Dear Professor Puzzler,

Pick a three digit number. Reverse the digits. Subtract the smaller from the larger. Now add the digits of your result, and you get 18. Every time. How does this work?

Yvonne from Georgia

Hi Yvonne,

First of all, it doesn't work every time. Let's look at a counter-example first, and then we'll look at why it mostly works.

In order to make this not work, I picked a number in which the first and last digits were the same:

525. Reverse the digits and you get the same number. Subtract these and you get zero. Add the digits of zero, and you get...zero!

So, maybe you should rephrase your instructions to say that the digits have to be different, or that the first and last digits can't be the same.

Okay, so now let's look at the math behind what is happening.

Suppose the number I picked was abc, with a representing the hundreds place, b the tens place, and c the ones place.

Then the value of my number is 100a + 10b + c. And if I reverse the digits, the new number's value is 100c + 10b + a.

Now, we don't know which number is larger, so we'll assume the original number was bigger. If it was the other way around, we'd just have to reverse the sign of our result.

100a + 10b + c - (100c + 10b + a) =
100a + 10b + c - 100c - 10b - a =
99a - 99c =
99(a - c)

Isn't that interesting! Your result will always be a multiple of 99! Which means it's a multiple of 9 and a multiple of 11.

We have a divisibility rule for nine (you can find several divisibility rules here: Reference Unit on Divisibility Rules) which says that if a number is divisible by 9, its digits add to a multiple of 9. And 18 is certainly a multiple of 9!

From here, the easiest thing to do is to list out the multiples of 99 that are 3 digits or fewer, and you'll see that in each case, their digits add to either 0 or 18:

0 ⇒ 0
99 ⇒ 18
198 ⇒ 18
297 ⇒ 18
396 ⇒ 18
495 ⇒ 18
594 ⇒ 18
693 ⇒ 18
792 ⇒ 18
891 ⇒ 18
990 ⇒ 18

And there you have it! Except for the case of a repeated first and last digit, which gives you zero, every other possibility gives you 18!

Thanks for a fun question, Yvonne!

Professor Puzzler

Hi Professor, I saw this riddle on a website: "The sum of three numbers is equal to their product. What are the numbers?" They gave an answer of "1,2, and 3". I wondered if this is the only answer. verity

Hi Verity,

In order to answer this question, we should start by writing the problem algebraically:

abc = a + b + c

What you should notice right away is that there are three unknowns (three variables) and only one equation. This makes it seem very likely that there will be multiple solutions. The first thing I tried with this equation was to solve for one variable. Shall we solve for a? Here are my steps:

abc = a + b + c 
abc - a = b + c  (subtract a from both sides)
a(bc - 1) = b + c (factor out an a)
a = ^{(b + c)}/_{(bc - 1)} (divide both sides by bc - 1)

You know what this looks like to me? It looks like we can pick any values we want for b and c, as long as b and c aren't reciprocals of each other, and we'll get a value for a. b and c can't be reciprocals of each other because then (bc - 1) would be zero, and we'd have a division by zero, which is illegal, immoral and socially unacceptable.

So let's pick a couple values. b = 3; c = 4.

a = ^{(3 + 4)}/_{(12 -1 )} = ⁷/₁₁.

Let's test our answer:

3 + 4 + ⁷/₁₁ = ⁸⁴/₁₁.

3(4)(⁷/₁₁) = ⁸⁴/₁₁.

Sure enough, they're the same! So the answer to your question is, no, their solution is not the only one; there are an infinite number of solutions.

I suspect that the site intended to make it more challenging by requiring you to use only integers, or maybe distinct integers. But that doesn't change the fact that there's an infinite number of solutions. For any integer x, the following three numbers work:

x, 0, -x

Someone told me that "most people have a higher than average number of bones in their body." That doesn't even make sense to me. How can most people be above the average?

Whoever told you that is probably not right, but I'm pretty sure I know what they were reasoning. So first I'll explore their reasoning, and then I'll present an alternate view of the statement.

You see, the 'normal" human has 206 bones in their body, but the average number of bones in the human body, if you were to take a survey of all humans currently alive in the world - is slightly less than 206.

