   SEARCH HOME Math Central Quandaries & Queries  Question from Peggy, a parent: What is the smallest number divisible by each of the first nine counting numbers? Hi Peggy,

Let me try a slightly smaller problem.

What is the smallest number divisible by each of 4, 5 and 6?

First I want to write 4, 5 and 6 as a product of primes

4 = 2 × 2
5 = 5
6 = 2 × 3

I need a way to refer to the number I am looking for, the smallest number divisible by each of 4, 5 and 6, so I am going to call it N. I am going to construct N in terms of its prime factors.

Since 4 = 2 × 2 divides N, N must have 2 as a prime factor twice.

First step: N = 2 × 2 × ?

5 must also divide N so 5 must be a prime factor of N.

Second step: N = 2 × 2 × 5 × ?

Finally 6 = 2 × 3 must divide N so 2 and 3 are prime factors of N. I already have a 2 in the construction of N so all I need to add is a 3

Third step: N = 2 × 2 × 5 × 3

Seeing N in this form it is easy to see that 4 = 2 × 2, 5 and 6 = 2 × 3 all divide N and leaving any of the 4 primes out of the construction of N will mean that one of 4, 5 or 6 will not divide N. Thus N = 2 × 2 × 5 ×3 = 60 is the smallest number divisible by each of 4, 5 and 6.

Now try this procedure with 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.