Ivars Peterson's MathTrek

July 6, 1998

### Prime Talent

Whole numbers have all sorts of curious properties.

Consider, for example, the integer 1998. It turns out that 1998 is equal to the sum of its digits plus the cubes of those digits (1 + 9 + 9 + 8 + 13 + 93 + 93 + 83). What's the largest number for which such a relationship holds? The answer is 1998. What about the smallest integer?

That was one of the playful challenges presented by number theorist Carl Pomerance of the University of Georgia in Athens to an audience that included the eight winners of the 27th U.S.A. Mathematical Olympiad, their parents, assorted mathematicians, and others. The occasion was the U.S.A.M.O. awards ceremony on June 8 at the National Academy of Sciences in Washington, D.C.

Each year, the Olympiad competition includes a problem involving the year in which it is held. Looking ahead, Pomerance pondered 1999--a prime number, evenly divisible only by itself and 1. In this case, the digits of 1999 add up to 28, which happens to be a perfect number. A perfect number is equal to the sum of all its divisors (see Cubes of Perfection). What's the smallest prime whose sum of digits is perfect? The answer is 1999.

The prime 1999 also comes up in another context: 1999 = 211 - 72. Pomerance calls any positive integer, n, that obeys the relationship n11 -7n a convenience store number. How many convenience store numbers are there? What's the smallest one?

Pomerance's address covered a wide range of topics in number theory, from the search for enormous primes (see Another Record Prime) and the difficulty of factoring large composite numbers to the Riemann hypothesis and the prime number theorem (see Prime Theorem of the Century).

Indeed, number theory offers a host of problems that are remarkably easy to state but fiendishly difficult to solve. For instance, primes often appear as pairs of consecutive odd integers: 3 and 5, 5 and 7, 11 and 13, 17 and 19, and so on. These so-called twins are scattered throughout the list of known prime numbers. Statistical evidence suggests there are infinitely many twin primes. No one, however, has yet proved that there is an infinite number of twin primes.

However, it turns out that if you add together the reciprocals of all twin primes (1/p + 1/(p + 2)), the sum has a specific numerical value now called Brun's constant. The fact that this sum converges to a certain number demonstrates the scarcity of twin primes--even though there are infinitely many of them!

It was in trying to determine the value of Brun's constant (1.902160577783278. . .) to a large number of decimal places that Thomas R. Nicely of Lynchburg College in Virginia uncovered a flaw in the Pentium microprocessor (see Pentium Bug Revisited).

Another series of problems involves iterated gaps between primes. Start with a list of prime numbers and calculate the difference between adjacent primes. Then calculate the differences between the differences, taking any negative values as positive. Continue the process. The result is a table that looks like this:

Mathematicians have conjectured that every row after the initial set of primes starts with 1. Using computers, Andrew M. Odlyzko of AT&T Labs in Florham Park, N.J., has verified that the conjecture is true for the first 3 x 1011 rows (covering primes up to 1013). No one, however, has yet proved the conjecture. Interesting patterns (and instances of chaos) also arise when both positive and negative differences are considered.

Playing with numbers serves up all sorts of "head games," Pomerance insists.

The students competing in the Math Olympiad tangled with a variety of mathematical puzzles. Here's one problem from this year's daunting 6-hour, 6-question exam:

Suppose that the set {1, 2, . . ., 1998} has been partitioned into disjoint pairs {ai, bi} so that for all i, equals 1 or 6. Prove that the sum (1 £ i £ 999) so that for all i, ½ aibi½ equals 1 or 6. Prove that the sum ½ a1b1½ + ½ a2b2½ + . . . + ½a999b999½ ends in the digit 9.

ends in the digit 9. The top six scorers on this exam earned places on the U.S.A.M.O. team, which will travel to Taipei later this year to compete in the International Mathematical Olympiad. For the first time, that team includes a high school girl. Melanie Wood of Park Tudor High School in Indianapolis not only made the team but also tied Sasha Schwartz of Radnor High School in Pennsylvania for the highest score.

The other team members are Reid Barton of Arlington, Mass., who was schooled at home, Gabriel Carroll of Oakland Technical High School, Calif., Kevin D. Lacker of Sycamore High School in Cincinnati, and Paul A. Valiant of Milton Academy, Mass. The two alternates are David T. Vickrey of Vermillion High School, S.D., and David E. Speyer of Choate Rosemary Hall in Wallingford, Conn.

There's more to mathematics than mental dexterity and quick thinking in mathematical competitions, however. To succeed in mathematics, you also need persistence, drive, and focus, Arthur M. Jaffe, American Mathematical Society president and Harvard mathematician, told the students. Original thinking plays a key role, he emphasized.

References:

Andreescu, T., E. Johnston, and C. Rousseau. 1998. Twenty-sixth annual USA Mathematical Olympiad-Problems and solutions. Mathematics Magazine 71(June):234.

______. 1998. Thirty-eighth annual International Mathematical Olympiad-Problems. Mathematics Magazine 71(June):238.

Peterson, I. 1998. The Mathematical Tourist: New and Updated Snapshots of Modern Mathematics . New York: W.H. Freeman.

Ribenboim, P. 1996. The New Book of Prime Number Records. New York: Springer.