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Bramasta asks, "Is it possible to have a number that is a multiple of 2 but not a multiple of 4 and is a perfect square number?"

That's a great question, Bramasta, and I'm guessing you've tried a few examples, and decided the answer is probably "no," but you're wondering if we can know it for sure.

Another way of describing "a number that is a multiple of 2 but not a multiple of 4" is to say "a number which is two more than a multiple of four." So if we can show that no perfect squares are two more than a multiple of four, we'll have answered your question. Ready?

If Y is a perfect square, that means its square root, X, is an integer. Consider that X must be either even or odd. In other words, for some value K,

X = 2K or

X = 2K + 1

Since Y = X2, Y = (2K)2 or Y = (2K + 1)2. Multiplying these out gives us the following:

Y = 4K2 or Y = 4K2 + 4K + 1 = 4(K^2 + K) + 1

Notice that in the first case, Y is a multiple of 4, and in the second case, it's one more than a multiple of four. In no case do we get a result that is either 2 more than a multiple of 4 or 3 more than a multiple of four. So we've actually proved MORE than you asked; if a number is either 2 more or 3 more than a multiple of 4, we can emphatically declare that it is NOT a perfect square.

Fifth grader Aaradhya asks: "How do we find the factors of 342, 450 and 540 without a calculator?"

That's a really good question! The best suggestion I can give you is to make sure you are familiar with some divisibility rules. These are very helpful tools in breaking a number down into their factors.

For example, we know that any number that ends with an even digit is a multple of 2. So, right off the bat, we can divide 342 by 2. Of course, without a calculator you'll need to do long division. But then you've got smaller numbers to work with. In this case, 342 divided by 2 is 171.

Does that have any factors? Well, here's where the divisibility rule for 3 might come into play. A number is divisible by 3 if the sum of its digits is divisible by 3. Since 1 + 7  + 1 = 9, and 9 is divisible by 3, then we know 171 is divisible by 3 also. So, another quick long division and we've got 57. Once again, 5 + 7 is divisible by 3, so we know 57 is also. We continue like this until we get down to a number that is prime - in this case, 19. Thus, the factors of 342 are 1, 2, 3, 19, and 342.

There are other useful divisibility rules - for example, if a number ends with a zero, that means it's divisible by 10. Thus, 450 is 10 times 45. Then we just break these down iunto smaller and smaller factors.

Similar to this rule, if a number einds in zero or five, it's divisible by 5.

There are other divisibility rules, but they get increasingly complicated. The rules for seven and thirteen are so cumbersome that I never use them. The divisibility rule for 11 requires you to alternately add and subtract the digits to see if the sum is a multiple of eleven.

For example, 14641 is a multiple of 11 because 1 - 4 + 6 - 4 + 1 = 0 , which is divisible by 11.

Without the divisibility rules, this process requires a lot of long division just to find out if a number is a factor. You still may need to do some of that, but the divisibility rules will help you keep that to a minimum!

The Professor Puzzler blog has been pretty quiet recently, so I'm breaking the monotony by asking myself a question.

Me: Professor Puzzler, what is a Double-Dactyl poem, and how do you write one?

Other Me: That's a great question! I hesitate to say that the Double-Dactyl is one of the silliest poetry forms in existence, but…well…it’s one of the silliest poetry forms in existence.

Let’s start with a Double-Dactyl example, and then we can dissect it to understand how the form works. This is a poem about Noah (yes, the guy with the ark).

Splishity splashity
Noah the Patriarch
Built a big boat out of
Gopher-wood trees

Finding some grace in the
Eyes of Jehovah, this
Antediluvian
Sailed o'er the seas

Before we get into the rules, what is a dactyl? A dactyl is a metrical foot which consists of one stressed syllable followed by two unstressed syllables. In other words, it sounds a bit like a waltz. OOOM-pah-pah OOOM-pah-pah. Okay, now let’s dive in.

