A famous equation:

$$\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial{t}}$$

Imagine that—mathematics on the Web!

Double-click for the LaTeX source. There’s more info in the tiny box you ought to find in the lower right hand corner of your browser window.

I’ll have more to say about all this in a few days.

A friend pointed me to Zimaths, a magazine for high school mathematics students, published in Zimbabwe in happier times. (The last issue available online is from 2005. The next-to-the-last issue has a helpful tutorial on inflation, which was then running at about 600 percent per year in Zimbabwe. The rate has gone up considerably since then, and lately the country has had even worse troubles.)

Looking back to the first issue of Zimaths, from October of 1996, I happened upon a list of problems from the Zimbabwe Maths Olympiad of that year. Problem 5 looked like something I might be able to do in my head:

The number a is randomly selected from the set {0, 1, 2, 3, …, 98, 99}. The number b is selected from the same set. What is the probabilty that the number 3^a + 7^b has a units digit equal to 8?

Clearly, all positive integer powers of 3 and 7 are odd numbers not divisible by 5, and so they must end in 1, 3, 7 or 9. Running through the first few of those powers (1, 3, 9, 27, …; 1, 7, 49, 343, …) suggested that all possible terminal digits appear with equal probability. The sum of two such odd numbers must be an even number. It seemed like a good bet that all even terminal digits would be equally likely—that is, each of 0, 2, 4, 6 and 8 would appear with probability 1/5. I checked this hypothesis by listing all possible pairs of the four allowed odd digits (with addition done modulo 10):

1 + 1 = 2
1 + 3 = 4
1 + 7 = 8
1 + 9 = 0
3 + 3 = 6
3 + 7 = 0
3 + 9 = 2
7 + 7 = 4
7 + 9 = 6
9 + 9 = 8

Sure enough, in this list of 10 combinations each even digit appears twice, confirming that the probability of seeing an 8 should be 0.2. If I had been competing in the Olympiad, I would have written down that answer and gone on to the next problem without a further thought.

Later in the day, though, I was looking for something I could use to test a new toy—the latest version of Mathematica. Here’s what I came up with:

f[] := Mod[3^RandomInteger[{0, 99}]
         + 7^RandomInteger[{0, 99}], 10]

BarChart[Part[#,2]&
    /@ Sort[Tally[Table[f[], {100000}]]]]

This generates 100,000 values of 3^a + 7^b mod 10 and tallies them up:

So I’ve discovered a bug in Mathematica, right? I know there’s nothing wrong with my program, and as for a mistake in my thinking—well, that’s just unthinkable.

Update: Thank you, commenters.

Svat’s question — “How do you think that error could have been avoided?” — goes right to the heart of the matter, but I need to start with a prior question: “How do I know which of my answers is the erroneous one?” When a computer program and my own reasoning yield different results, how should I resolve the conflict? Where should I look first for a mistake?

I don’t expect to find a universal and eternal answer to this question. On the contrary, everything depends on one’s personal equation. Some of us are better at analytical thinking and some of us are bit-players. Nevertheless, in this case I think it’s pretty clear that the computer calculation relies on fewer and simpler assumptions. It asks us to trust the random number generator and certain arithmetic operations, but if we grant that the implementation of those functions is correct, the program can hardly go wrong. Its structure follows directly from the problem statement. (Of course there’s also statistical uncertainty, but if we’re patient enough, we can reduce that to any level we please.)

The probability analysis depends crucially — as I discovered to my chagrin — on properly counting all the cases. My mistake was a very elementary one; I suspect it is also a very common one. It was when I saw the empirical results, with probability values clustered near 3/16, that I realized I should be looking at 16 cases rather than 10.

The point I want to make here isn’t about mathematical truths but about the art of problem solving — even when, as in this case, there’s no artistry involved. Barry’s variation on the problem (see the comments) offers me another sterling opportunity to make a fool of myself. So I present below, in first-person present-tense stream-of-consciousness diary style, a record of my wary engagement with that problem.

