So my approach is to check for all $$$v | n$$$ if there is exist such table.

Let call a square $$$(lx, ly)$$$ is a subtable $$$A[lx..rx][ly..ry]$$$ for

So there are $$$\frac{n}{v} \times \frac{n}{v}$$$ such subtables

Binary Table $$$B[][]$$$ will exist if and only if $$$A[x][y] = A[u][v]$$$ for all subtables $$$(lx, ly)$$$ and $$$lx \leq x, u \leq rx$$$ and $$$ly \leq y, v \leq ry$$$

int n;
bool a[LIM][LIM];
int b[LIM][LIM];
 
bool check(int x, int y, int u, int v) /// Check if this square is valid
{
    int sum = b[u][v] - b[u][y - 1] - b[x - 1][v] + b[x - 1][y - 1];
    return (sum == 0) || (sum == (u - x + 1) * (v - y + 1));
}
 
bool check(int scale) /// Check if the scaling is valid
{
    for (int lx = 1, rx = scale; lx <= n; lx += scale, rx += scale)
        for (int ly = 1, ry = scale; ly <= n; ly += scale, ry += scale)
            if (!check(lx, ly, rx, ry)) return false;
 
    return true;
}
 
int main()
{
    ios::sync_with_stdio(NULL);
    cin.tie(NULL);
 
    /// Input
    cin >> n;
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= n; j += 4)
        {
            char c;
            cin >> c;
 
            /// Decode a[i][j]
            int v = (c <= '9') ? c - '0' : c + 10 - 'A';
            a[i][j + 0] = v >> 3 & 1;
            a[i][j + 1] = v >> 2 & 1;
            a[i][j + 2] = v >> 1 & 1;
            a[i][j + 3] = v >> 0 & 1;
        }
    }
 
    /// Constructing partialsum table
    for (int i = 0; i <= n; ++i) b[i][0] = b[0][i] = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)
            b[i][j] = b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1] + a[i][j];
 
    /// Find answer
    for (int x = n; x >= 1; --x)
        if (n % x == 0 && check(x))
            return cout << x, 0;
 
    return 0;
}

int n;
bool a[LIM][LIM];
bool check(int scale) /// Check if this scaling is valid
{
    for (int lx = 1, rx = scale; lx <= n; lx += scale, rx += scale)
        for (int ly = 1, ry = scale; ly <= n; ly += scale, ry += scale)
            for (int x = lx; x <= rx; ++x)
                for (int y = ly; y <= ry; ++y)
                    if (a[x][y] != a[lx][ly])
                        return false;
 
    return true;
}
 
int main()
{
//    file("Test");
    ios::sync_with_stdio(NULL);
    cin.tie(NULL);
 
    /// Input
    cin >> n;
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= n; j += 4)
        {
            char c;
            cin >> c;
 
            /// Decode a[i][j]
            int v = (c <= '9') ? c - '0' : c + 10 - 'A';
            a[i][j + 0] = v >> 3 & 1;
            a[i][j + 1] = v >> 2 & 1;
            a[i][j + 2] = v >> 1 & 1;
            a[i][j + 3] = v >> 0 & 1;
        }
    }
 
    /// Find result
    for (int x = n; x >= 1; --x)
        if (n % x == 0 && check(x))
            return cout << x, 0;
 
    return 0;
}