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