# Discovering Square Roots with Long Division: Examples Included!

2023-05-16 07:32:30 - Grace Browns Grace Browns has been a lifestyle, fashion, and beauty writer for over 5 years, and she currently serves as a senior editor at 422346.com.

When we are presented with an integer X that is a perfect square, we need to determine its square root utilizing the long division method.

Examples:

Input: N = 484 Output: 22 2 = 484

Input: N = 144 Output: 12 2 = 144 To employ the long division method and discover the root of a square or any number, we must comprehend the following methodical approach:

1. Split the digits of the number into groups, beginning with the unit's position digit. If the number has an odd number of digits, the final digit forms its segment. For instance, 1225 is split into (12 25). 2. Beginning at the leftmost segment, select the greatest number that is equal to or lesser than it, which when squared, becomes the divisor. The corresponding quotient is also the divisor such that their product is the square. For example, as 9 is the nearest perfect square to 12, the first segment, its divisor and quotient are 9. 3. Subtract the square of the divisor from the first segment and add the next segment (if any) to the right of the remainder to form a new dividend. For instance, if the difference between 12 and 9 is 3, we then append 25, resulting in the new dividend being 325.  4. Calculate the new divisor by doubling the previous quotient (recall that 3 is the quotient of 9 and 12, as 32 = 9) and selecting a digit that, when appended to it, generates a divisor with the product similar or inferior to the new dividend. As an example, the doubling of 3 provides 6 as the new quotient. As a result, by appending 5, the new divisor is 65, and the product 65 times 5 equates to 325, which is closest to the new dividend. To arrive at the coveted square root of a given number, follow these steps in a cyclical fashion (namely, steps (2), (3), and (4)) until all segments have been accounted for. Upon completion, the quotient yielded will be the desired square root.

Here is how the above-mentioned approach can be put into action:

• #include
• using namespace std;
• #define INFINITY_ 9999999
• int longDivisionSquareRoot(int n)
• {
•     int i = 0, udigit, j;
•     int cur_divisor = 0;
•     int quotient_units_digit = 0;
•     int cur_quotient = 0;
•     int cur_dividend = 0;
•     int cur_remainder = 0;
•     int a = { 0 };
•     while (n > 0) {
•         a[i] = n % 100;
•         n = n / 100;
•         i++;
•     }
•     i--;
•     for (j = i; j >= 0; j--) {
•         cur_remainder = INFINITY_;
•         cur_dividend = cur_dividend * 100 + a[j];
•         for (udigit = 0; udigit = cur_dividend
•                                      - ((cur_divisor * 10 + udigit)
•                                         * udigit)
• ## Retaining Tags for Mathematical Operations

• ### Reimagining a Complex Equation

• If we want to find the quotient of two numbers, we can use a formula that involves dividing and subtracting. Let's explore a new way to represent these operations:
• while (cur_divisor < cur_dividend) {
•     int udigit = 0;
•     while (((cur_divisor * 10 + udigit) * udigit)
•                 <= cur_dividend) {
•         udigit++;
•     }
•     udigit--;
•     if (udigit < 0) {
•         break;
•     } else if ((cur_divisor * 10 + udigit)
•                 * udigit
•                 - ((cur_divisor * 10 + udigit) * udigit)
•                     >= 0) {
•         cur_remainder = cur_dividend - ((cur_divisor * 10
•                                                  + udigit)
•                                                 * udigit);
•         quotient_units_digit = udigit;
•     }
• }
• cur_quotient = cur_quotient * 10
•                     + quotient_units_digit;
• cur_divisor = cur_quotient * 2;
• cur_dividend = cur_remainder;
• return cur_quotient;
• Now we have a new way of approaching this problem that allows us to retain the necessary tags to ensure clarity in our mathematical expression.
• ### Testing Our Equations

• To test this new approach, let's set a variable for our starting number and output the quotient using our new formula:
• int main()
• {
•     int x = 1225;
•     cout << quotient(x) << endl;
•     return 0;
• }
• By transforming our equation and testing it using a starting value, we can have confidence in our new approach for calculating quotients.
• ## Retaining Tag Structures in Mathematical Computation

When performing mathematical computation, it is important to retain the structure of specific tags such as `<a>`, `<h2>`, `<h3>`, and `<strong>`. This ensures that the essence of the information remains intact despite significant rephrasing and creative linguistic divergence from the original.

### Code Example:

To illustrate this point, consider the following code:

``int cur_quotient = 0;int cur_dividend = 0;int cur_remainder = 0;int a[] = new int;while (n > 0) {    a[i] = n % 100;    n = n / 100;    i++;}i--;for (j = i; j >= 0; j--) {    cur_remainder = INFINITY_;    cur_dividend = cur_dividend * 100 + a[j];    for (udigit = 0; udigit = cur_dividend - ((cur_divisor * 10 + udigit) * udigit)                 && cur_dividend - ((cur_divisor * 10 + udigit) * udigit)                 >= 0) {        cur_remainder = cur_dividend - ((cur_divisor * 10 + udigit) * udigit)``

By retaining the original tag structure, we can ensure that the code remains clear and concise, even with significant linguistic divergence:

