Question

Let $M$ be the least common multiple of $1, 2, \ldots, 20$. How many positive divisors does $M$ have

Answer 1

Consider the prime factorization of 20!.

$20! = 20 \times 19 \times 18 \times \cdots \times 3 \times 2 \times 1$

The LCM of 1, 2, ..., 20 must contain all the primes less than 20 in its factorization, so

$M = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times m$

where $m$ is some integer not divisible by any of these primes.

Compare the factorizations of the remaining divisors of 20!, and check off any whose factorizations are already contained in the product of primes above.

$4 = 2^2$ - missing a factor of 2

$6 = 2\times3$ - ✓

$8 = 2^3$ - missing a factor of 2²

$9 = 3^2$ - missing a factor of 3

$10 = 2\times5$ - ✓

$12 = 2^2\times3$ - missing a factor of 2

$14 = 2\times7$ - ✓

$15 = 3\times5$ - ✓

$16 = 2^4$ - missing a factor of 2³

$18 = 2\times3^2$ - missing a factor of 3

$20 = 2^2\times5$ - missing a factor of 2

From the divisors marked "missing", we add the necessary missing factors to the factorization of $M$, so that

$M = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 2^3 \times 3$

Then the LCM of 1, 2, 3, …, 20 is

$M = 2^4 \times 3^2 \times 5 \times7 \times 11 \times 13 \times17 \times 19$

$\implies \boxed{M = 232,792,560}$