To solve this problem, we need to determine the number of integers (x) in the range ([1, n]) such that (x^2 \mod 5 = 1).

Approach

1. Pattern Recognition:

   • Compute (x^2 \mod 5) for (x) from 1 to 5:
     • (1^2 \mod5 =1)
     • (2^2 \mod5=4)
     • (3^2 \mod5=4)
     • (4^2 \mod5=1)
     • (5^2 \mod5=0)
   • The pattern repeats every 5 numbers: every 5 consecutive numbers have exactly 2 values of (x) where (x^2 \mod5=1).

2. Calculation:

   • Full Groups: For (n), the number of full groups of 5 is (n//5). Each group contributes 2 valid numbers.
   • Remainder: The remainder when (n) is divided by 5 ((n\%5)) gives the extra numbers beyond full groups. We precompute the count of valid numbers for each remainder (0-4):
     • Remainder 0: 0 valid numbers.
     • Remainder 1: 1 valid number (1).
     • Remainder 2:1 valid number (1).
     • Remainder3:1 valid number (1).
     • Remainder4:2 valid numbers (1 and4).

Solution Code

def count_valid_x(n):
    add = [0, 1, 1, 1, 2]
    q, r = divmod(n, 5)
    return q * 2 + add[r]

# Example usage:
n = int(input())
print(count_valid_x(n))

Explanation

• Full Groups: The number of full groups of 5 in (n) is (n//5), each contributing 2 valid numbers.
• Remainder Handling: Using the precomputed list add, we add the valid numbers from the remainder part. This list directly maps the remainder to the count of valid numbers in that remainder range.

This approach efficiently computes the result in constant time (O(1)) since it involves basic arithmetic operations and a lookup in a fixed-size list. This makes it suitable for very large values of (n).

Example:

• For (n=6): (6//5=1) (full group), remainder=1. So total valid numbers: (1*2 + add[1] =2+1=3) (1,4,6).
• For (n=9): (9//5=1), remainder=4 → (2+2=4) (1,4,6,9).

This solution is optimal and handles all cases correctly. (\boxed{count_valid_x(n)})

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