``<strong>Algorithm for computing the remainder and quotient of a number</strong><h3>Variables:</h3><p>cur_quotient = 0;</p><p>cur_dividend = 0;</p><p>cur_remainder = 0;</p><p>a[] = new int;</p><p>While n is greater than 0, perform the following:</p><ol>    <li>Set the value of a[i] to n % 100.</li>    <li>Divide n by 100.</li>    <li>Add 1 to i.</li></ol><p>Subtract 1 from i.</p><p>Starting from i and going down to 0, perform the following:</p><ol>    <li>Set cur_remainder to INFINITY_.</li>    <li>Multiply cur_dividend by 100 and add a[j].</li>    <li>Starting from udigit = 0 and going up to 9, perform the following:</li>    <ol>        <li>If cur_dividend is greater than or equal to (cur_divisor * 10 + udigit) squared, set cur_remainder to cur_dividend - ((cur_divisor * 10 + udigit) * udigit) and exit the loop.</li>    </ol></ol>``

Looking to compute square roots using a long division method, we can utilize various structural tags to ensure we retain the technical jargon associated with the code. Initializing values for variables, we set up a loop to get the remainder of our chosen number, with the end goal being the computation of the square root. We define a function that takes in a parameter "n," representing our chosen number. The variable "INFINITY_" is set to a large value of 9999999 for error handling purposes.

We start by defining variables for our current quotient, dividend, and divisor, and set them to 0 initially. We then set up a loop, whereby we look to perform long division to compute the square root of our chosen number. We get the digits of the number until we have completed the long division process.

We then utilize the structure of tags like ,

## , , and to convey our ideas. Utilizing a range of unique expressions and creative linguistic approaches, we reconstruct the code in a way that retains its meaning and essence but with a different and more diverse flair. The main function then calls up this multi-layered function by using "System.out.print(sqrtByLongDivision(x) +"\n");" Algorithm for Dividing Big Numbers If you are looking for an efficient and reliable method for dividing large numbers, here is an algorithm that may meet your needs. First, we initialize a variable named "i" to zero. Then, we perform a loop in which we extract the last two digits of the dividend, store them in an array named "a," and divide the dividend by 100. We continue this process until the dividend becomes zero, incrementing "i" by one after each iteration. Next, we decrement "i" by one to give us the index of the first non-zero element in the "a" array. Now that we have our "a" array, we can perform long division on the dividend and divisor. We begin by iterating from "i" to zero and performing the following steps: Multiply the current value of "cur_dividend" (which starts as the first element of "a") by 100 and add the next element in "a." Initialize "cur_remainder" to infinity. Iterate from 0 to 9 and perform the following steps for each digit: Calculate the new remainder if we subtracted the product of the (current divisor times 10 plus the current digit) from the current dividend. Check if this new remainder is less than or equal to the current remainder. If it is, update the value of "cur_remainder" to this new remainder. After iterating through all 10 digits, we divide the current dividend by the divisor times 10 plus the digit that provided us with the minimum remainder (i.e., the digit that gave us the update for "cur_remainder"). Append the digit to the quotient. Set the current remainder to "cur_remainder." By iteratively performing the above steps, we can obtain the quotient and remainder of the division of two large numbers in a time-efficient and practical manner. To calculate the square root of a number in C#, you can use long division. Here's how it works:First, set your initial values. We'll start with the number 1225.Then, create a loop that will run until your input number (n) is less than or equal to 0. Inside the loop, convert the number into an array of two-digit integers.Next, initialize the variables cur_remainder, cur_quotient, and cur_divisor to 0. Set cur_divisor to cur_quotient times 2, and set quotient_units_digit to the current unit digit of cur_quotient.Then, calculate the current quotient by multiplying cur_quotient by 10 and adding quotient_units_digit. Set cur_dividend to cur_remainder, and loop through the array of two-digit integers to calculate the current remainder.After the loop ends, return the final quotient.Here's an example code in C# using the long division algorithm:using System;class GFG{static readonly int INFINITY_ =9999999;static int sqrtBylongDivision(int n){int i = 0, udigit, j; int cur_divisor = 0;int quotient_units_digit = 0;int cur_quotient = 0;int cur_dividend = 0;int cur_remainder = 0;int []a = new int;// create array of two-digit integerswhile (n > 0) {a[i] = n % 100;// calculate current remaindercur_remainder = (cur_remainder * 100) + a[i];for (udigit = 9; udigit >= 0; udigit--) {j = cur_divisor + (2 * udigit + 1);if (j * udigit <= cur_remainder)break;}// calculate quotient units digitquotient_units_digit = udigit;// calculate current quotientcur_quotient = cur_quotient * 10 + quotient_units_digit;// update current divisorcur_divisor = (cur_divisor * 10) + (2 * quotient_units_digit + 1);// update current remaindercur_remainder -= udigit * cur_divisor;// set current quotientcur_quotient = cur_quotient * 10 + quotient_units_digit;// update array indexi++;// update input numbern = n / 100;}// return final quotientreturn cur_quotient;}static void Main(){int x = 1225;Console.WriteLine(sqrtBylongDivision(x));}}In the given algorithm, the value of n is divided by 100 until it becomes zero. During each iteration, the value of i is incremented. Once all the digits of the number have been stored in an array a, i is decremented. In the next loop, the remainder is initialized to a high value, and the dividend is set to the first element of the array. The digits of the divisor are iteratively determined. The units digit of the quotient is computed by considering the difference between the divisor and the dividend and selecting the largest suitable digit. This process is repeated for all the digits of the array, and the final quotient is calculated. Finally, the time complexity of the algorithm is O((log100n)2 * 10), and the auxiliary space required is O(1).To implement this algorithm, we first initialize INFINITY_ to a high value and declare all the variables required for the computation. We then use a while loop to obtain each digit of the input number and store them in an array. In the next for loop, we initialize the dividend and remainder, and iteratively determine the digits of the divisor and quotient. Finally, we update the values of the quotient, divisor, and dividend for each iteration of the loop. To test the implementation, we can set a value for x and output the square root obtained from the algorithm using document.write().

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