Why? Because some people were born without arms or legs. Others have had limbs amputated. So if you average the number of bones of all the people in the world, you'll end up with a number slightly lower than 206.

Let's just make up some numbers, so you can see what I mean.

We'll say that the population of planet earth is 8 billion people. Of these 8 billion people, we'll hypothesize that 0.5% of them have lost a limb (rough estimate based on the number of people in the U.S.A. with amputations), and that limb loss has cost them 30 bones (I'm completely making that number up).

So we have 7,960,000,000 people with 206 bones, and 40,000,000 people with 176 bones. How many bones is that in total?

7,960,000,000(206) + 40,000,000(176) = 1,646,800,000,000

That's a lot of bones! So to find the average number of bones per person, we divide by our 8 billion figure: 205.85.

Since most people have more than 205.85 bones, most people have a higher than average number of bones in their body.

Okay, so that's the reasoning your friend was using. But it's not quite right. Because, you see, when you are born you have about 300 bones in your body. These bones are connected by cartilage, and will eventually fuse together (the process by which cartilage turns to bone is called "ossification"). How long does it take this process to complete? Well, I don't know for sure. One site I looked at said the process is complete around the age of 20, another said 25. I didn't feel like tracking down more research on that, so let's go with the smaller number, and say the process is done at age 20.

This is important: I'm attempting to prove your friend wrong, which means that at every step, my rounding is going to be done in the direction that favors your friend's hypothesis. Because if he's wrong in the best-case scenario, then we know he's wrong for sure!

Now, I have no idea how many bones a child has at various ages, so here's what I'm going to do. I'm going to assume the best case scenario - every child under the age of 20 has 1 extra bone, leading to a huge chunk of the population having 207 bones instead of 206. Again, I know that I'm lowballing this massively, but my goal is to give your friend's hypothesis the benefit of the doubt.

So now we have to rework our numbers, and for this, I'm going to use 25% as the percentage of people in the world under the age of 20. This number is not completely out of my hat - I based it on some numbers for U.S.A. population, and rounded down a bit. 

In our previous calculations, we had 1,646,800,000,000 bones, but now I want to break it down by adult vs. child, so I'm going to multiply that by 0.75 (75%) to get the number of adult bones:

Adult bones = 1,646,800,000,000(0.75) = 1,235,100,000,000

For children's bones, we'll take the total in our previous calculation and multiply it by 0.25 (25%). However, this number is based on 206 bones per person, so we're going to then divide by 206 and multiply by 207. This number isn't going to come out even, so we'll round down, in fairness to your friend's hypothesis.

Child bones = 1,646,800,000,000(0.25)(207)/206 = 413,700,000,000.

Now let's add the adult bones and child bones together: 1,648,800,000,000. This is only slightly larger than the other number we obtained (because I did so much rounding down), but it's enough to put us on the other side of 206:

1,648,800,000,000/8,000,000,000 = 206.1

So it turns out that most people (the adults without amputated limbs) have a below average number of bones. Bear in mind that some of these numbers I pulled out of a hat without a lot of research, but I think it's safe to say I lowballed enough numbers that we can be reasonably sure of the conclusion. If you disagree, drop me a note with your reasoning!

I was told that the number 1,000,001, no matter what base you're working in, is a composite number. Is this true?

Hmm...that's an interesting question. Let's look at some sample bases to make sure it's a reasonable statement.

If this was base ten, then we'd have 1,000,001 is divisible by 101, so it's composite.

If this was base eleven, then the number would be equal to 1,771,562 (base ten), which is obviously composite, because it's even!

Actually, now that I think about it, any time the base is an odd number, 1,000,001 represents an even number, so it's definitely true for all odd bases. So let's focus on even numbered bases.

Let's try base eight. This number would be equal to 262,145 (base ten), which is a multiple of 5.

Base six? That's 46,657 (base ten), which is divisible by 37.

Now, I could keep trying more bases, but I just noticed something interesting. In our first example, the number is divisible by 101, which is one more than the square of 10, and in our last example, 46,657 is divisible by one more than the square of six. That's interesting. Your number may always be divisible by one more than the square of the base. Let's test that hypothesis.

