Question

Please It's a request don't give irRelevent AnswerS ​

Answer 1

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• Write a java program to check whether a number is a disarium number or not.

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• There are many approaches. I'm showing you one of them.
• Here is the source code.

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import java.util.*;

class x

{

public static void main(String s[ ])

{

Scanner sc = new Scanner(System.in);

System.out.print("Enter the number: ");

int n=sc.nextInt();

if(isDisarium(n))

System.out.println("Disarium Number.");

else

System.out.println("Not a Disarium Number.");

}

static boolean isDisarium(int num)

{

int x=num, s=0;

String s=Integer.toString(num);

int len=s.length();

while(x!=0)

{

int d=x%10;

s+=(int)Math.pow(d, len);

--len;

x/=10;

}

if(s==num)

return true;

else

return false;

}

}

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I hope you are satisfied with my answer.

Answer 2

Question:-

Write a Program to check whether a given Number is Disarium or Not.

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What is Disarium Number?

A number will be called Disarium if the sum of its digits powered with their respective position is equal with the number itself.

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Example:-

For example 175 is a Disarium number:-

${1}^{1} + {7}^{2} + {5}^{3} = 175$

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Logic applied in Coding:-

First we will count the Terms of Given number using Counter.Then using another loop we will do the sum of Term with their respective power(increasing by 1 till terms).

Then, we will check by using Condition wether the Number Matched or Not.

At last we will find the Output.

Remember:-

Before Doing any Calculation always try to make The copy of Real of Given number And do Calculation with that copy which will help full to You.

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CODE:-

class Disarium

{

public static void main(int n)

{

int m=n,sum,j=0;

for(;m>0;j++)

{

m=m/10;

}

for(int k=n;k>0;k=k/10,j=j-1)

{

sum=(int)sum+Math.pow(k%10,j);

}

if(sum==n)

System.out.println("Yes\n"+n+" is a Disarium Number");

else

System.out.println("No");

}

}

______________________________

Input:-

135

Output:-

Yes

135 is a Disarium Number