The sum of four consecutive whole numbers is a multiple of 5. Which of the following statements about these four numbers is always true?

(A) The sum of the numbers ends in a 5.

(B) The largest number ends in a 9.

(C) The smallest number is odd.

(D) None of the numbers are multiples of 5.

(E) One of the numbers ends in a 3.
Let's look at a couple of examples of four consecutive whole numbers that add to a multiple of 5, and see what possibilities we can eliminate.

First, we can look at 1, 2, 3, 4 (whose sum is 10).

Using this example, we can eliminate choice (A) (since the sum ends in a 0) and choice (B) (since the largest number ends in a 4).

With a bit more work, we can see that $6,7,8,9$ (whose sum is 30) is another example.

Using this second example, we can eliminate choice (C) (since the smallest number is even) and choice

(E) (since none of the numbers ends in a 3).

Therefore, the only remaining choice is (D).

Why is (D) always true? This is a bit more tricky to figure out, and requires some algebra.

Let the smallest of the numbers be $n$. Then the other three numbers are $n+1, n+2$, and $n+3$, and their sum is $n+n+1+n+2+n+3=4 n+6$.

We know that their sum is a multiple of 5 . Since the sum is also $4 n+6$, this sum is even and so must end in a 0 if it is also to be a multiple of 5 .

Since $4 n+6$ ends in a 0 , then $4 n$ ends in a 4 . What can the units digit of $n$ be? The only possibilities for the units digit of $n$ are 1 and 6 , and so the four numbers either end with 1,2,3, and 4, or 6, 7, 8, and 9 , and none of these four numbers is a multiple of 5.

ANSWER: (D)

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