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Lavinia from Italy writes: "Hi professor. I have just read your post about this year's number (2017). In this post you say that we can write 2017 as a the sum of one cube plus twice another. 2017 = 11^{3 }+ 2 **·** 7^{3} I was wondering how you managed manage to discover this. Did you use a formula or something? Thanks, Lavinia"

Thanks for asking, Lavinia. For those who missed it, this is a reference to the following blog post: Happy New Year 2017. In this blog post I make mention of the fact Lavinia pointed out:

**2017 = 11 ^{3 }+ 2 ·**

**7**

^{3}So how did I discover this odd fact? There wasn't a formula. I found it simply by playing around with the numbers to see what interesting things I found. I don't remember exactly what I was doing when I discovered that, but it probably involved a spreadsheet. I probably was thinking, "I wonder what results I'll get if I subtract perfect cubes from 2017!"

So I probably entered the following formula into cell A1 of a spreadsheet:

**= ROW()^3 (cube the row number)**

And then, in B1 cell put the following formula:

**= 2017 - A1**

Then I would have used the "fill down" feature to populate a bunch of rows with the results of those calculations:

Incidentally, this time as I looked at the rows of numbers, I realized that there's a perfect square in the second column: 289 = 17^{2}. So we can also write:

**2017 = 12 ^{3 }+ 17^{2}**

I probably did something very similar to find the perfect squares that add to 2017:

**A1 = ROW()^2
B1 = 2017 - A1
C1 = SQRT(B1)**

This would have given the following:

**2017 = 9 ^{2} + 44^{2}**

Incidentally, there was another row that had no decimal, further down the spreadsheet. It was row number 44, of course!

Excel (or any other spreadsheet application) can be a great help in exploring mathematical oddities. I realize that I'm a bit spoiled, having calculators and spreadsheets to use when playing with numbers. Imagine trying to figure out quirky number facts like these without even a calculator!

Steve from Cincinnati writes: "I believe this puzzle answer is 1089.

11 X 11 = 4

22 X 22 = 16

33 X 33 = ???

"The first two equations are wrong, but they trick us into looking for a pattern rather than apply the mathematical rules...many people assume an addition function exists between each of the two pairs of integers. Not so. Therefore, thirty three multiplied by thirty three is 1089."

Hi Steve! Of course, you are 100% correct. 33 x 33 = 1089, no matter how many incorrect equations you put before it. Now that I've said that, if I stop there, I've got a really short blog post, and no one wants that - so let's fix their notation in order to turn it into a more interesting puzzle. We'll treat it as a single variable function:

Find x if:

F*(*11) = 4

F(22) = 16

F(33) = x

It *still* isn't a very good puzzle, though, because there's not enough information to come up with a single solution. In that respect, it's very similar to the "Squares Problem" I wrote up last year. Here are just three possible answers to the puzzle.

**Solution #1**

We add the digits of the number in the function's domain, and then square it. (1 + 1)^{2} = 4; (2 + 2)^{2} = 16; (3 + 3)^{2} = **36**.

**Solution #2**

We square the number, and then add the digits: 11^{2} = 121 and 1 + 2 + 1 = 4; 22^{2} = 484 and 4 + 8 + 4 = 16; 33^{2} = 1089 and 1 + 0 + 8 + 9 = **18**.

**Solution #3**

We subtract 8 from the number, and then add the quotient the number and eleven: 11 - 8 + 11/11 = 4; 22 - 8 + 22/11 = 16; 33 - 8 + 33/11 = **28**.

I'm sure there are other solutions; those are just the ones that popped into my head on a quick glance. Can you come up with others? Send them my way and I'll add them to this post.

The following question comes to us from fifth grader Savannah, from Ruston. "How do you convert units with exponents using the method '*king henry doesn't usually drink chocolate milk'*?"

