Interesting Question?

Question

In General, what is a method to determine the smalles N, such that the first n natural numbers all divde N without remainder?
What would be N for n=10?

Answer 1

That's called the least common multiple (LCM). You need to factor the first n natural numbers into primes. Then find the LCM of 1 and 2, then the LCM of this and 3, then the LCM of this and 4, etc. .. up to ... the LCM of this and n.

For n=10, the answer is N = 2520.

Answer 2

Basically, you have to find the least common multiple
of the first n natural numbers. As you can imagine,
the factoring method would be pretty useless here!
For n = 10, the LCM is 9*8*7*5 = 2520.
Let's look at n.
Make a table of all primes <=n and find the largest
power of each less than or equal to n. Then multiply all these together.
Not a very practical method for large n, but good
for small n.
For 10, for example, the largest powers
of 2,3,5,7 are
3,2,1 and 1, respectively,
so the answer would be
2³*3²*5*7 = 2520, as before.
Let's try n = 100.
The answer would be
97*89*83*79*73*67*61*59*53*47
*43*41*37*31*29*23*19*17*
13*11*7²*5²*3^4*2^6.

Answer 3

If the number N is divided by number(P) less than or equal to n having highest power of the prime numbers less than or equal to n than the first n natural numbers all divide N without remainder.
For n = 10,
P=9,8,7,5
therefore, the smallest number divisible by these number i.e.
the multiple of these numbers;9*8*7*5= 2520 is the smallest number N for n=10

Answer 4

for 10 it would be 9x8x7x5=2520.