2025 AMC 10A Problem 11

The geometric sequence. Its first term is $1$, so its ratio is $p/1 = p$, an integer, and the sequence is $1, p, p^2, p^3$ with $z = p^3$. Strictly increasing forces $p \ge 2$.

The arithmetic sequence. With common difference $d$ the terms are $1, 1+d, 1+2d, 1+3d$, so
$$z = 1 + 3d, \qquad\text{that is}\qquad z \equiv 1 \pmod 3,$$
and $d \ge 1$.

So $z$ must be a cube $p^3$ with $p \ge 2$ and $p^3 \equiv 1 \pmod 3$. Since $p^3 \equiv p \pmod 3$ (check $p \equiv 0, 1, 2$ in turn), this says simply $p \equiv 1 \pmod 3$. Test in order:

• $p = 2$: $z = 8$, and $8 - 1 = 7$ is not a multiple of $3$. Fails.
• $p = 3$: $z = 27$, and $27 - 1 = 26$ is not a multiple of $3$. Fails.
• $p = 4$: $z = 64$, and $64 - 1 = 63 = 3 \cdot 21$. Works, with $d = 21$.

The smallest possible $z$ is therefore $64$, giving $1, 22, 43, 64$ and $1, 4, 16, 64$. Both are strictly increasing sequences of integers, so
$$x + y + z + p + q = 22 + 43 + 64 + 4 + 16 = 149.$$

The answer is $\boxed{\textbf{(E)}\ 149}$.

A geometric sequence whose first term is $1$ has its ratio sitting right there as the second term, so "integer terms" immediately means "integer ratio" and kills all fractional ratios at once. The two sequences meet only at $z$, so the whole problem is the intersection of two descriptions of $z$: a cube, and one more than a multiple of $3$. Turning each condition into a statement about $z$ alone, then searching upward, is the standard way to handle "as small as possible" problems.

Work from the answers. Each choice is $x + y + z + p + q$, and for the arithmetic part $x + y = 2 + 3d = 1 + z$, while for the geometric part $p + q = p + p^2$ with $z = p^3$. So the total is $1 + 2z + p + p^2$. Trying $p = 4$, $z = 64$ gives $1 + 128 + 4 + 16 = 149$; trying $p = 2$ or $p = 3$ gives $1 + 16 + 6 = 23$ or $1 + 54 + 12 = 67$, neither of which is offered, confirming that those values of $z$ are excluded.

• Taking the smallest cube z = 8 because 1, 2, 4, 8 is the smallest increasing integer geometric sequence, without checking that 1, x, y, z arithmetic needs z - 1 divisible by 3.
• Allowing a non-integer ratio such as $3/2$; then $p = 3/2$ is not an integer, so the sequence fails the "only integers" condition even though $z$ might be an integer.
• Summing only $x + y + z$ or forgetting that $z$ itself appears once in the requested total.