In base twelve, this number would be 2,985,985, and if our hypothesis is correct, it will be divisible by 12² + 1, or 145. Pull out your calculator...

It is!

So we have a reasonable extension of your conjecture: If 1,000,001 is a number written in base n, then it is divisible by n² + 1.

Ideally, it would be nice if we could prove this extension, because if we could, we'd be closer to proving your conjecture.

Let's write our number 1,000,0001 in terms of the base n:

1,000,0001 = 1·n⁶ + 1.

And suddenly I'm remembering one of my factoring rules - the rule for a sum of cubes:

n⁶ + 1 = (n²)³ + 1³ = (n² + 1)(n⁴ - n² + 1)

Sure enough, 1,000,0001 will always be divisible by one n² + 1!

So in order to finish proving your conjecture, we simply need to show that (n² + 1) can't be equal to 1, and it can't be equal to n. These are the two circumstances which could result in  n⁶ + 1 being prime (a number is composite if it has factor pairs other than one and itself).

n² + 1 = 1 has only one solution: n = 0. But we don't work in base zero, so this is irrelevant.

n² + 1 = n, or n² - n + 1 = 0, which has no real solutions, so YES! The conjecture we started with is true!

Thanks for asking - that was an interesting exercise!
Professor Puzzler

A student asked this question today: "Why is 0! (zero factorial) equal to one, instead of zero?"

Good question! Let's begin by making sure everyone knows what the "!" (factorial) notation means. n! means "the product of all the integers that lie between n and 1, inclusive."

Thus, 4! = 24, because 4! = 4(3)(2)(1) = 24.

6! = 6(5)(4)(3)(2)(1) = 720

The strange thing, though, is that 0! = 1, and that doesn't really seem to match our definition. After all, the integers between 0 and 1 inclusive are 0 and 1, and when you multiply them together, you get zero, not one!

Okay, so maybe our definition is flawed. We'll come back to that later.

The thing is, though, we don't want 0! to be equal to zero, because it's not useful. You see, we use factorials when we're calculating combinations of things, or when we're expanding a binomial to a power.

If you have something like (x + 1)⁵, the n^th term in the expansion of that is (₅C_n-1)x^6-n. That's the binomial theorem.

The problem is, that theorem doesn't work if we say that 0! = 0. Why? Because ₅C₀ = ^5!/_(0!·5!), and if 0! is zero, then we have a division by zero problem! On the other hand, if we say that 0! = 1, then this works out perfectly to ₅C₀ = 1.

And really, if you think about it, that makes sense: If you have five objects, in how many different ways can you choose none of them? Uh, one!

We can see that 0! = 1 makes sense using patterns, too. Consider this:

^7!/_6! = 7
^6!/_5! = 6
^5!/_4! = 5
^4!/_3! = 4
^3!/_2! = 3
^2!/_1! = 2

Now to continue this pattern, what do we need next? 

^1!/_0! = 1

Solve this equation for 0!, and you get: 0! = 1.

Okay, so it makes sense with the combination notation to say 0! = 1, and we can even see from patterns that it must equal 1. So the real problem is our definition. So maybe we should reword our definition a little bit.

I like this way of saying it: For all non-negative n, n! is the product of 1 with all the positive integers less than or equal to n.

Does that work? Sure! It keeps everything else the same, but since there no positive integers less than zero, we're left with 1.

That's one way of getting around it. Another way is to just say For all non-negative n, n! is the product of all the positive integers less than or equal to n. This way of defining it forces us to use the "empty product" definition, which says that multiplying together zero factors gives a result of one.

Or you can define it recursively by saying : 0! = 1, and for all integer n > 0, n! = n(n - 1)!.

Or, you can simply do this: For all positive n, n! is the product of 1 with all the positive integers less than or equal to n, and 0! = 1.

This last one defines n! when n>0, and then gives a special definition for 0!.

No matter how you choose to define it, the real point is that mathematicians chose to define it to be one instead of zero simply because it was of practical use to do so!