Using an array to find the multiples of two ints

Question

Im a programming student and im having alot of trouble with this question:

"complete a static method multiplesOf which takes two int parameters, number and count. The method body must return an int array containing the first count multiples of number. For example,

multiplesOf(5, 4) should return the array { 5, 10, 15, 20 } multiplesOf(11, 3) should return the array { 11, 22, 33 } multiplesOf(1, 15) should return the array { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 }

You must not use System.out.print or System.out.println in your method."

This is my current code:

public static void main(String[] args) {
    multiplesOf(5,4);
}

public static int[] multiplesOf (int number, int count) {
    int[] a = new int[count];
    a[0]= number;
    for(int i = 0; i<count; i++) {
        a[i] = (a[i]*count);
    }
    return a;
}

Ive been trying to figure out why the array "a" still only has the values 0,1,2,3

What does a[i] = (a[i]*count); do?

ajb
– ajb

2017-01-13 06:21:22 +00:00
Commented Jan 13, 2017 at 6:21 — ajb
– ajb, Commented Jan 13, 2017 at 6:21

thegauravmahawar · Accepted Answer · 2017-01-13 06:29:47Z

2

Try:

public static int[] multiplesOf(int number, int count) {
  int[] a = new int[count];
  for(int i = 1; i <= count; i++) {
    a[i - 1] = number * i;
  }
  return a;
}

answered Jan 13, 2017 at 6:29

thegauravmahawar

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Comments

wake-0 · Accepted Answer · 2017-01-13 06:41:29Z

1

a[0] = number;
for(int i = 1; i<count; i++)
{
      a[i] = (i+1)*number;
}

answered Jan 13, 2017 at 6:41

wake-0

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2 Comments

wake-0 Over a year ago

ah yeah you were faster :D but I think when I see all the wrong answers I can let the answer stay here :) I will upvote you

Scary Wombat Over a year ago

Pretty shocking really

Scary Wombat · Accepted Answer · 2017-01-13 06:43:14Z

1

try this

        public static int[] multiplesOf (int number, int count)
        {
            int[] a = new int[count];
            a[0] = number;
            for(int i = 1; i<count; i++)
            {
              a[i] = number * (i + 1);
            }
            return a;
          }

output

[5, 10, 15, 20]

edited Jan 13, 2017 at 6:43

answered Jan 13, 2017 at 6:36

Scary Wombat

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Comments

Pratik Ranjane · Accepted Answer · 2017-01-13 06:53:10Z

0

    public static int[] multiplesOf(int number, int count) {
    int[] a = new int[count];
    for (int i = 0; i < count; i++) {
        a[i] = (number * (i+1));
    }
    return a;
}

use this

edited Jan 13, 2017 at 6:53

answered Jan 13, 2017 at 6:27

Pratik Ranjane

161 bronze badge

5 Comments

CadaverousX Over a year ago

As OP says "You must not use System.out.print or System.out.println in your method."

msagala25 Over a year ago

must not use System.out.print or System.out.println in your method.

cst1992 Over a year ago

SOP isn't required; the array is supposed to just be returned, not printed.

Pratik Ranjane Over a year ago

i was just checking by printing..now changes are made.

Scary Wombat Over a year ago

Post you output then

Gopinath Langote · Accepted Answer · 2017-01-13 08:38:05Z

0

 public static int[] multiplesOf (int number, int count) {
     int[] a = new int[count];
     for(int i = 1; i<=count; i++) {
         //a[i-1] beacuse we started iterating array at i=1
         a[i-1] = (i*number); 
     }
     return a;
}

multiplesOf(5, 4) returns -> the array { 5, 10, 15, 20 }

answered Jan 13, 2017 at 8:38

Gopinath Langote

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