Hi Savannah. "King Henry Doesn't Usually Drink Chocolate Milk" is what we call a mnemonic; it's a phrase that's designed to help you remember something important. In this case, what we're trying to remember is the prefixes for metric units. The prefixes all begin with the same letters as the words of the mnemonic, as shown below, with an example unit:

King ⇒ kilo ⇒ kilometer

Henry ⇒ hecto ⇒ hectometer

Doesn't ⇒ deca ⇒ decameter

Usually ⇒ Unit without prefix ⇒ meter

Drink ⇒ deci ⇒ decimeter

Chocolate ⇒ centi ⇒ centimeter

Milk ⇒ milli ⇒ millimeter

So let's say I had 20 hectoliters. How would I convert that into milliliters? King Henry will help me remember the order of the units:

kilo ⇒ hecto ⇒ deca ⇒ unit (without prefix) ⇒ deci ⇒ centi ⇒ milli

If I want to go from hectoliters to milliliters, I'm moving 5 prefixes to the right, so I'm going to multiply by 10^{5}. Thus, I have 20 x 10^{5} = 2000000 milliliters.

If I wanted to convert 5 centimeters to decameters, I am moving the opposite direction through the units, and I'm moving 3 prefixes. So this time I *divide* (because I'm moving in the other direction) by 10^{3}. 5/10^{3} = 0.005 decameters.

One problem with this mnemonic is that both deca and deci begin with D, so you still have to remember which word goes with which prefix. I would try to remember that both "drink" and "deci" have the letter *i* in them, and that may help you out.

Good luck with your unit conversions!

"Can you explain terza rima and give an example?" ~Anon, grade 5

Terza Rima is an Italian phrase that means "third rhyme." It's a specific way of rhyming lines in a poem. I think of it as sort of a revolving door of rhymes. In each stanza of a Terza Rima poem, there are two lines that rhyme, and one line that does not. The line that *doesn't* rhyme provides the rhyming syllable for the next stanza. Even though it doesn't rhyme with other lines in that stanza, it provides a connection to the *next *stanza, thus building the whole poem into a progressive, seamless whole.

In a Terza Rima poem, the last stanza often has two rhyming lines (that's called a couplet).

In other words, the rhyme scheme looks something like this:

ABA - BCB - CDC - DD

If you wanted more than four stanzas, you could chain together as many stanzas as you want in this format.

If you have a hard time following that explanation, here's a silly poem I wrote just for you, that uses the Terza Rima rhyme scheme:

## Candy Land

I dreamed the world was made of cookie dough.

The skies were filled with cotton candy clouds,

And from them blew a storm of whipped-cream snow.

The fields of chocolate, farmers left unplowed;

The stalks of candy-cane grew everywhere,

And gum-drops grew on bushes, low but proud.

Oh, nothing in this world seemed quite so fair

As pine tree branches bowed with sugar cones -

Enough for all the hungry crowds to share.

A whiff of spearmint on the wind was blown

O'er milk-shake streams and maple syrup lakes.

I shouted from atop my candied throne:

"This world of ours, it really takes the cake -

If it's a dream, I do not wish to wake!"

^{Copyright 2017 by Douglas Twitchell}

*dough*and

*snow*rhyme. The word

*clouds*doesn't rhyme with anything in that stanza. However, it does rhyme with

*unplowed*and

*proud*in the next stanza. Similarly,

*everywhere*in the second stanza doesn't rhyme with anything else in that stanza, but it does rhyme with

*fair*and

*share*in the next stanza. Finally, the concluding couplet takes

*lakes*from the previous stanza and makes it the basis for the concluding rhyme.

Incidentally, Robert Frost wrote a terza rima sonnet titled "Acquainted with the Night." In addition, his poem "Stopping by Woods on a Snowy Evening" is not Terza Rima, but it's a very similar "chained" rhyme; each stanza has four lines. The third line doesn't rhyme with the others, but it does introduce the rhyme for the next stanza. The rhyme scheme looks like this:

AABA - BBCB - CCDC - DDDD

B.R. asks, "Can you explain what mechanical advantage is?"