• First, why the shift from {0..99} to {1..100}? Because including zeroth powers would contaminate each term in the sum with a 1. For example, 2ⁿ mod 10 is {2, 4, 6, 8} for n={1..100}, but for n={0..99} there’s a 1 percent chance of turning up a 1. That would make the sum much messier.
• 1ⁿ and 5ⁿ are easy. After a further moment’s thought: So is 6ⁿ. So every sum will include a term equal to (1 + 5 + 6) mod 10, or in other words 2.
• 4ⁿ is {4, 6} and 9ⁿ is {1, 9}. We can combine them — but carefully, now, let’s not make the same mistake again — to produce a single term of {3, 5, 5, 7}.
• We already know that 3ⁿ and 7ⁿ both yield {1, 3, 7, 9}. And 2ⁿ and 8ⁿ generate {2, 4, 6, 8}. Therefore the whole sum is
  {2}¹ + {3, 5, 5, 7}¹ + {1, 3, 7, 9}² + {2, 4, 6, 8}²,

  where the superscripts indicate the number of elements we are to choose (independently at random with replacement) from each of these sets. (Not sets, really. Bags.)

• Parity: There are an odd number of odd terms, so the sum will be odd.
• Is there some cute number-theoretical trick that’s going to make this easy? I don’t see it.
• Stuck. Back to the computer to do it the unclever way:

  g[]:=Mod[Sum[i^RandomInteger[{1,100}],{i,1,9}],10]

  Tallying 100,000 repetitions of this function produces the following result:

  

  Repeating the computation a few times more suggests that the slight excess of 7s is not statistical noise. So now I think I know the answer, but I don’t know where it came from.

• Back to the probability calculation. The 4 × 4 sum matrix for the original problem was easy (once I knew what to do with it). This one is 1 × 4 × 4 × 4 × 4 × 4. There must be a shortcut; that’s too much to do with pencil and paper.
• [After a break.] Aha: The sum matrix for {2, 4, 6, 8}² is exactly the same as the matrix for {1, 3, 7, 9}² (after permuting some columns and rows). Thus the sum can be expressed as:

  {2}1 + {3, 5, 5, 7}1 + {1, 3, 7, 9}4.

  However, this still looks like a 4 × 256 matrix, which exceeds my patience.

• Back to the computer again, this time to do all those little sums. What I want is something along the lines of:

  Outer[PlusMod, {1, 3, 7, 9}, {1, 3, 7, 9}],

  where Outer is like a cross product; but instead of multiplying it applies the PlusMod procedure, which I define to do addition modulo 10. The result returned by the expression above is the answer to the original Zimbabwe Olympiad problem:

  {{2, 4, 8, 0},
   {4, 6, 0, 2},
   {8, 0, 4, 6},
   {0, 2, 6, 8}}
  

  For Barry’s more-elaborate variation the full program (without shortcuts) is:

  Sort[Tally[Flatten[
    Outer[PlusMod, {1}, {2, 4, 6, 8}, {1, 3, 7, 9},
       {4, 6}, {5}, {6}, {1, 3, 7, 9},
       {2, 4, 6, 8}, {1, 9}]]]]
  

  The Outer[] expression yields 1,024 values, which are then tallied up. The returned value is:

  {{1, 204}, {3, 204}, {5, 205}, {7, 206}, {9, 205}}

  This appears to be our answer. Barry asked about the frequency of the digit 5: It is 205/1024. The 7 digit is most common by a slight edge, at 206/1024.

• Going back to the simulation program and running 1,024 × 100,000 iterations to get better statistics yields this result (about an hour later):

  {{1, 20400852},
   {3, 20396224},
   {5, 20501004},
   {7, 20601096},
   {9, 20500824}}

• In the end I still wonder if there’s an easier way. Barry?

It’s the custom among mathematicians that when you reach age 60, you’re put on an ice floe with a day’s supply of walrus meat and set adrift.

But I’m not a mathematician.

Computer scientists get a few years’ grace. Because they count in binary, for them the dreaded end of productive intellectual life is put off until age 2⁶.

I’m not a computer scientist either. I’m a writer. This is no dispensation, however; the literary world also worships precocity. Byron said it: “Is there anything in the future that can possibly console us for not being always twenty-five?” I took an even harder line in my own youth. At age 16, when I ran away from home to write a novel, I vowed that I would be a famous author by 22 or find an ice floe and make an end of it.

And today I am 60–1, and I find I am not yet quite ready to be set out to sea. I look back to the poets who seemed so intensely young when I was young, and I find they have aged well. Here’s George Herbert: