The first computer is the Eniac which was 50 tons!
Answer:
- public class Main {
-
- public static void main (String [] args) {
-
- for(int i = 2; i < 10000; i++){
- if(isPrime1(i)){
- System.out.print(i + " ");
- }
- }
-
- System.out.println();
-
- for(int i = 2; i < 10000; i++){
- if(isPrime2(i)){
- System.out.print(i + " ");
- }
- }
- }
-
- public static boolean isPrime1(int n){
-
- for(int i=2; i <= n/2; i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
-
- public static boolean isPrime2(int n){
-
- for(int i=2; i <= Math.sqrt(n); i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
- }
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Explanation:
Firstly, create the first version of method to identify a prime number, isPrime1. This version set the limit of the for loop as n/2. The for loop will iterate through the number from 2 till input n / 2 and check if n is divisible by current value of i. If so, return false to show this is not a prime number (Line 22 - 26). Otherwise it return true to indicate this is a prime number.
In the main program, we call the isPrime1 method by passing the i-index value as an argument within a for-loop that will iterate through the number 2 - 10000 (exclusive). If the method return true, print the current i value). (Line 5 - 9)
The most direct way to ensure all the prime numbers below 10000 are found, is to check the prime status from number 2 - 9999 which is amount to 9998 of numbers.
Next we create a second version of method to check prime, isPrime2 (Line 31 - 40). This version differs from the first version by only changing the for loop condition to i <= square root of n (Line 33). In the main program, we create another for loop and repeatedly call the second version of method (Line 13 - 17). We also get the same output as in the previous version.
Answer:
C = 3
Explanation:
The maximum number of physical memory accesses that may be caused by this single instruction is ; 3
This is because In the case of loading two values into R2 and R3 there is need for their memories to be accessed and after execution R1 value there will be a write back into memory hence another memory is also accessed and that makes the number of Physical memories accessed to be three(3) in Total
Search engines enable you to locate web pages related to a specific subject.
The answer is Field research
