Two Numbers and Show A Number Which is Greater Than The Largest Number And Equal To Sum Smallest Number

For Details of The Program: CLICK HERE

class Q1

{

int n,m, no, digit, i;

BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

Q1()

{

digit=0;

}

void accept( ) throws Exception

{

System.out.print("\nEnter 'M':");

m=Integer.parseInt(br.readLine());

System.out.print("\nEnter 'N':");

n=Integer.parseInt(br.readLine());

check();

}

void check( )

{

if(n >100 ||m<100 m="">10000)

System.out.println("Invalid Input.");

else

{

while(true)

{

no=0;

for(i=m;i >0;i=i/10)

no=no+i%10;

if(no==n)

{

System.out.println("\nThe required number:"+m);

System.out.println("\nTotal number of digits:"+digit(m));

break;

}

m++;

}

}

}

int digit(int x)

{

int c=0;

for(int j=x;j >0;j=j/10)

c++;

return c;

}

public static void main(String args[]) throws Exception

{

Q1 one=new Q1();

one.accept();

}

}

Variable description

Type	Variable Name				Purpose			Scope
int	n				To store the lower range			Entire program
Int		m				To store the upper			Entire program
						range
int		no				To store the			Entire program
						nearest required
						number
Int		digit				Count the digit			Entire program
int		i				Loop control			Entire program
						variable
int		x				To hold the value			digit function
						of m
Int		c				counter			digit function
BufferedReader			br			Input object			Entire program
int			j			Loop control			check function
						variable
Algorithm

 

Step 1: Create int variables n,m,digit,no,i

Step 2: Take lower range and upper range values and store in n and m

Step 3: Check whether n and m are within specified range

Step 4: Repeat steps 5 and 6 until the required number is reached

Step 5: Take the sum of the digits of ‘m’ and check it with ‘n’. If match found count the number of digits in it, display both and go to step 7 

Step 6: Increase ‘m’ by 1 

Step 7: End

2015 Computer Practical Paper: CLICK HERE