2026 AMC 10A Problem 23

Even numbers come every $2$ steps and multiples of $3$ every $3$ steps, so together they repeat every $6$ steps. Build everything around blocks of $6$.

Step 1: what a block of 6 contains.

1. Any $6$ consecutive integers contain exactly $3$ even numbers and exactly $2$ multiples of $3$. For example, in $7, 8, 9, 10, 11, 12$ the even numbers are $8, 10, 12$ and the multiples of $3$ are $9, 12$. Starting somewhere else just shifts the pattern, so the counts stay $3$ and $2$.

2. Any list of consecutive integers can be cut into a short leftover piece of $0$ to $5$ numbers at the start, followed by $m$ full blocks of $6$ (where $m$ can be $0$). The blocks contribute $3m$ even numbers and $2m$ multiples of $3$. So
$$(p, q) = (3m, 2m) + (\text{what the leftover piece contributes}).$$
The list can start anywhere, so the leftover piece can start at any point of the repeating pattern.

Step 2: what the leftover piece can contribute.

1. Only the starting point within the pattern matters, so try leftover pieces starting at each of $1, 2, 3, 4, 5, 6$. For each length, write down the pair (even numbers, multiples of $3$). For example, the five numbers $4, 5, 6, 7, 8$ contain the even numbers $4, 6, 8$ and the single multiple of $3$, namely $6$, giving $(3, 1)$.

1. Collect the different extras: $(0,0)$, $(1,0)$, $(0,1)$, $(1,1)$, $(2,1)$, $(2,2)$, $(3,1)$ and $(3,2)$. The last one adds exactly one more block's worth, so $(3m, 2m) + (3, 2)$ is the same as the $(0,0)$ extra with $m + 1$ blocks. Nothing new there.

5. So for each $m = 0, 1, 2, \ldots$ we get a group of $7$ possible pairs:
$$(3m, 2m),\ (3m{+}1, 2m),\ (3m, 2m{+}1),\ (3m{+}1, 2m{+}1),\ (3m{+}2, 2m{+}1),\ (3m{+}2, 2m{+}2),\ (3m{+}3, 2m{+}1).$$
Every one of these really happens: use $m$ full blocks plus the right leftover piece. (For $m = 0$ the pair $(0, 0)$ comes from a one-number list such as $1$, since the list cannot be empty.) As a check, the pair $(6, 3)$ in group $1$ comes from the $11$ numbers $4, 5, \ldots, 14$: even numbers $4, 6, 8, 10, 12, 14$ and multiples of $3$ $6, 9, 12$.

Step 3: no pair is counted twice.

6. Give each pair a "fingerprint": the number $2p - 3q$. For a pair in group $m$,
$$2(3m + x) - 3(2m + y) = 2x - 3y,$$
so the $m$ cancels and the fingerprint depends only on which of the seven patterns it is. The seven patterns have fingerprints $0, 2, -3, -1, 1, -2, 3$, all different.

1. So two pairs from different patterns have different fingerprints, and two pairs from the same pattern but different groups have different values of $p$. Every pair appears only once.

Step 4: count the pairs with $p$ and $q$ at most 100.

1. In group $m$ the largest $p$ is $3m + 3$ and the largest $q$ is $2m + 2$, which is smaller. So $p$ is the one that runs into the cap.

1. Groups $m = 0$ to $32$. The largest $p$ is $3 \cdot 32 + 3 = 99$ and the largest $q$ is $66$, so all $7$ pairs fit. That gives $33 \times 7 = 231$ pairs.

10. Group $33$. Its pairs are
$$(99, 66),\ (100, 66),\ (99, 67),\ (100, 67),\ (101, 67),\ (101, 68),\ (102, 67).$$
The first four fit; the last three have $p$ above $100$. That gives $4$ pairs.

1. Groups $34$ and up. Every $p$ is at least $3 \cdot 34 = 102$. Nothing fits.

The total is $231 + 4 = 235$.

The answer is $\boxed{\textbf{(D)}\ 235}$.

Divisibility by $2$ and by $3$ repeats every $6$ numbers, so a long list behaves exactly like a short one plus some full blocks. That reduces an endless question to a table of $6$ small cases. The fingerprint $2p - 3q$ measures how far the two counts are from being perfectly in step, and it is what keeps the listing honest. In problems like this, spend most of your care on the edge of the range: here the answer choices differ by $1$.

Describe all possible pairs with one inequality, then count row by row.

1. In a list of $L$ numbers, the number of even numbers is $\frac{L}{2}$ rounded down or up, so $2p$ is within $1$ of $L$. The number of multiples of $3$ is $\frac{L}{3}$ rounded down or up, so $3q$ is within $2$ of $L$. Hence $2p$ and $3q$ differ by at most $3$.

1. The other direction also holds: every pair of whole numbers with $2p - 3q$ between $-3$ and $3$ appears. Each such fingerprint belongs to one of the seven patterns $(x, y)$ of step 5. Then $2(p - x) = 3(q - y)$, so $p - x$ is a multiple of $3$, say $3m$, and $q - y = 2m$; checking the seven patterns shows that $p$ and $q$ being at least $0$ forces $m \ge 0$. So $(p, q)$ is in group $m$.

3. So count whole numbers $p, q$ from $0$ to $100$ with $3q - 3 \le 2p \le 3q + 3$. For each $q$, look at which $p$ fit:
- $q$ odd: $3q - 3$ and $3q + 3$ are even, so $p$ runs over $4$ values. For example $q = 1$ gives $p = 0, 1, 2, 3$.
- $q$ even (not $0$): $p$ runs over $3$ values, for example $q = 2$ gives $p = 2, 3, 4$.
- $q = 0$: $p$ can only be $0$ or $1$ ($p = -1$ is not allowed), so $2$ values.
- $q = 67$: $p$ could be $99, 100, 101, 102$, but only $99$ and $100$ are allowed, so $2$ values.
- $q = 68$ or more: $p$ would have to be at least $101$. None.

1. From $q = 1$ to $66$ there are $33$ odd values and $33$ even values, giving $33 \cdot 4 + 33 \cdot 3 = 231$. Adding the two ends: $2 + 231 + 2 = 235$.