Grace writes a sequence of 20 numbers. The first number is 43 and each number after the first is 4 less than the number before it, so her sequence starts $43,39,35, \ldots$. How many of the numbers that Grace writes are positive?
(A) 11
(B) 9
(C) 13
(D) 15
(E) 12
Solution 1

We write out the numbers in the sequence until we obtain a negative number:

$$
43,39,35,31,27,23,19,15,11,7,3,-1
$$

Since each number is 4 less than the number before it, then once a negative number is reached, every following number will be negative.

Thus, Grace writes 11 positive numbers in the sequence.

## Solution 2

The $n$th number in the sequence is $4(n-1)$ smaller than the first number, since we will have subtracted 4 a total of $n-1$ times to obtain the $n$th number.

Therefore, the $n$th number is $43-4(n-1)=47-4 n$.

The first positive integer $n$ for which this is expression negative is $n=12$.

Therefore, Grace writes 11 positive numbers in the sequence.

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