2026 AMC 10A Problem 18

The rule says: take half of the previous term, then add $2$. Before doing any algebra, try it on a number and watch what happens.

1. Try a starting value. Start at $a_0 = 20$:
$$20 \to 12 \to 8 \to 6 \to 5 \to 4.5$$
The sixth term is already a fraction, so $20$ is far too small. Notice that the terms are creeping down toward $4$.

1. Why $4$ is special. If a term equals $4$, the next one is $\frac12 \cdot 4 + 2 = 4$ again. A sequence that reaches $4$ is stuck there forever.

1. Look at the distance above $4$. Here is the example again, with each term's distance above $4$ underneath:

Each distance is exactly half of the one before. This always happens. If a term is $4 + d$, the next term is
$$\frac12(4 + d) + 2 = 2 + \frac{d}{2} + 2 = 4 + \frac{d}{2}.$$
So the rule simply halves the distance above $4$.

4. What the two conditions say about the distance.
- Every term is a whole number, so every distance is a whole number.
- The terms must go down. If the distance were $0$, the terms would all equal $4$ and stay put. If a term were below $4$ (a negative distance), halving would bring it up toward $4$. So the distance must be positive at every step.

1. Count the halvings. From $a_0$ to $a_{10}$ there are $10$ steps, not $11$. So the distance at $a_{10}$ equals the starting distance divided by $2^{10} = 1024$. That final distance must be a positive whole number, so it is at least $1$. Therefore the starting distance is at least $1024$.

6. The smallest start. A starting distance of exactly $1024$ works: the distances are $1024, 512, 256, \ldots, 2, 1$, all positive whole numbers. That gives $a_0 = 4 + 1024 = 1028$, and the sequence is
$$1028,\ 516,\ 260,\ 132,\ 68,\ 36,\ 20,\ 12,\ 8,\ 6,\ 5.$$
That is $11$ positive integers, each smaller than the one before. So $N = 1028$.

1. Digit sum. $1 + 0 + 2 + 8 = 11$.

The answer is $\boxed{\textbf{(D)}\ 11}$.

A rule of the form "multiply by a number, then add a constant" always has one value it leaves unchanged; here it is $4$, found by solving $x = \frac{x}{2} + 2$. Measuring every term from that value turns the rule into plain halving, which is easy to follow for many steps. Next time you see a rule like this, find the unchanged value first and measure from it.

Run the rule backwards. Undoing "halve, then add $2$" means "subtract $2$, then double", that is, $a_{k-1} = 2a_k - 4$.

1. Every term must be bigger than $4$ (see step 4 above), so the last term $a_{10}$ is at least $5$.
2. Running the rule backwards keeps whole numbers whole, and a bigger last term gives a bigger first term. So the smallest $a_0$ comes from the smallest last term, $a_{10} = 5$.
3. Double and subtract $4$, ten times:
$$5 \to 6 \to 8 \to 12 \to 20 \to 36 \to 68 \to 132 \to 260 \to 516 \to 1028.$$

So $N = 1028$, and its digit sum is $11$.