What is the smallest number that is divisible by both 420 and 5,250?

A. 2,205
B. 5,250
C. 7,428
D. 10,500
E. 12,240

The first step in solving this problem is to break up each number in to its prime factors:
420 = 42 × 10 = 6 × 7 × 2 × 5 = 2 × 3 × 7 × 2 × 5
and
5,250 = 525 × 10 = 7 × 75 × 10 = 7 × 3 × 5 × 5 × 2 × 5.

The smallest number that is divisible by both is the number that has all of their prime factors. In order for the number to be divisible by 5250, it must contain the factor string: 7 × 3 × 5 × 5 × 2 × 5, the only thing that is missing from this string that will allow it to also be divisible by 420 is a factor of 2. Notice that 5250 has only a single 2 in its factor string whereas 420 has two 2`s. So we simply multiply 5,250 by 2 to get the answer (D) 10,500.
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In the explanation the factor string of 10 was not really clear and what was the basis of choosing two as the multiplier, why not 3 or 7 or 5?

For a number to be divisible by 5,250 it must contain a factorization string of 5,250 (7 × 3 × 5 × 5 × 2 × 5) in its own factorization string.

So the factorization string of the number we're looking for is 7 × 3 × 5 × 5 × 2 × 5 × .....

Let us see which prime factors of 420 it already contains:
7 × 3 × 5 × 5 × 2 × 5 × ....
420 = 2 × 3 × 7 × 2 × 5

We see that it contains 2 × 3 × 7 × 5, but doesn't contain another 2. So we must add at least one more prime factor "2" to the factorization string: 7 × 3 × 5 × 5 × 2 × 5 × 2.

Since we are looking for the smallest number, we must add no more factors to it.
7 × 3 × 5 × 5 × 2 × 5 × 2 = (7 × 3 × 5 × 5 × 2 × 5) × 2 = 5,250 × 2 = 10,500.

If we added a multiplier 3, 5 or 7 to the string instead of 2, then the resulting number would NOT be divisible by 420, because it would NOT contain enough prime factors of 420.

This question is solely about finding the least common multiple of two numbers.