write an algorithm for read 25 number, find the odd number s and print them.?
917997
Simply loop through the list of numbers and compute:\n\n'number' modulo '2'\n\nIn C, that would be:\n\nnumber % 2\n\nA modulus of 2 would return a '1' for odd numbers, and '0' for even numbers (since even numbers divide by 2 with no remainder, and odd numbers divide by 2 with a remainder of 1).\n\nExample:\n\nint index;\nint numArray[25];\n\n...(store the 25 numbers in the array)...\n\nfor(index = 0; index < 25; index++) {\n     if(numArray[i] % 2) {\n          printf("%d&#92;n", numArray[i]);\n     }\n}
94104
u can do it by trial and error. ok the possible digits for ones are just 1,3,5,7,9 since the statement says it is an odd number. ok then try to add 2 and equate it to 12. if it is = that's is the answer. 7+5= 12. so and 7 is greater than 5 by 2. the answer is 75 degrees.
240615
I am giving the solution in C++ using recursion..\n\n!!\n\n#include<iostream.h>\nvoid printstars(int n)\n{\nfor(int k=0;k<n;k++)\n     cout<<"*";\ncout<<endl;\n}\nvoid print1(int n)\n{\nif(n<=1)\n   printstars(n);\nelse\n   printstars(n-2);\n}\nvoid print2(int m,int n)\n{\nif(m==n)\n   cout<<endl;\nelse \n{\n   printstars(m);\n   print2(m+2,n);\n}\n}\n\nmain()\n{\nint n,m=1;\ncout<<"Enter an odd number n";\ncin>>n;\nprint2(m,n);\nprint1(n);\n}\n\nThis produces the output u needed and also uses recursion.
906203
Assuming you cannot reuse a digit unless there is more than one copy in the list...\n\nYou could do it this way...\n5 one-digit numbers: 2, 5, 6, 7, 9\n22 two-digit numbers: 22, 25, 26, 27, 29, 52, 56, 57, 59, 62, 65, 66, 67, 69, 72, 75, 76, 79, 92, 95, 96, 97\netc.\n\nBut it would take a while. The answer is: at least a few.
936111
2 different ways:\n\nint n = 5;\nchar buffer[10]; /* long enough to hold your number + null */\nsprintf(buffer, "%i", n);\n\n\nor\n\nitoa(number,string,radix) \n\nradix - 10 for decimal, 16 for hex...
326729
nested loops and if..then jump\n\nfor I = 1 to 42\nfor J = I+1 to 43\nfor K = J+1 to 45\nfor L = K+1 to 47\nfor M = L+1 to 48\nfor N = M+1 to 49\nif K-I=2 or L-J=2 or M-K=2 or N-L = 2 then [skip next line]\nprint I,J,K,L,M,N\nnext\nnext\nnext\nnext\nnext\nnext\n\nSomething like that;\nTranslate into your language's syntax
1051900
1 2 3 4 5 6 10 12 15 20 30 60\n-1 -2 -3 -4 -5 -6 -10 -12 -15 -20 -30 -60
274854
The difficult thing here is reading and writing the numbers. They must be transformed from ASCII and then the result transformed back to ASCII.I found a program that calculates the sum of 2 numbers.I hope it helps.
411292
static_cast<int>("A") = 65\n\ngives you the ASCII value of the character\n\nWhole program:==============================================\n\nchar array[25]={'A','B',.....'Z'}\n\ncout << "Enter number: ";\ncin >> number\n\nif(number <= 10){\n  cout << number;\n}\nelse\n cout << array[number- 10];\n\nThat is the basic program you need to write it will work just dress it up to look pretty
49819
Hold on....10....20......25.....29..30....25..29\n\n25
45582
10,7,8,4,9,8,25=4,7,8,8,9,10,25 if you put them in order from least to greatest, eliminate the numbers like ths: 7,8,8,9,10       \n            8,8,9\n              8\n          so its 8
496127
For the number to be odd, the second digit must be 1, 3, or 5. Therefore, there are three choices for the second digit. The first digit can be any of the six digits, but it cannot be the digit used for the second digit. Therefore, there are 6 - 1 = 5 choices for the first digit.\n\nSo, 5 * 3 = 15.\n\nTo verify, the numbers are: E = 13, 15, 21, 23, 25, 31, 35, 41, 43, 45, 51, 53, 61, 63, 65. So, N(E) = 15 and the solution is correct.
