dividesSelf challange

Hi,

I am working on below challeng



I have not undersood the description.



We'll say that a positive int divides itself if every digit in the number divides into the number evenly. So for example 128 divides itself since 1, 2, and 8 all divide into 128 evenly. We'll say that 0 does not divide into anything evenly, so no number with a 0 digit divides itself. Note: use % to get the rightmost digit, and / to discard the rightmost digit.

dividesSelf(128) → true
dividesSelf(12) → true
dividesSelf(120) → false

why if 0 then doe not devide and false?

please advise
LVL 7
gudii9Asked:
Who is Participating?
 
DonnaConnect With a Mentor Commented:
128 / 1 = 128 remainder 0 (remainder zero, means "evenly"
128 / 2 = 64 r-0 (divides evenly)
128 / 8 = 16 remainder 0, so 128 divides by 1, 2, and 8 evenly, with  "no remainder" ...so for the value 128 the function needs to evaluate to TRUE.

65 evaluates to false because 65/6 = 10.833333 , which is not "evenly"

and zero is ALWAYS false (any value with a zero anywhere) because anytime you divide by zero, it's an irrational number.

does that help at all?

so the criteria of the problem is this:
#1 determine the number of digits
#2 you need to divide the number by each of it's digits and if the remainder is zero, that part evaluates to TRUE. (and you can use the "modulus" operator...In Java, the % means modulus...and all it does is determine the remainder. So for example, 65%6=10.833333 , and should return FALSE. 65%5 = 0 and should return TRUE.  Modulus says "divide the 1st number by the second number and return the remainder of the operation that "does not" go evenly. So given any number x and y,
if x % y = 0{
   return TRUE;
}
else
   return FALSE;   //BUT you have to do that for each digit in the number.
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CPColinConnect With a Mentor Senior Java ArchitectCommented:
The challenge is declaring that a zero digit results in false, so you don't really need to know why. I'm betting the reason is so you don't have to worry about divide-by-zero errors, though.
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d-glitchCommented:
Do you have a calculator?
Try dividing 120 by 1, 2, and then 0.
When that doesn't work, try dividing by some number close to zero, like 0.001
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gudii9Author Commented:
128 divides itself since 1, 2, and 8 all divide into 128 evenly.

what it means by 1, 2 and 8 devides evenly for 128?
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gudii9Author Commented:
128
are they saying 128 consist of 3 digits which are 1 then 2 then 3??
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d-glitchConnect With a Mentor Commented:
128 consist of 3 digits which are 1 then 2 then  8.

You can use % to determine if a division is even.

You can't hope to solve this until you understand the has One challenge.
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d-glitchCommented:
what it means by 1, 2 and 8 devides evenly for 128?

Divide evenly means that the remainder is zero.
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gudii9Author Commented:
Divide evenly means that the remainder is zero with  1 separately


Divide evenly means that the remainder is zero with 2 separately

Divide evenly means that the remainder is zero with 8 separately

or
Divide evenly means that the remainder is zero with 1, 2, 3 combinedly i.e 1*2*8
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d-glitchCommented:
Separately.
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gudii9Author Commented:
public boolean dividesSelf(int n) {
  
  for (; n>0;) {
			if (n % n == 0)
				return true;
			n = n / 10;

			
		}
		return false;
}

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i am failing some tests
Expected      Run            
dividesSelf(128) → true      true      OK      
dividesSelf(12) → true      true      OK      
dividesSelf(120) → false      true      X      
dividesSelf(122) → true      true      OK      
dividesSelf(13) → false      true      X      
dividesSelf(32) → false      true      X      
dividesSelf(22) → true      true      OK      
dividesSelf(42) → false      true      X      
dividesSelf(212) → true      true      OK      
dividesSelf(213) → false      true      X      
dividesSelf(162) → true      true      OK      
other tests
X      
Correct for more than half the tests
please advise
0
 
gudii9Author Commented:
Here is the URL of the challenge

http://codingbat.com/prob/p165941
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gudii9Author Commented:
pseudo code is:
1. loop the given number.
2. if the reminder by dividing by 10 is 0 then return false
3. else if divide by 10 to go to each digit and if all those digits evenly divide by giving reminder of 0 then return true
4, return false
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gudii9Author Commented:
package com.solution;

public class DevideItself {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		System.out.println("valiue is-->"+dividesSelf(213));

	}

	public static boolean dividesSelf(int n) {
		int k = 1;
		boolean temp=false;
		while (n > 0) {
			if (n % 10 == 0)
				return false;
			else {
				k = (n % 10);

				if (n % k != 0)
					return false;
				else if(n%k==0&&!temp){
				  temp=true;
					//return true;
				}
				//n = n / 10;
				//return false;
				// k=n/10;
			}
		}
		return false;
	}

}

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tried something like above going to infinite loop
public boolean dividesSelf(int n) {
		int k = 1;
		boolean temp=false;
		while (n > 0) {
			if (n % 10 == 0)
				return false;
			else {
				k = (n % 10);
				if (n % k != 0)
					return false;
			//	else if(n%k==0&&temp){
			else if(n%k==0){
				  //temp=true;
					return true;
				}
				//n = n / 10;
				//return false;
				// k=n/10;
			}
		}
		return false;
	}

Open in new window


above code fails 3 tests
Expected      Run            
dividesSelf(128) → true      true      OK      
dividesSelf(12) → true      true      OK      
dividesSelf(120) → false      false      OK      
dividesSelf(122) → true      true      OK      
dividesSelf(13) → false      false      OK      
dividesSelf(32) → false      true      X      
dividesSelf(22) → true      true      OK      
dividesSelf(42) → false      true      X      
dividesSelf(212) → true      true      OK      
dividesSelf(213) → false      true      X      
dividesSelf(162) → true      true      OK      
other tests
OK      

please advise
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gudii9Author Commented:
public boolean dividesSelf(int n) {
		int k = n;
		if (n % 10 == 0)
			return false;
		// boolean temp=false;
		while (n > 0) {
			if (k % (n % 10) != 0) {
				return false;
			}
			n = n / 10;
		}

		return true;
	}

Open in new window


above passes all tests. Any improvements/alternate approaches?
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gudii9Author Commented:
something like recursion or string approach?
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