In this time, we see the programs like armstrong number,binary to decimal,decimal to binary,fibonacci series,greatest common divisor, least common divisor,neon number,perfect number,spy number, sum of digits.
ARMSTRONG NUMBEER
It is the number with sum of each digits multiplied itself three times and its equal to the given number.
Ex:
153
3–> 3*3*3=27
5–>5*5*5=125
1–>1*1*1=1
total=27+125+1=153
package mars1;
public class Armstrong {
public static void main(String[] args) {
// TODO Auto-generated method stub
int no=153,rem=0,sum=0,armstrong=0,no2=no;
while(no>0)
{
rem=no%10;
armstrong=armstrong+(rem*rem*rem);
no=no/10;
}
if(no2==armstrong)
{
System.out.println(“ARM”);
}
}
}
RESULT
ARM
BINARY TO DECIMAL
It converts given binary number to decimal.
Ex:1001
1–>1*2^0=1
0–>1*2^1=0
0–>1*2^2=0
1–>1*2^3=8
total=1+0+0+8=9
package mars1;
public class BinaryToDecimal {
public static void main(String[] args) {
// TODO Auto-generated method stub
int no=1001,power=0,dec=0, rem=0;
while(no>0) {
rem=no%10;
dec=(int)(dec+(rem*Math.pow(2,power)));
no=no/10;
power++;
}
System.out.println(dec);
}
}
RESULT
9
DECIMAL TO BINARY
It converts decimal number into binary.
Ex:no=4
rem=no%2,no=no/2;
rem=4%2=0,no=4/2=2;
rem=2%2=0,no=2/2=1;
rem=1%2=1
package mars1;
public class DecimalToBinary {
public static void main(String[] args) {
// TODO Auto-generated method stub
int no=4;
String rem=””;
while(no>0)
{
rem=no%2+rem;
no=no/2;
}
System.out.println(rem+””);
}
}
RESULT
100
FIBONACCI SERIES
Fibonacci series moving their series from left to right by adding the first and second digit to form the third digits and neglect the first digit. The new first digit is the second one and second digit is the third one.
Ex:011
Here below F is first,S is second,T is third
F—S—T
0—1—1
1—2—3
2—3—5…,
package mars1;
public class FibonacciProblems {
public static void main(String[] args) {
// TODO Auto-generated method stub
int first=0,second=1,count=0,third=1;
while(true)
{
first=second;
second=third;
third=first+second;
count++;
if(third==233) {
System.out.println(“I got”);
break;
}
else if(third>223){
System.out.println(“not get”);
}
}
}
}
RESULT
I got
FIBONACCI SERIES USING FOR LOOP
Same concept used in above forloop used instead of while.
package mars1;
public class FibonacciSeries {
public static void main(String[] args) {
// TODO Auto-generated method stub
int first=0,second=1;
for(int i=2;i<=10;i++)
{
int third=first+second;
first=second;
second=third;
System.out.println(third);
}
}
}
RESULT
1
2
3
5
8
13
21
34
55
FIBONACCI SERIES WITHOUT USING THIRD VARIABLE
In this program we eliminate the third variable and use
first=second
second=first+second
package mars1;
public class FibonacciSerieswithoutThirdVariable {
public static void main(String[] args) {
// TODO Auto-generated method stub
int first=0,second=1;
for(int i=2;i<=10;i++)
{
first=second;
second=first+second;
System.out.println(second);
}
}
}
RESULT
2
4
8
16
32
64
128
256
512
GREATEST COMMON DIVISOR
In greatest common divisor
30–>15,10,6 ,5,3,2
18–>18,9,6 ,3,2
both have greatest common is 6.
package mars1;
public class GreatesCommonDivisor {
public static void main(String[] args) {
// TODO Auto-generated method stub
int no1 = 30, no2 = 18;
int small = no1 < no2 ? no1 : no2;
int big = no1 > no2 ? no1 : no2;
while (small >= 2) {
if ((no1 % small == 0) && (no2 % small == 0)) {
System.out.println(“GCD IS ” + small);
break;
}
small–;
}
}
}
RESULT
GCD IS 6
LEAST COMMON MULTIPLE
example:
multiples of 2 and 3.
2–>2,4,6,8,12,14,16,18,20
3–>3,6,9,12,15,18,21
both have least common is 6.
package mars1;
public class LeastCommonMultiple {
public static void main(String[] args) {
// TODO Auto-generated method stub
int no1 = 8, no2 = 3;
int small = no1 < no2 ? no1 : no2;
int big = no1 > no2 ? no1 : no2;
int bigcount=big;
while (true) {
if (big % small == 0) {
System.out.println(“LCM IS ” + big);
break;
}
big=big+bigcount;
}
}
}
RESULT
LCM IS 24
NEON NUMBER
The example of neon number given below.
9*9=81
8+1=9
the product of given number twice produce a number have its sum of digits equal to given number.
package mars1;
public class NeonNumber {
public static void main(String[] args) {
// TODO Auto-generated method stub
int no=9,neon=0,rem=0;
int no2=no*no;
while(no2>0) {
rem=no2%10;
neon=neon+rem;
no2=no2/10;
}
if(neon==no) {
System.out.println(“neon”);
}
else {
System.out.println(“not neon”);
}
}
}
RESULT
neon
PERFECT NUMBER
It is number of perfect divisor equal to the given number.
Ex:6
1,2,3,4,5,6 only 1,2,3 exactly divide number 6 so add the number of exact divisors 1+2+3=6.
package mars1;
public class PerfectNumber {
public static void main(String[] args) {
// TODO Auto-generated method stub
int sum=0;
int no=8;
for(int i=1;i<no;i++)
{
if (no%i==0) {
sum=sum+i;
}
}
if(sum==no) {
System.out.println(“PERFECT NUMBER”);
}
else {
System.out.println(“NOT PERFECT NUMBER”);
}
}
}
RESULT
NOT PERFECT NUMBER
SPY NUMBER
spy number means the sum of digits and product of digits of a give number to be equal.
package mars1;
public class SpyNo {
public static void main(String[] args) {
// TODO Auto-generated method stub
int no=1234,sum=0,prod=1;
while(no>0)
{
int rem=no%10;
sum=sum+rem;
prod=prod*rem;
no=no/10;
}
System.out.println(“SUM ”+sum);
System.out.println(“PROD “+prod);
if(sum==prod) {
System.out.println(“spy number”);
}
else {
System.out.println(“not spy number”);
}
}
}
RESULT
SUM 10
PROD 24
not spy number
SUM OF DIGITS
In this program we need sum of digits in a single digit number so we use do while condition and the range is below 9 will give single digit. so, we choose it.
package mars1;
public class SumOfDigits {
public static void main(String[] args) {
// TODO Auto-generated method stub
int no=123,sum=0;
do {
while(no>0)
{
int rem=no%10;
sum=sum+rem;
no=no/10;
}
no=sum;
}while(sum>9);
System.out.println(sum);
}
}
RESULT
6
selvakrish 