What is the lowest common multiple for 2 3 and 9?

Answer 1

We see that we must find a number n such that it satisfies the condition:

n ≡ 0 (mod 2) ≡ 0 (mod 3) ≡ 0 (mod 9)

Since 9 is a multiple of 3, we can forget about the 0 (mod 3). Since 2 and 9 are relatively prime, the Chinese Remainder Theorem states that there indeed exists a number n such that it satisfies n ≡ 0 (mod 2) ≡ 0 (mod 9). Now let 2K represent some multiple of 2, and set it congruent to 0 (mod 9):

2K ≡ 0 (mod 9)

This is a particularly easy case; 2K would have to equal some multiple of 9 for it to satisfy this expression. Therefore, K = 9 and n must = 18c, where c is an arbitrary multiplier. This is your new modulus:

n ≡ 0 (mod 18)

Any n that satisfies this condition will also satisfy n ≡ 0 (mod 2) ≡ 0 (mod 3) ≡ 0 (mod 9).