C# get last 3 digits of int
WebSolution: Declare two variables ( n and lastDigit ). Read the user input from the console. (int.Parse (Console.ReadLine ());). Find the last digit of the number by the formula: int … WebMar 11, 2024 · I tested with some inputs, which are given below. int Result1 = takeNDigits (666, 4); int Result2 = takeNDigits (987654321, 5); int Result3 = takeNDigits (123456789, 7); int Result4 = takeNDigits (35445, 1); int Result5 = takeNDigits (666555, 6); The output is shown below. Result1: 666 Result2: 98765 Result3: 1234567 Result4: 3 Result5: 666555
C# get last 3 digits of int
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WebNov 15, 2005 · How do I extract the last 3 characters of a string i.e. string s = "000001" I want "001" Probably overkill for your current problem, but you can use an Regular Expression [RE] to select the part of the string that you want, in this case, it would be @".{3}$". You would be amazed at what can be accomplished with RE's ! WebFeb 9, 2014 · The value % 10 will give the last digit, use this value as the last character of the result (with a space prefix) the " {1}" part of the string format. The problem is now reduced to doing the same thing all over again, this time with a value without the last digit (value / 10) and placing the result in the " {0}" part of the format string.
Webint num, last; // Reading number Console.Write("Enter any number: "); num = Convert.ToInt32(Console.ReadLine()); last = num % 10; Console.WriteLine("The last … WebDec 19, 2012 · Solution 3 C++ bool removeLastDigit (float& number) { int32_t tempNum = static_cast< int32_t > (number); tempNum = tempNum % 10 ; if ( 0 == tempNum ) return false; //No digit to remove number = number - tempNum; return true ; } Won't work for real number though. :) Posted 20-Dec-12 3:28am Aswin Waiba Updated 20-Dec-12 3:32am …
WebJan 21, 2014 · Sample: C#. … WebMar 13, 2024 · Below is the solution to get the product of the digits: C++ Java Python3 C# PHP Javascript #include using namespace std; int getProduct (int n) { int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product; } int main () { int n = 4513; cout << (getProduct (n)); } Output 60
WebIt's just a simple integer division. So select (123456/100). But that only valid is the column is integer type. The example provided is big. If the column is integer then using a simple case logic is suffice. First check if the value is bigger than …
WebTo access the last 3 characters of a string, we can use the built-in Substring () method in C#. Here is an example, that gets the last 3 characters ris from the following string: using … goldberg vs the fiendWebJan 27, 2024 · Approach: Follow the steps below to solve the problem: Traverse the array and check for each array element, whether it is possible to convert it to a pronic number.; For each array element, apply all the possible rotations and check after each rotation, whether the generated number is pronic or not. goldberg vs shawn at youtubeWebJun 9, 2024 · Approach: For large numbers it is difficult to rotate and divide each number by 8. Therefore, ‘divisibility by 8’ property is used which says that a number is divisible by 8 if the last 3 digits of the number is divisible by 8. Here we do not actually rotate the number and check last 8 digits for divisibility, instead we count consecutive sequence of 3 digits … goldberg vs the fiend super showdownhbo-programma basis software engineeringWebMar 17, 2024 · To find last digit of a number, we use modulo operator %. When modulo divided by 10 returns its last digit. Suppose if n = 1234 then last Digit = n % 10 => 4 To find first digit of a number is little expensive … hbo programs scheduleWebMay 29, 2024 · The problem can easily be solved by using counting. Firstly, loop through numbers less than n and for each number count the frequency of the digits using count array. If all the digits occur only once than we print that number. The answer always exists so there is no problem of infinite loop. goldberg vs the fiend 2020WebAug 30, 2009 · int value = 1234; int newVal = value / 10; If you instead want 234, then: int value = 1234; int newVal = value - value / 1000 * 1000; At this point newVal can be printed or whatever. Aug 28, 2009 at 11:17am Gregor (147) webjose what are you doing for 234, makes no sense for 234: int newVal = value%1000; general: get rid of last n digits in … goldberg washington pa