How to find the number of values in a given range that has a certain remainder when divided by a given number?

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Given four integers N, L, R and Rem. I have to find the the number of values between L and R(inclusive) that gives remainder Rem when divided by N.

For example: If N = 3, L = 2, R = 10 and Rem = 1 then the numbers with remainder 1 when divided by 3 in this range are {4, 7, 10}. So, the answer is 3.

Here is the brute force approach I coded:

int main() {
    int N, L, R, Rem;
    cin >> N >> L >> R >> Rem;

    int Ans = 0;
    for (int i = L; i <= R; i++) {
        if (i % N == Rem)
            Ans++;
    }
    cout << Ans << endl;
}

What would be a better approach to solve this problem?

Count of distinct remainders when N is divided by all the numbers , It contains well written, well thought and well explained computer Given an integer N, the task is to find the count of total distinct remainders which can be obtained when N is divided by every element from the range [1, N]. of N the number of distinct remainders will be N / 2 and for odd values of N it will� It's quite simple: test which of n = 5 q + 1 = 1, 6, 11 has the desired remainder when divided by 3. Since the remainders mod 3 are 1 → 1, 6 → 0, 11 → 2, it is 11 with sought remainder = 2. The same quick test works if we replace 5 by any m not divisible by 3.

TLDR it is roughly (R-L+1)/N give or take +- 1

for instance L=2 R=10 N=3 REM=0

the numbers are 3,6,9 (R-L+1)/N = (10-2+1)/3 = 9/3 = 3

here's an accurate solution with explanation, that requires no loops

find the first number greater or equals to L that divides nicely by N

L = (L % N == rem)? L : L + (REM - L%N)

find the last number smaller or equals to R that divides nicely by N

R = (R % N == rem)? R : R - (N - (REM - R%N))

the result is

int result = ((R-L)/N) + 1

Minimum integer such that it leaves a remainder 1 on dividing with , It contains well written, well thought and well explained computer science and Given an integer N, the task is to find the minimum possible integer X such that X % M = 1 Now we know that lcm is the smallest number which is divisible by all the elements from the range [2, N] and @geeksforgeeks, Some rights reserved. The quotient remainder theorem. This is the currently selected item. Modular addition and subtraction. Practice: Modular addition. Modulo Challenge (Addition and

(R - L) / N 
+ 1 if (R - L) % N ≥ (rem - L) % N

Finding a number given its remainder when divided by other numbers, In particular, I don't want to use "congruences" or modulo arithmetic that I List the numbers that give remainder 1 on division by 5, looking for remainder test which of n=5q+1=1,6,11 has the desired remainder when divided by 3. For 2nd stat., the possible values are= 1 11 . . . now for the given number, the remainder is� The number of groups of this size that can be formed, c, is the quotient of a and b. - is nothing more than the number of integers within range/interval ]0..a] (not including zero, but including a) that are divisible by b. so by definition: Y/Z - number of integers within ]0..Y] that are divisible by Z and

The quotient remainder theorem (article), When we want to prove some properties about modular arithmetic we often make We can see that this comes directly from long division. Have you proven the Quotient Remainder Theorem previously? Given any real number A and any positive real number B if I divide A by B, I will The only integer in that range is 0. Calculate the range of the data set by subtracting the smallest from the largest number given in Step 3. The range for the example is 12 - 1 = 11. Practice the method outlined in Step 2 through Step 4 to find the range of the following test scores: 55, 60, 75, 80, 85, 90 and 100.

How to use the Excel MOD function, The Excel MOD function returns the remainder of two numbers after division. number - The number to be divided. divisor - The number to divide with. In the example shown, conditional formatting has been applied to the range B4:G11 To get the number at a specific place value you can use a formula based on the� Method 3 : [Efficient] Let B = b * M and A = a * M The count of numbers divisible by 'M' between A and B will be equal to b - a. Example: A = 25, B = 70, M = 10. Now, a = 2, b = 7. Count = 7 - 2 = 5. It can be observed that, if A is divisible by M, ‘b – a’ will exclude the count for A, so the count will be less by 1.

Multiples, Factors and Powers, What particular property of odd and even numbers does each illustrate? The number 8 has two rectangular arrays and a three-dimensional array: If we divide any of these multiples by 6, we get a quotient with remainder zero. 2 place-value system, so computer programmers need to know the powers of 2, and must be� Print all numbers in range divisible by a given number in python. Then the user must enter the number to be divided from the user. Whenever the remainder of the number divided by i is

Comments
  • Hint: If you have two numbers, A and B and you divide them by N and they have the same remainder then A + NC = B for some integer C
  • You have to reduce loop, instead i++ use i += N
  • Obviously, it doesn't work for negative values. The OP specified input as integers, not positive integers.
  • @ciamej, added a few words about negatives.
  • Obviously, it doesn't work for negative values. The OP specified input as integers, not positive integers.