1. There are two sets of four lines. These sets of four lines are called “quatrains.”
2. In each quatrain, the first three lines are each two dactyls. The last line of each quatrain is a single dactyl followed by one accented syllable (OOOM-pah-pah-OOOM)
3. The only rhymes we need are at the end of the two quatrains.
4. The first line must be nonsense words. Since they’re nonsense, they don’t have to have anything to do with the subject of the poem, but it can be fun – as I did in my Noah poem – to make them similar to real words that connect to the subject of the poem.
5. The second line must be a proper noun. This would most commonly be a person’s name (or name and title), but could also be a place, organization name, or even something like “U.S.S. Nautilus.” In the example above, NO-ah the PA-tri-arch has the accents on NO and PA, with the other syllables unaccented.
6. Somewhere in the poem – preferably in the second quatrain, and usually the sixth line of the poem (but the location is not a hard-and-fast rule), there must be a line that consists of a single six-syllable word that fits the dactyl rhythm. In the Noah poem, that word is “antediluvian” (which means “before the flood”).

There you have it – now you’re ready to write your own Double-Dactyl. On my YouTube channel (link: Doug's ventiloquism, music, and teaching) my puppets (yes, I’m a ventriloquist!) and I will posted a series of three Double-Dactyls. The third in the series is actually sung instead of recited; if you learn the tune, that may help you write your own Double-Dactyls!

Wiffity Woffity (a poem about a timberwolf and a dodo bird)

Axity Waxity (a poem about George Washington and a cherry tree)

Offity Scoffity (a poem about Alice in Wonderland)

Note that in my poems, my nonsense words all end in “-ity” but that is not a requirement; I just like the way it sounds. I guess I just got stuck in a nonsense rut.

Finally, if you’re interested in other kinds of poetry, one of my puppets shares his own rendition of Robert Burns’ “My Love Is Like a Red, Red Rose,” and several of my puppets have done limericks: Jeorge’s Limerick, Jeffrey’s Limerick, Professor Jameson’s Limerick, Doctor Jonas’ Limerick.

And for those who like Star Wars, be on the lookout for some Star Wars themed Double Dactyls late in 203!

This blog post is about the "Monty Hall Problem" - which you can find a blog post about here: Ask Professor Puzzler about the Monty Hall Problem, and a game simulation here: Monty Hall Simulation.

Larry from Louisiana writes: "Relating to the 3 door problem the answer given is just plain silly. Consider everything the same except you now have two players, one selects door one and the other selects door two. Now door three is shown to have a goat so, according to the given solution both players switch doors and each now has a 66% chance of winning? This may be possible in the new math but it is not in the old math that I learned."

Hi Larry, thanks for the message.

Before responding, I'd like to mention, in case anyone is confused by the "goat" part of your message, that in different versions of the game either two doors are empty, or they contain a donkey, or a goat, or some other silly prize. In the blog post here on this site I refer to the door as being empty, but to make my answer consistent with Larry's comment, I'll refer to the empty doors as containing a goat.

I'd also like to add that when I'm writing math problems for math competitions, my proofreader and I always comment - half joking (which means half serious!) that we hate probability problems, and wish we didn't have to write them. Probability problems can be very tricky, and it's easy to overlook assumptions we make that completely change our understanding of the problem.

In the case of the Monty Hall problem, there are a couple hidden assumptions that can be overlooked.

1. Since the game show host always shows the contestant an empty door, that implies (even though this is not always stated outright) that he knows which door contains the prize. His choice is dependent on his own knowledge.
2. Since only one door has been opened, the game show host has a choice, and it is always possible for him to choose an empty door.

It may be helpful to think of the game show host as a "player" or "participant" in the game. If the game show host cannot operate under the same conditions within the game, it is not the same game.

The game you proposed is an entirely different game with conditions that violate the conditions stated above. Specifically, even though in your stated scenario the game show host shows a goat, we need to understand that it is not always possible for him to choose a goat door; if the contestants have each picked a goat door, there is no door left for him to show except the one with the prize. In other words, the game show host's role is completely different. While he still knows which door contains the prize, he no longer has a choice, and (just as importantly) it is not always possible for him to choose an empty door. Another way of saying this: in your game, the host is no longer a participant in any meaningful way.

