Write down the numbers 2,3,4,...10. Start from the top, ie. 10. Now since 10 = 2*5, cross these two out. Then there's 9 = 3*3. So you can cross out 3. Also you can cross out 6, since you have a 2 from 10 and a 3 from 9. Then you come to 8 = 4*2. Since you already have the 2 (from 10) you only need 4. So you write down 4 instead of 8 and cross out the 4. Then there's 7, which is prime (and uncrossed-out) so you need it. All lower numbers have been thrown out as redundant, so that's it :10*9*4*7 = 2520.
The other way of doing it is by using only the prime numbers (see Nexus' solution) and making sure all other numbers are covered through the various products of these prime numbers.
#7
Nexus[Free-DC]
37
0
You can prove, somehow, that to find the lowest number divisible by all numbers less than N, all you have to do is list all the primes not less than N and find the highest powers of those primes not less than N.
Take N=16 for example. The primes below 16 are 2,3,5,7,11 and 13 and the corresponding highest powers are 2^4,3^2,5,7,11,13. So the lowest number divisible by 2 to 16 is 2^4 x 3^2 x 5 x 7 x 11 x 13 = 720720.