Problem

A number N is automorphic if its square ends with the same digits as N itself. Given a non-negative integer N, return true if it is automorphic and false otherwise.

Input / Output

• Input: integer N.
• Output: boolean (or the string "Automorphic" / "Not Automorphic").

Examples

N	N²	Ends with N?	Result
5	25	yes	true
6	36	yes	true
25	625	yes	true
76	5776	yes	true
376	141376	yes	true
7	49	no	false
10	100	no	false

Constraints

• 0 <= N <= 10^9 (use 64-bit arithmetic for the square).

Approach

Digit comparison. Compute sq = N * N. Then compare the last k digits of sq with N, where k is the number of digits in N:

while N > 0:
    if N % 10 != sq % 10: return false
    N /= 10; sq /= 10
return true

This peels one digit from each number at a time and stops as soon as a mismatch appears.

Modular check. Let p = 10^k. N is automorphic iff sq % p == N. Computing p is O(k).

String method. Convert both to strings and test str(sq).endswith(str(N)). Simplest to write, slightly slower.

Edge cases: N = 0 and N = 1 are automorphic (0² = 0, 1² = 1).

Complexity

• Time: O(k) where k is the number of digits, i.e. O(log N).
• Space: O(1).

Background

Automorphic numbers are also called circular numbers. There are only two infinite families ending in 5 and 6; each longer automorphic number extends a shorter one (5 → 25 → 625 → 0625…, 6 → 76 → 376 → 9376…). Interviewers use the problem to test overflow awareness — N = 10^9 gives a square of 10^18, which does not fit in 32 bits.

Reference solution (Python)

def is_automorphic(n):
    sq = n * n
    t = n
    while t > 0:
        if t % 10 != sq % 10:
            return False
        t //= 10
        sq //= 10
    return True

Test cases

Input	Expected
5	true
76	true
7	false
0	true