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 SEPTEMBER/OCTOBER 2018

THE MATH COLUMN
Think Before Counting

By Alfred Posamentier, Ph.D.

Very often a problem situation seems so simple that we plunge right in without first thinking about a strategy to use. This impetuous beginning for the solution often leads to a less elegant solution than one that results from a bit of forethought. Here are a few examples of simple problems that can be made even simpler by thinking before working on them in the traditional fashion.

Find all pairs of prime numbers whose sum equals 999.

Some of you will begin by taking a list of prime numbers and trying various pairs to see if they obtain 999 for a sum. This is obviously very tedious as well as time consuming, and you would never be quite certain that you had considered all the prime number pairs.

Let’s use some logical reasoning to solve this problem. In order to obtain an odd sum for two numbers (prime or otherwise), exactly one of the numbers must be even. Since there is only one even prime, namely 2, there can be only one pair of primes whose sum is 999, and that pair is 2 and 997. That, now, seems so simple.

Here is another problem to consider.

Suppose your school is running a basketball tournament with 25 teams competing. This single-elimination tournament will be run in one gymnasium. The question is how many games will be required to be played in this gymnasium in order to get one champion?

The traditional solution is to set up a chart where you simulate games beginning with 12 teams playing a second group of 12 teams, with one team drawing a bye. This will result in the gymnasium being used 12 times. Continuing along this way, of the remaining 13 teams, have 6 teams playing another 6 teams, with one team drawing a bye. This will add another 6 games required to use the gymnasium. Continuing along this way, and counting the number of games required to get a champion will take a few minutes.

However, looking at this problem from another point of view can give you the answer in a flash. Asking yourself: how many losers must there be in order to get one champion from the 25 teams? The answer is there would be 24 losers, which requires 24 games to be played in the gymnasium. And so, you have the answer to the question in one spurt.

Another problem where pre-planning, or some orderly thinking makes sense is as follows:

A palindrome is a number that reads the same forwards and backwards, such as 747 or 1991. How many palindromes are there between 1 and 1,000 inclusive?

The traditional approach to this problem would be to attempt to write out all the numbers between 1 and 1,000, and then see which ones are palindromes. However, this is a cumbersome and time-consuming task at best, and one could easily omit some of them.

Let’s see if we can look for a pattern to solve the problem in a more direct fashion.

Range            Number of        Total Number
Palindromes

1 – 9                9                           9

10 – 99              9                          18

100 – 199            10                         28

200 – 299            10                          38

300 – 399            10                          48

There is a pattern. There are exactly 10 palindromes in each group of 100 numbers (after 99). Thus, there will be 9 sets of 10, or 90, plus the 18 from numbers 1 to 99, for a total of 108 palindromes between 1 and 1,000.

Another solution to this problem would involve organizing the data in a favorable way. Consider all the single digit numbers (self-palindromes). There are nine such. There are also nine two-digit palindromes. The three-digit palindromes have 9 possible “outside digits” and 10 possible “middle digits”; so, there are 90 of these. In total there are 108 palindromes between 1 and 1,000, inclusive.

The motto is think first, then begin to work on a solution! #

Alfred S. Posamentier, Ph.D., is the Executive Director for Internationalization and Sponsored Programs, Professor Emeritus, Mathematics Education and Former Dean, CCNY-City University of New York.