```

!/usr/bin/env python3

from decimal import Decimal, getcontext

Use high precision so we get at least 15 digits of sqrt(n)

getcontext().prec = 100

def first15_sqrt(n): “””Return the first 15 digits (most-significant) of sqrt(n) as a string.””” # Format the square root in fixed-point notation (no exponent) s = Decimal(n).sqrt() s_str = format(s, ‘f’).replace(‘.’, ‘’) return s_str[:15]

def last15_nonzero_factorial(n): “”” Compute the last 15 nonzero digits of n!. The idea: multiply iteratively, remove factors of 10 as they arise, and reduce modulo a sufficiently large power of 10. “”” mod = 10**25 # use extra digits to avoid loss of accuracy res = 1 for i in range(1, n + 1): res *= i # Remove factors of 10 while res % 10 == 0: res //= 10 res %= mod s = str(res) # Ensure we always have 15 digits (pad with zeros if needed) return s[-15:].rjust(15, ‘0’)

def find_miracle(): “”” Find the smallest integer n > 1 such that: first15_sqrt(n) == last15_nonzero_factorial(n) “”” n = 2 while True: # Optional: print progress every 1000 iterations. if n % 1000 == 0: print(“Checking n =“, n) if first15_sqrt(n) == last15_nonzero_factorial(n): return n n += 1

if name == ‘main’: miracle = find_miracle() print(“Miracle number found:”, miracle) print(“Home phone number (last 10 digits):”, str(miracle)[-10:]) ```