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Solution for
Programming Exercise 3.2
THIS PAGE DISCUSSES ONE POSSIBLE SOLUTION to the following exercise from this on-line Java textbook.
Exercise 3.2: Which integer between 1 and 10000 has the largest number of divisors, and how many divisors does it have? Write a program to find the answers and print out the results. It is possible that several integers in this range have the same, maximum number of divisors. Your program only has to print out one of them. One of the examples from Section 3.4 discussed divisors. The source code for that example is CountDivisors.java.
You might need some hints about how to find a maximum value. The basic idea is to go through all the integers, keeping track of the largest number of divisors that you've seen so far. Also, keep track of the integer that had that number of divisors.
Let's use a variable named maxDivisors to keep track of the largest number of divisors we have seen so far and use numWithMax to store the number that had that many divisors. We have to compute the number of divisors of each integer from 1 to 10000. Whenever we find a larger number of divisors than maxDivisors, we have to make note of that fact by changing the values of maxDivisors and numWithMax. At the end of the process, maxDivisors will be the absolute maximum number of divisors and numWithMax will be a number that had that many divisors. These are the values we want to print out. We can express this with a pseudocode algorithm
for each integer N from 1 to 10000:
Count the number of divisors of N
If that number is greater than maxDivisors:
Let maxDivisors = the number of divisors of N
Let numWithMax = N
Output maxDivisors and numWithMax
However, there is a problem here: The very first time maxDivisors is used in the test "If that number is greater than maxDivisors," the variable maxDivisors hasn't yet been assigned a value. The computer will report this as an error. This can be fixed by assigning a value to maxDivisors before the beginning of the for loop. One way to do this is to handle N=1 as a special case before the loop and then to let N go from 2 to 10000 in the for loop. We know that N=1 has just 1 divisor:
Let maxDivisors = 1 // number of divisors of 1
Let numWithMax = 1
for each integer N from 2 to 10000:
Count the number of divisors of N
If that number is greater than maxDivisors:
Let maxDivisors = the number of divisors of N
Let numWithMax = N
Output maxDivisors and numWithMax
Here's a curious thing: If you leave out the line "numWithMax = 1" from the program, the computer will report a syntax error where you try to output the value of numWithMax. It will say that the variable numWithMax might not have been initialized. That is, it might never have been assigned a value. Now, you know that it will be assigned a value (since when N=2 is processed, numWithMax will become 2). However, when the computer compiles the program, it doesn't know whether the body of the if statement will ever be executed, so it doesn't know whether numWithMax will ever be assigned a value. The syntax rule is that every variable must be "definitely initialized" before its value is used. This means that it is initialized on every possible execution path through the program.
We still have to expand the step "Count the number of divisors of N." This was already done in Section 3.4 for the example program CountDivisors.java. This step requires another for loop, so we have here an example of one for loop nested inside another. Here is a complete algorithm, which can be translated into a program:
Let maxDivisors = 1 // number of divisors of 1
Let numWithMax = 1
for each integer N from 2 to 10000:
Let divisorCount = 0
for each D from 1 to N:
if D is a divisor of N:
add 1 to divisorCount
If divisorCount is greater than maxDivisors:
Let maxDivisors = the number of divisors of N
Let numWithMax = N
Output maxDivisors and numWithMax
This can be translated pretty much directly into a program. By the way, the maximum number of divisors is 64. There are two numbers between 1 and 10000 that have 64 divisors, 7560 and 9240. The program will output the first of these. (It would output the second if the test "if (divisorCount > maxDivisors)" were changed to "if (divisorCount >= maxDivisors)". Do you see why?)
public class MostDivisors {
/* This program finds the integer between 1 and 10000 that has
the largest number of divisors. It prints out the maximum
number of divisors and an integer that has that many divisors.
*/
public static void main(String[] args) {
int N; // One of the integers whose divisors we have to count.
int maxDivisors; // Maximum number of divisors seen so far.
int numWithMax; // A value of N that had the given number of divisors.
maxDivisors = 1; // Start with the fact that 1 has 1 divisor.
numWithMax = 1;
/* Process all the remaining values of N from 2 to 10000, and
update the values of maxDivisors and numWithMax whenever we
find a value of N that has more divisors than the current value
of maxDivisors.
*/
for ( N = 2; N <= 10000; N++ ) {
int D; // A number to be tested to see if its a divisor of N.
int divisorCount; // Number of divisors of N.
divisorCount = 0;
for ( D = 1; D <= N; D++ ) { // Count the divisors of N.
if ( N % D == 0 )
divisorCount++;
}
if (divisorCount > maxDivisors) {
maxDivisors = divisorCount;
numWithMax = N;
}
}
System.out.println("Among integers between 1 and 10000,");
System.out.println("The maximum number of divisors is " + maxDivisors);
System.out.println("A number with " + maxDivisors + " divisors is " + numWithMax);
} // end main()
}
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