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Happy number is similar to magic number, but instead of obtaining sum of the digits we obtain sum of square of the digits. EXAMPLE-1 19 =1²+9²=82 82=8²+2²=68 68=6²+8²=100 100=1²+0²+0²=1 EXAMPLE-2 7=7²=49 49=4²+9²=97 97=9²+7²=130 130=1²+3²+0²=10 10=1²+0²=1 EXAMPLE-3 13=1²+3²=10 10=1²+0²=1 A java program on happy number can be made by extracting digits of the number then adding the square of the digits. If the sum result to 1 then it is a happy number and if it not results to 1 then if the sum is greater than 10 then it digits are re-extracted and then again added it's square of digits. JAVA CODE- import java.util.*; public class happy { int n; happy() {     n=0; } void getNum(int nx) {   n=nx;   } int sum_sq_digits(int d) {     int sum=0;     while(d>0)     {         sum=sum+((d%10)*(d%10));         d=d/10;     }     int s;int c=0;     while(sum>9)     {         s=sum;         while(s>0)         {             c=c+(s%10);             s=s/10;         }         sum=c;     }  

Emirp is simply the reverse prononciation of word "prime". An emirp number is the number which is prime and whose reverse is also a prime number. EXAMPLE-1 17  Reverse of the number = 71 Both the numbers are prime so 17 is an emirp number. EXAMPLE-2 37 Reverse of the number = 73 Both the numbers are prime so 37 is an emirp number. EXAMPLE-3 13 Reverse of the number = 31 Both the numbers are prime so 13 is an emirp number. Emrip number can be easily made in java with the help of methods/functions. Firstly, take an input of number, then check it is prime or not, then reverse the number and again check it is prime or not, if both are true, print the suitable output. ALGORITHM- STEP 1 : start STEP 2 : print "enter a number" STEP 3 : take input of number in e STEP 4 : int n=num, int rev =0 STEP 5 : repeat STEP to STEP until n>0 STEP 6 : int d=n%10 STEP 7 : rev=(rev*10)+d STEP 8 : n=n/10 STEP 9 :  int p= input value from STEP 13 STEP 10 : int d=0, x=num STEP 11 : int i

A number is said to be disarium if the sum of its digit powered with their positions is equal to the number itself. EXAMPLE- 135 = (1^1) + (3^2) + (5^3)                             = 1 + 9 + 125                             = 135 A simple java program to check disarium number can be made by firstly calculating the length of the number and then adding digits of the number raised to the power of its position in the number. ALGORITHM - STEP 1 : Start STEP 2 : print "enter a number" STEP 3 : taking input of integer in n STEP 4 : int l= Integer.toString().length() STEP 5 : int no=n STEP 6 : int s=0 STEP 7 : repeat STEP 8 to STEP 11 until (no!=0) STEP 8 : int d=no%10 STEP 9 : s=s+(int)(Math.pow(d,l) STEP 10 :no=no/10 STEP 11 : l-- STEP 12 : if  (s==n) then goto STEP 13 else goto STEP 14 STEP 13 : print "the number is disarium" STEP 14 : print "the number is not disarium" STEP 15 : end JAVA CODE- import java.util.*; public class disarium {     public static void m

Hi everyone,  In  this article I am going to tell you how to complete the whole syllabus for exam in one day. I will tell you a full one day plan to complete all your syllabus without having any doubt and also prepare for that subject. To follow this, firstly you have to divide the syllabus of your subjects into 3 parts-  The first part is for easy chapters second part is for moderate chapters and third parties for difficult chapters in easy chapters you have to put it at topics for which you can answer all the question.  In moderate you have to put that chapters in which you have very less doubt. In the difficult you have to put that topic whose questions you are not able to answer. HERE IS THE TEMPLATE FOR YOUR PLANNING- PLAN starting in the morning  Prepare a chart of the whole day time table. Begin your day at 8:00 a.m.  In the first three hours revise or learn all the chapters which are easy for you remember after every 50 minutes you have to take 10 minute break .once all the cha

A number is said to be magic number if the repetitive sum of digits of number is equal to 1 or a single digit number . If the repetitive sum ends up  as 1 then the number is a magic number, else not. For example- n= 289;.   2+8+9 = 19      19;.      1+9 = 10      10;.      1+0= 0       1 Since the number ends in 1, hence 289 is a magic number. Magic number in java is the java program to check whether a number is magic number or not . For this we do repetitive addition of the digits of the number till the number is a two or more digit number. If after repetitive adding digits of the number ends up to 1 then the loop stops and a message is displayed. For this program we need to extract digits of the number for this we use while loop and add this digit . Code snippet to get digits of the number - while(n!=0) {        int d = n%10;         s=s+d;        n=n/10; } ALGORITHM - STEP 1: start STEP 2: print "enter a number" STEP 3: taking input of integer in n STEP 4: rep

A number is said to be palindrome if it reads both forward and backwards same. Example- 175571 454 A program for palindrome number can be made by simply extracting digits of the number in reverse order and adding them with their place value and then checking is it same as the original number or not. To extract a digit in revers order and adding with place value we simply follow the below code. while(n>0) { int d=n%10; rev=(rev*10)+d; n=n/10; } To explain this let us take an example 657 Then  For first iteration =  n=657                                     d=657%10=7                                     rev=(0*10)+7=7 For second iteration = n=65                                         d=65%10=5                                         rev=(7*10)+5=75 For third iteration = n=6                                      d=6%10=6                                      rev=(75*10)+6=756 Hence, we get the reverse of the number. ALGORITHM - STEP 1: Start STEP 2: print "enter a n

A number is said to be niven or harshad if the number is divisible by the sum of its digits. EXAMPLE-135 The sum of digits are: 1 + 3 + 5                                                                  135/9=15 A simple java program to check whether a number is harshad / niven or not can be made with the help of while loop. For this, firstly we extract the digits of the number and then we sum up all the digits.After all the digits are summed then we check whether rhe number when divided by the sum of its digits leaves any remainder or not. If the remainder is zero then number is niven / harshad number else not. ALGORITHM - STEP 1: Start STEP 2 : print "enter a number" STEP 3 : taking input of integer in n STEP 4 : int s = 0 STEP 5 : int no = n STEP 6 : repeat STEP 7 to STEP 9 until no!=0 STEP 7 : int d=no%10 STEP 8 : s=s+d STEP 9 : no=no/10 STEP 10: if (n%s==0) then goto STEP 11 else goto STEP 12 STEP 11 : print n+"is a harshad number" STEP 12 : print n+"is n