## Archive for May 3, 2011

### MathCounts National Competition

Three days from now, more than 200 mathletes will compete in the MathCounts National Competition in Washington, DC. The 228 mathletes — four each from the 50 U.S. states and various territories — will descend on the Renaissance Washington DC Downtown Hotel and attempt to solve 48 problems in three hours. If you happen to be in the nation’s capital and can spare a few hours, it’s worth attending the event. The students who attend are so talented, mature and well educated, it’s easy to forget that they’re only in middle school. (Coincidentally, the hotel has a wonderfully numerical address: 999 Ninth Street NW, Washington, DC 20001. Four 9’s in the street address and three 0’s in the ZIP code — how awesome!) As a member of the MathCounts Question Writing Committee, I’ve authored 170 problems and reviewed more than 700 problems in the past eight months. A subset of those questions will appear on the 2012 MathCounts school, chapter, state, and national competitions.

To show their appreciation for our work, the MathCounts Foundation recently sent a shirt to all members of the QWC. Here it is: On the front, it says:

I CREATE PROBLEMS

On the back, it says:

SEE IF YOU CAN SOLVE THEM
MathCounts Question Writing Committee 2010–11

Among the questions that I reviewed this year are some absolute gems that I’d love to share, but for obvious reasons, I cannot. However, I can share a few of my favorite problems from the book The All-Time Greatest MathCounts Problems, which I edited with Terrel Trotter, Jr.

The following problem appeared on the Sprint Round of the 1992 National Competition. During the Sprint Round, students solve 30 problems in 40 minutes, which is an average of just 1 minute, 20 seconds per problem.

What is the greatest number of bags that can be used to hold 190 marbles if each bag must contain at least one marble, but no two bags may contain the same number of marbles?

Think about it for a second — or 80 seconds, like the competitors do — then read on.

When this problem was originally placed on the 1992 National Competition, the members of the QWC believed the answer to be 19 bags: 1 marble in the first bag, 2 marbles in the second bag, 3 marbles in the third bag, and so on, with 19 marbles in the nineteenth bag (1 + 2 + 3 + … + 19 = 190). At the competition, one student protested, claiming that the answer was 190 bags. On the protest form, he drew a picture like this: The student explained: Put one marble in the first bag. Then put one marble and the first bag with its marble inside the second bag; hence, the second bag now has two marbles. Continue in this manner for 190 bags.

Consequently, the alternate answer of 190 bags was accepted by the judges. This is especially noteworthy, since the problem had been reviewed by nearly 40 people, including university professors, secondary math teachers, and engineers, and none of them had considered this alternate solution.

The next problem is from the 1999 National Competition. It appeared on the Team Round, which means that it was one of 10 problems that four students attempted to solve in 20 minutes.

As shown below, a square can be partitioned into four smaller squares, or nine smaller squares, or even into six or seven smaller squares provided the squares don’t have to be congruent. (Note that overlapping squares are not counted twice.) What is the greatest integer n such that a square cannot be partitioned into n smaller squares? Any square can be divided into four smaller squares, thus increasing the total number of squares by three. The first figure shown above is divided into four smaller squares; when the lower left square within that figure is divided into four even smaller squares, the result is the last figure shown above, which is divided into seven squares. By dividing one of the smaller squares of that one into four would yield a figure divided into 10 smaller squares, then 13, 16, 19, …, accounting for all values of n ≡ 1 mod 3 for which n > 1.

Also shown above is a square divided into six smaller squares. Dividing one of the smaller squares into four would yield a figure divided into 9 smaller squares, then 12, 15, 18, 21, …, accounting for all values of n ≡ 0 mod 3 for which n > 3. That leaves only the case for n ≡ 2 mod 3, for which there are no examples shown above. But producing a square divided into eight smaller squares is easy enough to do: Dividing one of the smaller squares into four would yield a square divided into 11 smaller squares, then 14, 17, 20, 23, …, accounting for all values of n ≡ 2 mod 3 for which n > 5.

The only number of squares not on our list are 2, 3, and 5. Consequently, n = 5 is the greatest integer such that a square cannot be partitioned into n smaller squares.

The final round of a MathCounts competition is the Countdown Round, a Jeopardy!‑like event in which two students compete head-to-head to answer questions that are projected on a screen. Students are given just 45 seconds per problem. Because the questions are expected to be answered in a short amount of time, they are not as challenging as those in other rounds. That said, they’re not exactly easy, either! In 1995, the final question of the Countdown Round had a counterintuitive answer, so it garnered a lot of attention:

Out of 200 fish in an aquarium, 99% are guppies. How many guppies must be removed so that the percent of guppies remaining in the aquarium is 98% ?

Originally, there must have been 200 × 0.99 = 198 guppies. If x guppies are then removed, the ratio of guppies remaining is (198 – x) / (200 – x). Since 98% of the remaining fish must be guppies, the following equation results:

(198 – x) / (200 – x) = 98/100

Solving for x yields x = 100.

Although the algebra is not difficult, most folks find it counterintuitive that 50% of the fish have to be removed from the aquarium to reduce the percent of guppies by just 1%.