Modify the CountByFives application so that the user enters the value to count by. Start each new line after 10 values have been displayed. public class CountByAnything
{
// Modify the code below
public static void main (String args[])
{
final int START = 5;
final int STOP = 500;
final int NUMBER_PER_LINE = 50;
for(int i = START; i <= STOP; i += START)
{
System. out. print(i + " ");
if(i % NUMBER_PER_LINE == 0)
System. out. println();
}
}
}

The modified program is as follows:

import java.util.*;

public class CountByAnything{  

public static void main (String args[]){

   Scanner input = new Scanner(System.in);

final int START = 5;

final int STOP = 500;

int countBy; int count = 0;

System.out.print("Count By: ");

countBy = input.nextInt();

final int NUMBER_PER_LINE = 10;

for(int i = START; i <= STOP; i += countBy){

System.out.print(i + " ");

count++;

if(count == NUMBER_PER_LINE){

System.out.println();

count = 0;} } } }

Explanation:

To solve this, we introduce two variables

(1) countBy --> This gets the difference between each value (instead of constant 5, as it is in the program)

(2) count --> This counts the numbers displayed on each line

The explanation is as follows:

final int START = 5;

final int STOP = 500;

This declares countBy and count. count is also initialized to 0

int countBy; int count = 0;

This prompts the user for countBy

System.out.print("Count By: ");

This gets value for countBy

countBy = input.nextInt();

final int NUMBER_PER_LINE = 10;

This iterates through START to STOP with an increment of countBy in between two consecutive values

for(int i = START; i <= STOP; i += countBy){

This prints each number

System.out.print(i + " ");

This counts the numbers on each line

count++;

If the count is 10

if(count == NUMBER_PER_LINE){

This prints a new line

System.out.println();

And then set count to 0

count = 0;}

answer: internal fragmentation is the difference between the (fixed) size of a memory hole and the data within in that memory. for example, if a 1,024 byte block of memory holds 800 bytes of data, then the 224 bytes difference is considered lost to internal fragmentation.

external fragmentation occurs when the sum of memory fragments is large enough to satisfy a request, but the holes are non-contiguous and the request cannot be satisfied. this can occur when memory is dynamically allocated and freed.

explanation:

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