## Solution for Programming Exercise 3.6

This page contains a sample solution to one of the exercises from Introduction to Programming Using Java.

### Exercise 3.6:

Exercise 3.2 asked you to find the
number in the range 1 to 10000 that has the largest number of divisors. You
only had to print out one such number. Revise the program so that it will
print out **all** numbers that have the maximum number of divisors. Use an
array as follows: As you count the divisors for each number, store each
count in an array. Then at the end of the program,
you can go through the array and print out all the numbers
that have the maximum count. The output from the program should look
something like this:

Among integers between 1 and 10000, The maximum number of divisors was 64 Numbers with that many divisors include: 7560 9240

**Discussion**

This is a fairly straightforward exercise in using arrays. We need to save 10000 numbers in an array. The numbers are the divisor counts for the numbers 1 through 10000. The numbers that we want to store in the array are of type int, so int is the base type of the array. In the program, I use an array named saveCount of type int[] to store the data. It seems natural to store the divisor count for N in array element saveCount[N], but note that in that case, we don't use array element number 0, and we need an array element number 10000. Ignoring element 0 is fine, but for saveCount[10000] to exist, the length of the array must be at least 10001. I use an array of length 10001:

saveCount = new int[10001];

(An alternative would be to store the advisor count for N in saveCount[N-1]. In that case, we wouldn't need the extra array element, and the length of the array would be 10000.)

After computing each divisor count, we store the count in the array. At the same time, we are keeping track of the maximum number of divisors. After the end of the for loop that does the counting, we know the divisor count for each number and we know the maximum number of divisors. We just have to go through the array and print out every integer N for which the divisor count is equal to the maximum:

System.out.println("Numbers with that many divisors include:"); for ( N = 1; N <= 10000; N++ ) { if ( saveCount[N] == maxDivisors ) { System.out.println( " " + N ); } }

Note that this code checks whether the number in the array, saveCount[N], is equal to the maximum, but it prints the array index, N, which is the integer that has that many divisors.

**The Solution**

New code that involves the array is shown in red.

import textio.TextIO; /** * This program counts the number of divisors for integers in the range * 1 to 10000. It finds the maximum divisor count. It outputs the * maximum divisor count and a list of all integers in the range that * have the maximum number of divisors. */ public class MostDivisorsWithArray { 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[] saveCount; // Store the number of divisors for each number saveCount = new int[10001]; maxDivisors = 1; // Start with the fact that 1 has 1 divisor. saveCount[1] = 1; /* Process all the remaining values of N from 2 to 10000, and store all the divisor counts in the array. Update the value of maxDivisor whenever we find a value of N that has more divisors than the current value. */ for ( N = 2; N <= 10000; N++ ) { int D; // A number to be tested to see if it's 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++; } saveCount[N] = divisorCount; // Record the count for N in the array if (divisorCount > maxDivisors) { maxDivisors = divisorCount; } } System.out.println("Among integers between 1 and 10000,"); System.out.println("The maximum number of divisors was " + maxDivisors); System.out.println("Numbers with that many divisors include:"); for ( N = 1; N <= 10000; N++ ) { if ( saveCount[N] == maxDivisors ) { System.out.println( " " + N ); } } } // end main() }