you and the hint, here are some fraction jokes:

How is sex like a fraction?
It’s improper for the larger one to be on top.

Which king invented fractions?
Henry the Eighth.

There’s a fine line between the numerator and denominator.
(And it’s called a vinculum.)

Okay, you’ve waited long enough. Here’s your hint. The items in the table are a fird (fish + bird), wooden forts, bottles of whiskey, the Sith lord Darth Maul, wraiths, and tents. (By the way, thanks to for the cool drawing of the fird!) Hope that helps.

To put some space between the hint and the answer, here are some more fraction jokes:

A student once told me, “To prove to you that I understand equivalent fractions, I only did three-sevenths of my homework.”

I was scared half to death… twice.

What is one-fifth of a foot?
A toe.

Okay, you’ve waited (and endured) enough. Without further adieu, the answer is 2.5. The images in the table are:

  • one fird
  • two forts
  • four fifths
  • one Sith
  • four wraiths
  • two tents

the sum of which is

\frac{1}{3} + \frac{2}{4} + \frac{4}{5} + \frac{1}{6} + \frac{4}{8} + \frac{2}{10} = \frac{300}{120} = 2\frac{1}{2}.

You’re welcome.

Thanks for stopping by. Have a great day!

November 11, 2016 at 11:11 am Leave a comment

Number Words and Learned Helplessness

How about some number word puzzles? Here’s a well-known puzzle that you’ve likely seen before:

What is the first positive integer that, when spelled out, contains the letter a?

And here’s a modification of that puzzle that you may find a little more difficult:

What is the first positive integer that, when spelled out, contains the letter c?

And taking it one step further:

What letters are never used in the spelling of any positive integer?

Who says that math isn’t useful in English class?

One more problem in a similar vein:

Pick any positive integer you like, and count the letters when that number is spelled out. Now count the letters when the resulting number is spelled out. Continue ad infinitum. What do you get?

Maybe those weren’t your cup of tea. Perhaps anagrams are more to your liking, so here are two (related) puzzles for you.

Try to make an anagram for each of the following three words.

  • whirl
  • slapstick
  • cinerama

Too tough? Then try these three words instead.

  • bat
  • lemon
  • cinerama

If you had trouble with the first set, you’re in good company. There are no anagrams for the words whirl or slapstick.

These two sets of words were used by Charisse Nixon, a pyschologist at Penn State–Erie, who gave the first set of words to half her class and the second set of words to the other half. She instructed them to find an anagram of the first word on their list; those students who had received the second set were successful. Nixon then instructed them to find an anagram of the second word on their list; again, those students who had received the second set were successful. When she then instructed them to find an anagram of the third word on their list — of which there is exactly one, American — those who hadn’t found anagrams for the first two words were less successful than their peers, even though the final challenge was identical.

Afterwards, students who received the first set of words admitted to feeling confused, rushed, frustrated, and stupid.

Nixon was studying learned helplessness, a condition in which a person suffers from a sense of powerlessness, often arising from persistent failure.

This has implications the math classroom. Students who perform at a fourth-grade level but are asked to participate in an eighth-grade class are surely as confused and frustrated as the subjects in Nixon’s experiment. Students need to occasionally feel success, or else they’ll shut down. If you’re a teacher, you don’t need me or a psychological research study to tell you that. So the question is, how can you get students to feel success? That is, what can you do to prevent learned helplessness?

My suggestion is to look for acceptable and accessible entry points.

Consider the following problem, which might be seen in a middle school classroom:

What is the maximum possible product of a set of positive integers whose sum is 20?

As written, that problem contains three words — maximum, product, and integers — that may confound some students. For middle school students who do understand the terminology, finding an appropriate strategy might be daunting.

In my opinion, the following is a better way to present this problem so that all students have an entry point:

Find some numbers with a sum of 20. Now, multiply those numbers together. Compare your result with a partner. Whose result was greater? Can the two of you work together to find a product that’s greater still?