961194
you can use /, % operator to extract digits. \nyou also have to make two void functions. which will translate numbers into digits. \n\nvoid data(int num)\n{\n  switch(num)             \n{\ncase 0: \n         break;\ncase 1: cout<<"One ";\n         break;\ncase 2: cout<<"Two ";\n         break;\ncase 3: cout<<"Three ";\n         break;\ncase 4: cout<<"Four ";\n         break;\ncase 5: cout<<"Five ";\n         break;\n}\nand just keep going till twenty.\nand then another function which convert numbers from 20-100 into words. \n\nvoid getdata(int num)\n{\n  if(num>=90)                     \n                                  \n{\n cout<<"Ninety ";\n less20(num-90);\n}\nelse if(num>=80)\n{\n cout<<"Eighty ";\n less20(num-80);\n}\nyou have to complete it.\nexample to extract the digits for first and tenth position.\ndigit=number%10;                     \nlast_two=digit;\nnumber=number/10;\ndigit=number%10;\nlast_two=(digit*10)+last_two;\n\nnow just figure out the positions. \nand you are good.
233108
For your question:\n"design the logic for a module that would print every number from 1 through 10 along with its square and cube .design the pseudocode"\n\nHere's my pseudocode:\n\nfor i=1 to 10 do\n  print i\n  print i * i\n  print i * i * i\nend of for loop\n\nFor the following question:\n"A mail order company sends multiple packages per order ,for each customer order, print mailing labels to use on each of the separate boxes that will be mailed .the mailing labels contain the customers complete name and address along with a box number in the form box 9 of 9. for example an order that requires three boxes produces three labels box 1 of 3 box 2 of 3 and box 3 of 3.\nThe file description is as follows SHIPPING FILE DESCRIPTION file name ; ORDERS field description  positions data type decimals\n\nORDERS file:\nFIELD DESCRIPTION: TITLE, FIRST NAME, LAST NAME, STREET ,CITY, STATE, BOXES ,BALANCE DUE"\n\nMy pseudocode is as follows:\nOpen ORDERS file\nRead one record in ORDERS file\nWhile there are still records in the ORDERS file Loop through the following\n  For i=1 to record.BOXES\n    Println record.TITLE, record.FIRSTNAME, record.LASTNAME\n    Println record.STREET\n    Println record.CITY\n    Println record.STATE\n    Println 'Box ', i, ' of ', record.BOXES\n  End of For loop\n  Read another record in ORDERS file\nEnd of while loop\n\n\nFor the following question:\n"A secoundhand store is haveing a seven day sale during which the price of any unsold item drops 10% each day the inventory file includes an item number,description, and original price on day one. for example an item costs $10.00 on the first day costs 10% less or $9.00 0n the secound day on the third day the same item is 10% less than $9.00 0r $ 8.10 produce a report that showes the price of the item on each day one through seven. write the pseudocode for this program"\n\nMy pseudocode is as follows:\nOpen INVENTORY file\nRead one record in INVENTORY file\nWhile there are still records in the INVENTORY file Loop through the following\n  price = record.ORIGINALPRICE\n  For i=1 to 7\n    Println 'Price for day', i\n    Println record.ITEMNUMBER, record.DESCRIPTION, price\n    price = price * 0.9\n  End of For loop\n  Read another record in INVENTORY file\nEnd of while loop
352166
8\ntake away o from top and you're left with o on bottom - half - nothing\nor i guess 10 could work, too (or any double digit ending in 0) take away the first number and you're left with 0
52919
in many cases, you need innovative thinking.. Consider this:\n\n250*844 = 250*(4*411) = (250*4)*411 = 1000*411 = 411000\n251*844 = (250+1)*844 = 250*844 + 1*844 = 411000 + 844 = 411844\n\n37*123 = 37*3*41 = 111*41 = (100+10+1)*41 = 4100+410+41 = 4551\n\nsometimes, do not oversimplify...\n105*598 = (100+5)(600-2) = 100*600 - 100*2 + 5*600 - 5*2 = 60000 - 200 + 3000 -10 = 62890 (not good)\n\n105*598 = 105*(600-2) = 63000 - 210 = 62890 (better)\n\nThe key is, you have gotta practice a lot.\nFor example, when I see 111*41, I wrrite down in a peice of paper like this:\n\n(* = space)\n41**\n*41*\n**41\n----\n4554
470294
Press Alt and the numbers 164 = ñ\n\nPress Alt and the numbers 165 = Ñ