Since your game has conditions that violate the conditions of the Monty Hall Problem, it is not the same game, and we shouldn't expect it to have the same probability analysis. In fact, it doesn't, and you correctly observed that it wouldn't make sense for it to work out to the same probabilities.

I would encourage you to try out the simulation I linked at the top of this page; if you're willing to trust that I haven't either cheated or made a mistake in programming it, you'll see that the probability really does work out as described. And if you don't want to put your trust in my programming (which is fine - I think it's good for people to be skeptical about things they see online!), I'd encourage you to get a friend and run a live simulation. The setup looks like this:

1. Take three cards from a deck, and treat one of them as the "prize" (maybe the Ace of Spades?)
2. You (as the game show host) spread the three cards face down on the table. Before you do, though, remember that you're going to have to flip one of the cards that isn't the Ace of Spades, which means you need to know which card is the Ace of Spades. So before you put the cards on the table, look at them.
3. Now ask your friend to pick one of the cards (but don't look at it).
4. Since you know which card is the Ace of Spades, you know whether he has selected correctly. If he's selected correctly, you need to just randomly pick one of the other two cards to show him. If he hasn't picked the Ace of Spades, you know which one is the Ace of Spades, so you show him the other.
5. Now ask him if he wants to switch. You'll need to agree beforehand whether he'll always switch or never switch, but he should be consistent and make the same choice each time.
6. Do steps 2-5 about 100 times. If your friend always chooses to switch, you'll find that he wins about 2/3 of the time, while if he never switches, he'll win about 1/3 of the time.

I've had classes of high school math students break up into pairs and run this simulation, and the results always come out as described above. Happy simulating!

William from Arizona asks, "Myself and a fellow math teacher of 20+ years each got into a discussion about extraneous solutions for a particular problem. The problem is this: 1/(x-a) = x/(x-a) The provided answer to this problem is that x = a is always an extraneous solution. However, when you solve it using the LCD method to multiply both sides, the x = a solution candidate does not present itself. (x-a) * 1/(x-a) = x/(x-a) *(x-a) 1 = x cancelling x-a The only candidate after cancelling is x = 1. The counter argument was that you solve by cross multiplying and setting equal getting: x - a = x(x - a) x - a = x^2 - ax 0 = x^2 - ax - x - a 0 = x^2 - (a+1)x - a Using quadratic formula: x = a or x = 1 therefore, x = a is extraneous. Using the cross multiply method, I run into x = a and need to call it extraneous. Using LCD multiply to both sides, it doesn't present as a candidate. So, is it true to say x = a is always an extraneous solution? Thank you for considering"

Hi William, thanks for asking. This is an interesting question. An extraneous solution is generally defined as "a solution to a transformed equation which is not a solution of the original equation." So, for example, if you square both sides of an equation, you've transformed it, and therefore you've introduced the possibility of solutions that might not be solutions to the original equation.

In this case, the transformation that produces an extraneous solution is cross multiplying, which removes the (x - a) from the denominator (and therefore including x = a in the domain).

So an extraneous solution is a solution that arises because of a transformation. If the transformations you performed did not result in a new equation that has an expanded domain, then it doesn't produce extraneous solutions. In other words, no, I don't think you have an extraneous solution using the first method.

The phrasing here is odd; if x = 1 was an extraneous solution, you wouldn't say "x = 1 is always an extraneous solution." I suspect that what they're trying to say is, "it doesn't matter what the value of 'a' is; x = a will never be a solution." So I don't think they were trying to suggest that there's no way to avoid x = a as an extraneous solution, but rather that if it does arise as a solution, it is always extraneous, regardless of the value of 'a'.

As one last note, I'd like to point out that you can avoid using the quadratic formula in your second method:

(x - a) = x(x - a)
0 = x(x - a) - (x - a)  //subtract (x - a) from both sides
0 = (x - a)(x - 1) //use the distributive property to rewrite the previous equation
x =a or x = 1