 Quandaries and Queries Name: Simon Who is asking: Student Level of the question: Secondary Question: determine smallest positive integer that is divisible by each of the first ten counting numbers Hi Simon, I'm going to find the smallest positive integer N that is divisible by each of the first six counting numbers. The first six counting numbers are 1, 2, 3, 4, 5 and 6 but since every number is divisible by 1 I am only going to work with 2, 3, 4, 5 and 6. The first step is to find the prime decompositions. 2 = 21 3 = 31 4 = 2 2 = 22 5 = 51 6 = 21 31 The primes that appear in these decompositions are 2, 3 and 5. Since each of these counting numbers divide N, the three primes 2, 3 and 5 must divide N. Since these are the only primes that divide these counting numbers and since I want N to be as small as possible, the only primes that divide N are 2, 3 and 5. Thus the prime factorization of N is N = 2? 3? 5? All that is left is to determine what valued go in for the question marks. I want these values to be as small as possible to keep N small. 5 only appears once in the factorizations above (5 = 51) the power of 5 in the factorization of N is 1. N = 2? 3? 51 3 appears twice in the factorizations, (3 = 31 and 6 = 21 31) but in each case the power of 3 is 1 so the power of 3 in the prime factorization of N is 1. N = 2? 31 51 Finally, 2 appears 3 times and in one of these (4 = 22 ) the power of 2 is 2. Thus if this number is to divide N the power of 2 in the prime factorization of N must be at least 2 also. Since we want N as small as possible a power of 2 will suffice. Hence N = 22 31 51 = 60 Now you try the first ten counting numbers. Penny   