1. The problem asks for the lowest common multiple (LCM) of 9 and 4, the shortest fence length Zara and Chester can both build using their panel lengths, the largest number of boxes Joel can split notebooks and pens into with equal counts, and the time when trains on two routes next stop together. 2. The LCM of two numbers is the smallest positive integer divisible by both numbers. 3. To find the LCM of 9 and 4, list multiples: - Multiples of 9: 9, 18, 27, 36, 45, ... - Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, ... The smallest common multiple is 36. 4. For Zara and Chester's fences: - Zara's panels are 6 feet long. - Chester's panels are 4 feet long. The shortest fence length both can build is the LCM of 6 and 4. Multiples of 6: 6, 12, 18, 24, 30, 36, ... Multiples of 4: 4, 8, 12, 16, 20, 24, ... The LCM is 12 feet. 5. Joel wants to split 15 notebooks and 20 pens into boxes with the same number of notebooks and pens per box. This is a greatest common divisor (GCD) problem. 6. The GCD of 15 and 20 is found by prime factorization: - 15 = 3 × 5 - 20 = 2² × 5 Common factor is 5. So, the largest number of boxes is 5. 7. For the trains stopping problem: - One train stops every 9 minutes. - The other every 6 minutes. They both stop at midday (time 0). We find the LCM of 9 and 6 to find when they next stop together. 8. Multiples of 9: 9, 18, 27, 36, ... Multiples of 6: 6, 12, 18, 24, 30, ... The LCM is 18 minutes. Final answers: - LCM(9,4) = 36 - Shortest fence length = 12 feet - Largest number of boxes = 5 - Trains next stop together in 18 minutes