1. The problem is to find the largest prime factor of 999936. 2. Start by factorizing 999936. Notice that it is even, so divide by 2: $$999936 \div 2 = 499968$$ 3. Continue dividing by 2 repeatedly: $$499968 \div 2 = 249984$$ $$249984 \div 2 = 124992$$ $$124992 \div 2 = 62496$$ $$62496 \div 2 = 31248$$ $$31248 \div 2 = 15624$$ $$15624 \div 2 = 7812$$ $$7812 \div 2 = 3906$$ $$3906 \div 2 = 1953$$ We divided by 2 eight times, so $$999936 = 2^8 \times 1953$$ 4. Now factor 1953. Check for divisibility by 3: $$1 + 9 + 5 + 3 = 18$$ Since 18 is divisible by 3, 1953 is divisible by 3. $$1953 \div 3 = 651$$ 5. Factor 651 by 3 again: $$6 + 5 + 1 = 12$$ Since 12 is divisible by 3, divide: $$651 \div 3 = 217$$ So the factors so far: $$999936 = 2^8 \times 3^2 \times 217$$ 6. Factor 217. Check if divisible by 7: $$217 \div 7 = 31$$ Yes, since 7 × 31 = 217. 7. Both 7 and 31 are primes, so the prime factors are: $$2, 3, 7, 31$$ Among them, the largest prime factor is $$31$$. **Final answer:** The largest prime factor of 999936 is **31**.