Hi B.R., I'll give it a shot. Not knowing how much Physics background you have, I'll try to explain it in simple terms.

First of all, I'd like you to consider the following arrangement, which is a person pulling on a rope, which runs over a pulley, and down to an object with a weight of 400 Newtons (that's about 90 pounds, if you're used to using English units).

Now, when we talk about someone pulling on something, we're using the concept of a *force*. Forces are measured in Newtons (or pounds). If the object has a weight of 400 Newtons, that means the person holding the rope has to pull with a force of 400 Newtons in order to keep the object from speeding up or slowing down. If he pulled with a force greater than 400 Newtons, the object would begin accelerating upward. If he pulled with a force less than 400 Newtons, the object would begin to accelerate downward.

This system has a mechanical advantage of 1. This means that on one end, the person applies a force that is equal to the force at the other end of the system. 400 / 400 = 1.

So let's take our pulley system and make it just a little more complicated. Here I've put a second pulley at the bottom, and attached the rope to the ceiling. The object hangs from the lower pulley.

Now, one of the things that's important to understand is that if the man pulls with a force of 400 Newtons, there is *tension* in the rope, and the tension is the same everywhere along the rope. That tension is also measured in Newtons, and it is 400 Newtons. But here's where it gets interesting. Look at that lower pulley (that the object is hanging off). What is the tension in the rope to the left of the pulley? 400 N, right? What is the tension in the rope to the right of the pulley? It's *also* 400 Newtons. This gives a total of 400 + 400 = 800 Newtons. The upward force on the object is *double* the object's weight. This object is going to accelerate upward, even though in the first scenario, it would not have accelerated.

So if the man pulls with a force of 400 N, the output force is 800 N. The mechanical advantage is 800 / 400 = 2.

Incidentally, this means that the man could hold up an 800 Newton weight by applying a 400 Newton force. That's pretty impressive!

You could add more pulleys to the system, as shown here:

In this case, if the man is pulling with a force of 400 N, that means that there is 400 N of tension all through the rope, and so *each* of the lower pulleys has an upward force of 800 N, which means that the object (which is being supported from those two pulleys) has an upward force of 1600 N applied!

So the mechanical advantage is: 1600 / 400 = 4.

It seems like this is a win-win situation; you get better results by adding more and more pulleys. So why not add five hundred pulleys? If you did, you could hold up a 400 Newton object by applying a tiny, tiny force (less than 1 Newton!). So is it really a win-win situation? Not exactly. You see, there's a tradeoff for mechanical advantage. The tradeoff is the distance you pull the object. Take a look at the first diagram. If the man pulls the rope down 1 meter, the object will rise 1 meter.

But what about the second diagram? What will happen there? Well, if the man pulls the rope down 1 meter, that meter is evenly distributed on either side of the lower pulley. Which means that pulley will only rise 0.5 meters. And therefore, if he pulls down a meter, the object only rises half a meter.

What about the third diagram? In that scenario, if the man pulls the rope downward 1 meter, that meter gets evenly distributed among four sections of the rope (one on either side of the two lower pulleys). In other words, the object only moves 0.25 meters.

So, if you had 500 pulleys, you'd be in the very interesting situation that you could easily lift a 400 Newton object, but it would take you a long time to do it; for every meter you pulled the rope down, the object would only move 2 millimeters!

There are many other kinds of machines: screws, wedges, levers, axle-and-wheels, and inclined planes. For every machine, there's a tradeoff; you can make the work easier, but in doing so, you end up extending that effort over a longer distance. An inclined plane is another easy one to picture. Imagine trying to lift a grand piano 1 meter straight off the ground. Can't do it, right? But if I made a long ramp, you could push the piano up. Much easier than lifting, but the tradeoff is that you have to push it a whole lot more than 1 meter!

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