can u pls write a java program that compute the sum, difference, product, quotient of the 2 integer number?
961655
Yes, I can.  Quite simple actually.\n\nBut ask yourself if you know what these words mean:\n\n- integer\n- sum\n- difference\n- product  (think mathematics, not retail)\n- quotient
992739
You mean the sum of the first integers, starting with 2?\nWrite 2 in A1\nWrite 5 in A2\nWrite =2*A2-A1+1 in A3\nWrite =2*A3-A2+1 in A4\nWrite =2*A4-A3+1 in A5\nand so on. It can be continued by dragging.\n\nIf you only want the sum of 2 to N you can use\n= 0.5*N*N+1.5*N
253284
#include <stdio.h>\n\nint main(void) {\n\nint i, j, count; //C Sucks, wait till you take java\n\nfor(i=1; i<=1000; i++){&#92;&#92;goes through all numbers 1 to 1000\n    count=0;\n    for(j=2; j<i; j++)&#92;&#92;goes through all numbers 2 to i\n    {\n     if(i%j==0)&#92;&#92;checks to see if it i is evenly div. by i\n        count++;\n    }//nested for\n    if (count ==0)\n        printf("%d&#92;n", i);\n}//for\n\nreturn 0;\n}//main\n\n\n//Hope that helps!
536743
The sum of two numbers is 30\nx+y = 30\n\nThier difference is 6\nx-y = 6\n\nEliminate Y through addition (it's practically eliminated itself)\n\nx+y = 30   (+)\nx-y = 6\n_________\n2x = 36\n\nx = 18\n\nnow \nx + y = 30 right?\n18 +y = 30 (x = 18 so 18 replaces x)\n-18    -18\n___________\ny = 12
630943
1) Let n = the first integer, then\nn+1 = second integer\nThe product is n(n+1)\n2) Let n = the first even integer, then\nn+2 = second even integer and\nn+4 = third even integer.\nThe sum would be n+(n+2)+(n+4)
840708
Remove the trailing zeroes, do the multiplication, and add the removed zeroes to the product.\nFor example,\n530*47000\n=53*47 (add 4 zeroes later)\n=24910000\n=24,910,000
1082750
Find the GCD and then add/ subtract the numerator
642129
ok, here is one declaration, of many;\n\nfloat Doc(int arr[100], int NumCt)\n{\nfloat sum = 0.0;\nfor (int i = 0; i < NumCt; i++)\nsum += arr[i];\n\nif (!NumCt)\nreturn sum;\n\nreturn sum / NumCt;\n}
926880
(1+1)-2*3/2^2+2\n\nWithout useing Order of Operations: 2\n\nWith Order of Operations: 2.5
797941
My Dear Aunt Sally is how I learned it.\n\nMultiplication Division Addition Subtraction
619527
1st step: try for n = 1\n\nThen 1*2^1 = 2 and\n(1-1)*2^(1+1)+2 = 2\n\n2nd step: Assume true for n, show true for n+1.\n\nThen sum(i=1 to n) = (n-1)*2^(n+1)+2\nThen for n+1:\n\n(n-1)*2^(n+1)+2 + (n+1)2^(n+1)\n= (n-1 + n+1)2^(n+1)+2\n= 2n2^(n+1)+2\n= n2^(n+2)+2\n= ((n+1) - 1)*2^((n+1)+1) + 2\n\nThus it works.
55565
Your computer has a calculator built in.  Failing that, try long division.  You're gonna have to learn how to do it anyway.\n\nSheesh.
1013929
t's actually pretty easy. When you take a number, jumble around it's digits and then subract the new number from the old, you are always going to get the sum of the digits of the difference is divisible by 9 (ignoring any negative). \n\nExample: 12345-54321 =-41976 ----> 4+1+9+7+6= 27=9•3\n\n\nWhen you give all but one of the digits (and knowing that the last one is not 0), the computer just takes the sum of the digits you gave it, and subtracts it from a multiple of 9 (so that the remainder is between 1 and 9). \n\nLet's take 1234 and 4321 for example:\n\n1234-4321=-3087\n\nNow choose a non-zero number:\nif the number was 3 then the sum of the rest would be 15, and 18-15=3\nIf the number was 8, then the sum of the rest would be 10 and 18-10=8\nIf the number was 7, then the sum of the rest would be 11 and 18-11=7\n\n(now, the number you subract rom will not always be 18, but it will always be a multiple of 9)\n\nThe number that you picked will always be the remainder.\n\n\nWhy can't you pick 0?\n\nLets look at this example:\n\nChoosing 0, we have the sum or the remaineder is 3+8+7=18.  Therefore, the computer doesn't know if it should subtract 18 from 18, and the number you picked was 0, or subract 18 from 27 and the number you picked was 9.
730496
iteration1: a = 3, b = 2\niteration2: a = 13, b = 8\niteration3: a = 51, b = 30\nloop exit:  a = 81, b = 30 // condition a = a + b executed before exit
737081
If you want to know visit www.java.com\n\nHope ds is d best answer!
226471
get it sorted 1st, \nthen get the differences of adjacent number pairs, \nget the min of this differences
373216
any number N1 can be converted to N2 BY\naddtion:\n  N1 + N2 + (-N2 ) \ndivision:\n 1/[(N1/N2)/N1]\nMULTIPLICATION\n N1 * N2 * N1^-1\nSUBTRACTION\n[N1-(N1-N2]