-------ASIDE--------------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----NOW ONTO THE QUESTION-----------------

What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?
Let K = that lowest positive integer

This means that there's a 2 "hiding" within the prime factorization of K, a 3 "hiding" within the prime factorization of K, a 4 "hiding" within the prime factorization of K, etc.

So, let's begin with a 2 "hiding" within the prime factorization of K.
This means that K = (2)(other numbers)

Also, if there's a 3 "hiding" within the prime factorization of K, then we need to add a 3 like so: K = (2)(3)

There's a 4 "hiding" within the prime factorization of K.
Since 4 = (2)(2), then we need to add a SECOND 2 to get: K = (2)(2)(3)

There's a 5 "hiding" within the prime factorization of K, so we'll add a 5 to get: K = (2)(2)(3)(5)

There's a 6 "hiding" within the prime factorization of K.
Since 6 = (2)(3), we can see that we ALREADY have a 6 "hiding" in the prime factorization: K = (2)(2)(3)(5)

There's a 7 "hiding" within the prime factorization of K, so we'll add a 7 to get: K = (2)(2)(3)(5)(7)

We have now ensured that K is divisible by every integer from 1 to 7 inclusive. This means that we have found the LEAST possible value of K that satisfies the given conditions.
So, K = (2)(2)(3)(5)(7) = 420

Answer: A

Cheers,
Brent