2025 AMC 10A Problem 11

P11-15~4 minPrint

Sequences & Series Modular Arithmetic Organized listing / direct enumeration Set up the equation/formula and compute; no special trick needed

The sequence $1,x,y,z$ is arithmetic. The sequence $1,p,q,z$ is geometric. Both sequences are strictly increasing and contain only integers, and $z$ is as small as possible. What is the value of $x+y+z+p+q$ ?

Show solution without answering

Key insight

The geometric sequence 1, p, q, z makes z a perfect cube. The arithmetic one makes z one more than a multiple of 3. Test cubes 8, 27, 64.

Solution

The two sequences share only the number $z$ . So find out what each sequence says about $z$ , then find the smallest $z$ that satisfies both.

1. The geometric sequence $1, p, q, z$ . Each term is the one before times the same ratio. Starting from $1$ , the ratio is $p$ itself, so the terms are
$1,\ p,\ p^2,\ p^3.$
Since $p$ is a term, it is a whole number, and "increasing" means $p$ is at least $2$ . So $z = p^3$ is a perfect cube: $8, 27, 64, \ldots$

2. The arithmetic sequence $1, x, y, z$ . Each step adds the same whole number $d$ . So the terms are $1,\ 1 + d,\ 1 + 2d,\ 1 + 3d$ , and
$z = 1 + 3d.$
That means $z - 1$ must be a multiple of $3$ .

Test cubes from smallest up.

$p$	$z = p^3$	$z - 1$	multiple of $3$ ?
$2$	$8$	$7$	no
$3$	$27$	$26$	no
$4$	$64$	$63$	yes, $63 = 3 \cdot 21$

So the smallest possible $z$ is $64$ .

Write out both sequences. With $d = 21$ the arithmetic one is $1, 22, 43, 64$ . With $p = 4$ the geometric one is $1, 4, 16, 64$ . Both are increasing and made of whole numbers.

5. Add up what is asked.
$x + y + z + p + q = 22 + 43 + 64 + 4 + 16 = 149.$

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

Why this works

A geometric sequence that starts at $1$ shows its ratio as the second term, so "all whole numbers" forces a whole-number ratio and makes the last term a cube. When two sequences share one term, turn each sequence into a rule about that term, then search upward for the smallest number that obeys both rules. That is the usual way to handle "as small as possible" questions.

Alternative approach

Find a formula for the total first. In the arithmetic sequence, $x + y = (1 + d) + (1 + 2d) = 2 + 3d$ , which is $1 + z$ . In the geometric one, $p + q = p + p^2$ . So
$x + y + z + p + q = 1 + 2z + p + p^2.$
For $p = 2$ this is $1 + 16 + 2 + 4 = 23$ , and for $p = 3$ it is $1 + 54 + 3 + 9 = 67$ . Neither is a choice, which agrees with those cubes failing. For $p = 4$ it is $1 + 128 + 4 + 16 = 149$ , choice (E).

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(E)}\ 149}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)