Even a struggling middle school student could start this activity. Surely he could find some numbers with a sum of 20. Certainly, he could multiply them without a problem.

Why is this a better presentation? The wording is simplified. There is encouragement to work with a partner. It feels more like a collaborative game than a traditional math problem. It sounds — dare I say it? — like fun.

When a struggling student is able to get into a problem, and they’re able to make some strides in the right direction, and they’re rewarded by your positive encouragement, they attain some level of success. Maybe they won’t solve the problem entirely, but who cares? For many students, trying is progress.

And for students who are having trouble finding any success, perhaps the following words of encouragement will help.

If at first you don’t succeed, call it version 1.0.

If at first you don’t succeed, destroy all evidence that you ever tried.

If at first you don’t succeed, blame someone else and seek counseling.

If at first you don’t succeed, then skydiving is not for you.

If at first you don’t succeed, get new batteries.

If at first you don’t succeed, try two more times so your failure is statistically significant.


January 8, 2015 at 7:42 am Leave a comment

Ring Me Up!

Cash RegisterWhen my college roommate contracted crabs, he went to CVS to buy some lice cream. As you can imagine, he didn’t want to announce to the world what he was buying or why, so he put the box on the counter with a notepad, a bottle of aspirin, a pack of cigarettes, a bag of M&M’s, and a tube of toothpaste — hoping the cream would blend in. The attractive co-ed clerk at the register rang him up without a second look.

As he walked out of the drug store thinking he had gotten away with it, he opened the cigarettes, put one to his lips, and realized he had nothing with which to light it. He returned to the checkout and asked the clerk for a pack of matches.

“Why?” she asked. “If the cream doesn’t work, you gonna burn ’em off?”


My luck with clerks wasn’t much better. At a grocery store, I placed a bar of soap, a container of milk, two boxes of cereal, and a frozen dinner on the check-out counter. The girl at the cash register asked, “Are you single?”

I looked at my items-to-be-purchased. “Pretty obvious, huh?”

“Sure is,” she replied. “You’re a very unattractive man.”

I did, however, have an exceptional experience at a convenience store. This is what happened.

I walked into a 7-11 and took four items to the cash register. The clerk informed me that the register was broken, but she said she could figure the total using her calculator. The clerk then proceeded to multiply the prices together and declared that the total was $7.11. Although I knew the prices should have been added, not multiplied, I said nothing — as it turns out, the result would have been $7.11 whether the four prices were added or multiplied.

There was no sales tax. What was the cost of each item?

As you might have guessed, that story is completely false. (The one about me being called ‘unattractive’ is a slight exaggeration. The one about my roommate, sadly, is 100% true.) The truth is that I learned this problem from other instructors when teaching at a gifted summer camp.

It may not be true. It is, however, one helluva great problem.

But it has always bothered me that the problem is so difficult. I’ve always wanted a simpler version, so that every student could have an entry point. Today, I spent some time creating a few.

Use the same set-up for each problem below… walk into a store… take some items to check-out counter… multiply instead of add… same total either way. The only difference is the number of items purchased and the total cost.

I’ve tried to rank the problems by level of difficulty. Below, I’ve given some additional explanation — but not the answers… you’ll have to figure them out on your own.

  • (trivial) Two items, $4.00.
  • (easy) Two items, $4.50.
  • (fun) Two items, $102.01.
  • (systematic) Two items, $8.41.
  • (perfect) Three items, $6.00.
  • (tough) Three items, $6.42.
  • (rough) Three items, $5.61.
  • (insane) Four items, $6.44.
  • (the one that started it all) Four items, $7.11.

