The product of integers w, x, y, and z is 210, and w > x > y > z > 0.

Let’s break down 210 as a product of two positive integers:

210 = 1 x 210 = 2 x 105 = 3 x 70 = 5 x 42 = 6 x 35 = 7 x 30 = 10 x 21 = 14 x 15

For each of the expressions except 1 x 210, we can write each factor as the product of two distinct factors. For example:

210 = 2 x 105 = (1 x 2) x (3 x 35) = (1 x 2) x (5 x 21) = (1 x 2) x (7 x 15)

From the above, we see that the sum w + x + y + z could be 41, 29, and 25, respectively. Thus, we can eliminate choice E.

Now let’s use 210 = 3 x 70:

210 = 3 x 70 = (1 x 3) x (2 x 35) = (1 x 3) x (5 x 14) = (1 x 3) x (7 x 10)

From the above, we see that the sum w + x + y + z could be 41, 23, and 21, respectively. Thus, we can eliminate choice C.

Now let’s use 210 = 5 x 42:

210 = 5 x 42 = (1 x 5) x (2 x 21) = (1 x 5) x (3 x 14) = (1 x 5) x (6 x 7)

From the above, we see that the sum w + x + y + z could be 29, 23, and 19, respectively. Thus, we can eliminate choice B.

Now let’s use 210 = 6 x 35:

210 = 6 x 35 = (1 x 6) x (5 x 7) = (2 x 3) x (1 x 35) = (2 x 3) x (5 x 7)

From the above, we see that the sum w + x + y + z could be 19, 41, and 17, respectively. Thus, we can eliminate choice A.

We can stop here since we’ve eliminated choices A, B, C, and E. Thus, D is the correct answer.

Answer: D

When you're factoring any number that ends in 0, a good first rule is to break that number into something-times-10. 210 therefore becomes 21×10.

From there, you'll want to break each of 21 and 10 even further. 21 = 3×7, and 10 = 2×5. Since those are prime numbers, you're done factoring there.

The prime factors of 210 are 2, 3, 5, and 7.

If you sum 2 + 3 + 5 + 7, you get 17, so you can confidently say that choice A is incorrect, as 17 could be the sum of four factors of 210.

How are the other choices formed? Recognize that 1 is a factor of all integers, so you can include 1 in the factorization, too. In order to do that and have exactly four integers w, x, y, and z, however, you will need to combine two of the other primes. For example, if you combined 2 and 3 to multiply to 6, your terms would be:

1×(2×3)×5×7. This is then 1×6×5×7 = 210, so you can sum 1 + 6 + 5 + 7 to see that 19 is also a possible sum. Choice B is also then eliminated.

Going further, you can combine 2 and 5 as a single term, making the factorization 1×3×(2×5)×7, or 1×3×10×7 = 210, and the sum of those terms is 1 + 3 + 10 + 7, which is 21. Choice C is also eliminated.

Choice E is eliminated by combining larger prime factors. If you take 1×2×3×(5x7), you get 1×2×3×35 = 210, and the sum 1 + 2 + 3 + 35 = 41. So choice E is eliminated, leaving only choice D.

Choice D is therefore correct.