1. **State the problem:** Find the largest prime factor of 999936. 2. **Start with factorization:** Recognize that 999936 is close to 1000000, which is $10^6 = 2^6 \times 5^6$. Let's factor 999936 step-by-step. 3. **Check divisibility by 2:** 999936 is even, so divide by 2: $$999936 \div 2 = 499968$$ 4. **Continue dividing by 2 repeatedly:** $$499968 \div 2 = 249984$$ $$249984 \div 2 = 124992$$ $$124992 \div 2 = 62496$$ $$62496 \div 2 = 31248$$ $$31248 \div 2 = 15624$$ $$15624 \div 2 = 7812$$ $$7812 \div 2 = 3906$$ $$3906 \div 2 = 1953$$ Now 1953 is odd, so stop dividing by 2. We divided by 2 eight times, so $$999936 = 2^8 \times 1953$$ 5. **Factor 1953:** Check divisibility by 3: Sum of digits of 1953 = 1+9+5+3=18, which is divisible by 3. $$1953 \div 3 = 651$$ 6. **Continue factoring 651 by 3:** Sum of digits of 651 = 6+5+1=12 (divisible by 3) $$651 \div 3 = 217$$ So, $$1953 = 3^2 \times 217$$ 7. **Factor 217:** Check primes 7 and 11: $$217 \div 7 = 31$$ exactly, because $7 \times 31 = 217$ Both 7 and 31 are primes. 8. **Combine all factors:** $$999936 = 2^8 \times 3^2 \times 7 \times 31$$ 9. **Largest prime factor:** Among the prime factors 2, 3, 7, and 31, the largest is 31. **Answer:** The largest prime factor of 999936 is $\boxed{31}$.