Editor’s Notes

trivial — C’mon, now… even my seven-year-old sons figured this one out!

easy, fun, systematic — All of these are systems of two equations in two variables. Should be simple enough for anyone who’s studied basic algebra. All others can use guess-and-check.

perfect — Almost as easy as trivial, and the name is a hint.

tough — But not too tough. Finding one of the prices should be fairly easy. Once you have that, what’s left reduces to a system of equations in two variables.

rough — Much tougher than tough. None of the prices are easy to find in this one.

insane — Gridiculously hard, so how ’bout a hint? Okay. Each item has a unique price under $2.00. If you use brute force and try every possibility, that’s only about 1.5 billion combinations. Shouldn’t take too long to get through all of them…

the one that started it all — As tough as insane, and not for the faint of heart. But no hint this time. Good luck!


September 5, 2014 at 7:11 am 2 comments

Two Simple Math Games

In his article “What Is the Name of This Game?” author John Mahoney discussed the mathematics of the following game:

Cards numbered 1-9 are placed face up on a table. Two players alternate picking up one card at a time. The winner is the first player who has exactly three cards with a sum of 15.

You can play this game with nine cards removed from a deck of cards, or you can play online by going to The online version is a one-player game, but it has modifications that use different numbers of cards, different values on the cards, and different required sums.

Can you find the winning strategy for this game? (Hint: The strategy is described in the linked article above.)

Here’s a modification of the game that seems interesting, too.

Use cards with the following numbers: 1, 2, 3, 4, 6, 9, 12, 18, 36. The winner is the first player who has exactly three cards with a product of 216.

The optimal strategy for this game is different than the strategy from the original game. Can you find it?

Note: For the original game, there are eight sets of three cards with a sum of 15:

{1, 5, 9}, {1, 6, 8}, {2, 4, 9}, {2, 5, 8},
{2, 6, 7}, {3, 4, 8}, {3, 5, 7}, {4, 5, 6}

For the modification, there are 36 sets of three cards with a product of 216.

Perhaps that fact will help you identify the optimal strategy.

March 3, 2013 at 12:12 pm Leave a comment

Upside-Down Tooth Numbers

Alex and Eli know that 15 ÷ 3 = 5, that 63 ÷ 7 = 9, and, given a little time, could figure out that 104 ÷ 8 = 13. That’s not bad for five-year-olds.

As we were discussing sharing a treat the other day, we naturally happened upon a situation in which it would be good to know what 1 ÷ 2 is.

Alex suggested, “Two.”

I explained that while the order of the numbers in multiplication doesn’t matter — for example, 2 × 3 = 3 × 2 — order does matter with division. I used the word commutative, but I also tried to explain it with plainer language, too.

I took a rectangular piece of paper and ripped it in half. “How big is each piece?” I asked. The both knew it was one-half. “So there you have it: 1 ÷ 2 = 1/2.”

But why stop there? I divided one of the halves in half again, and I asked, “How big is this piece?” They both knew it was one-quarter. “So that shows that 1/2 ÷ 2 = 1/4.”

They saw that if we continued in this manner, we would get 1/8, 1/16, 1/32, and so on, a pattern they called the upside-down tooth numbers, because the numbers in the sequence are the reciprocals of the powers of two. (For them, tooth = 2th.)

Looking at the pieces of paper on the table, I asked a more advance question. “What do you think we’d get if we added 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + …, and we kept adding half as much each time?”

Eli thought for a few seconds. “I think it would be one whole,” he offered.

I was surprised. “Why?” I asked.

“Well,” he said, “I’m thinking about a circle.”

I looked at the rectangular pieces of paper on the table, which had nothing to do with a circle. Sure, he got the right answer, but his reasoning is way off base, I thought.

He explained further. “If you fill half a circle, then a quarter more, then an eighth, and keep going, you’ll eventually fill the whole thing. That’s why I think it’s one.”

Visually, Eli’s argument would look something like this…

Infinite Series - One Half

Holy sh*t, I thought. That’s pretty good for a five-year-old.

I don’t even care that he doesn’t understand this joke:

An infinite number of mathematicians walk into a bar. The first asks for half a beer. The second asks for a quarter of a beer. The third asks for an eighth of a beer. The bartender interrupts, pours one beer for them, and says, “You guys don’t know your limits.”

