What is the least positive integer greater than 1 that leaves a remainder of 1 when divided by each of 2, 3, 4, 5, 6, 7, 8 and 9?
This problem is asking for the smallest number that is one more than a multiple of all the numbers from 2 to 9. One way to find such a number is to find the least common multiple of all the numbers from 2 to 9, and then add 1 to it. To find the least common multiple of a set of numbers, I can use the prime factorization method. I write each number as a product of prime factors, using exponents to indicate repeated factors. For example, 2 = 2, 3 = 3, 4 = 2^2, 5 = 5, 6 = 2 * 3, 7 = 7, 8 = 2^3, and 9 = 3^2. Then, to find the least common multiple, I take the highest power of each prime factor that appears in any of the numbers, and multiply them together. For example, the highest power of 2 that appears is 2^3, the highest power of 3 that appears is 3^2, the highest power of 5 that appears is 5, and the highest power of 7 that appears is 7. So, the least common multiple is 2^3 * 3^2 * 5 * 7 = 2520. Therefore, the smallest number that is one more than a multiple of all the numbers from 2 to 9 is 2520 + 1 = 2521. 