I think I was so impressed with Eli’s solution because infinite series can be difficult, even for mathy professionals:

A mathematician will call an infinite series convergent if its terms go to zero. An engineer will call it convergent if the first term is finite.

And let us not forget the Eilenburg swindle, which proves that 1 = 0 since:

1 − 1 + 1 − 1 + … = 1 + (−1 + 1) + (−1 + 1) + … = 1


1 − 1 + 1 − 1 + … = (1 − 1) + (1 − 1) + … = 0

Related to all this, there is a perhaps apocyphal story in which it is rumored that a student once posed the following problem to John von Neumann:

Two trains begin a mile apart and head towards each other at 60 miles an hour. A fly on one train flies at 120 mph to the other train, and when it touches the other train, it immediately turns around and flies back to the first train, and so on, flying back and forth between the two trains until it gets squashed in the middle. How far does the fly travel?

von Neumann thought about it a moment and said, “One mile.” The student said to von Neumann that most people don’t realize that the problem can be figured out easily: the trains meet in 30 seconds, and the fly can travel one mile in half a minute; yet most people think they have to add up the infinite series to figure out how far the fly travels. As the story goes, von Neumann replied, “But that’s how I did it.”

November 30, 2012 at 1:45 pm Leave a comment

What (Math) is in a Name?

One of my favorite online tools is the Mean and Median app from Illuminations. This tool allows you to create a data set with up to 15 elements, plot them on a number line, investigate the mean and median, and consider a box-and-whisker plot based on the data. Perhaps the coolest feature is that you can copy an entire set of data, make some changes, and compare the modified set to the original set. For example, the box-and-whisker plots below look very different, even though the mean and median of the two sets are the same.

Mean and Median

It’s a neat tool for learning about mean and median, and I plan to use this tool in an upcoming presentation.

For classroom use, I like to use this app with real sets of data. However, the app requires all elements of a data set to be integers from 1-100. Can you think of a data set with a reasonable spread that has no (or at least few) elements greater than 100? If so, leave a comment.

Recently, and rather accidentally, I found a data set that works well. Do the following:

Assign each letter of the alphabet a value as follows: A = 1, B = 2, C = 3, and so on. Find the sum of the letters in your name; e.g., BOB → 2 + 15 + 2 = 19.

Now imagine that every student in a class finds the sum of the letters in their first name. For a typical class, what is the range of the data? What is the mean and median?

The name with the smallest sum that I could find?

ABE → 1 + 2 + 5 = 8

The name with the largest sum?

CHRISTOPHER → 3 + 8 + 18 + 9 + 19 + 20 + 15 + 16 + 8 + 5 + 18 = 139

The Social Security Administration provides a nice resource for investigation, Popular Baby Names. Using a randomly selected set of 2,000 names and an Excel spreadsheet, I found the mean name sum to be 62.49, and 96% of the names had sums less than 100. Of the 80 names with sums greater than 100, many (such as Christopher, Timothy, Gwendolyn, Jacquelyn) have shortened forms (Chris, Tim, Gwen, Jackie) for which the sum is less than 100.

As it turns out, the frequency with which letters occur in first names differs from their frequency in common English words. The most common letter in English words is e, but the most common letter in names is a. The chart below shows the frequency with which letters occur in first names.

Letter Frequency

Because of this distribution, the average value of a letter within a first name is 10.54, which is slightly less than the 13.50 you might expect. This is because letters at the beginning of the alphabet, which contribute smaller values to the name sum, occur more often in names than letters at the end of the alphabet.

The chart below shows the distribution for the number of letters within first names. The mean number of letters within first names is 5.92 letters, and the median is 6. (In the data set of 2,000 names from which this chart is derived, no name contained more than 11 letters.)

Letters In Names

Do you know a name that has more than 11 letters or has a name sum greater than 139 or less than 8? Let me know in the comments.

October 3, 2011 at 12:12 pm 4 comments

About MJ4MF

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

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Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.

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