AMC 10 · 2025 · #11
Grade 6 algebraThe 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?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two four-term sequences both start at $1$ and end at the same value $z$. One is arithmetic (add a fixed amount each step); the other is geometric (multiply by a fixed amount each step). Both are strictly increasing and every term is an integer. Make $z$ as small as it possibly can be, then compute $x + y + z + p + q$.
Givens: The sequence $1, x, y, z$ is arithmetic — each term adds the same fixed amount; The sequence $1, p, q, z$ is geometric — each term multiplies by the same fixed amount; Both sequences are strictly increasing and every term is an integer; The two sequences share the same first term $1$ and the same last term $z$; Among all valid choices, $z$ is made as small as possible; Answer choices: (A) $66$, (B) $91$, (C) $103$, (D) $132$, (E) $149$
Unknowns: The value of the sum $x + y + z + p + q$
Understand
Restated: Two four-term sequences both start at $1$ and end at the same value $z$. One is arithmetic (add a fixed amount each step); the other is geometric (multiply by a fixed amount each step). Both are strictly increasing and every term is an integer. Make $z$ as small as it possibly can be, then compute $x + y + z + p + q$.
Givens: The sequence $1, x, y, z$ is arithmetic — each term adds the same fixed amount; The sequence $1, p, q, z$ is geometric — each term multiplies by the same fixed amount; Both sequences are strictly increasing and every term is an integer; The two sequences share the same first term $1$ and the same last term $z$; Among all valid choices, $z$ is made as small as possible; Answer choices: (A) $66$, (B) $91$, (C) $103$, (D) $132$, (E) $149$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #14 Extreme Principle, #6 Guess and Check
Nothing is numbered yet, so I name the two hidden rules. Tool #4 (Introduce a Variable): let $d$ be the amount the arithmetic sequence adds and $r$ the amount the geometric one multiplies by. Then every term of both sequences is a formula in $d$ or $r$, and the one fact tying them together — they end at the same $z$ — becomes a single equation $r^3 = 1 + 3d$. That equation is the whole game. Tool #14 (Extreme Principle) handles the phrase "$z$ as small as possible": since $z = r^3$ climbs as $r$ climbs, the smallest $z$ comes from the smallest usable $r$, so I only need the first ratio that keeps every term a whole number. Tool #6 (Guess and Check) then tests $r = 2, 3, 4, \dots$ against that whole-number condition until one passes.
Execute — Answer: E
6.EE.A.2 Step 1 Name the arithmetic step
- The arithmetic sequence starts at $1$ and adds the same amount $d$ each time.
- So the four terms are $1$, $1+d$, $1+2d$, $1+3d$.
- Matching these to $1, x, y, z$ gives every arithmetic term as a formula in the single unknown $d$.
💡 One repeated step means the last term is just the start plus three steps.
6.EE.A.1 Step 2 Name the geometric ratio
- The geometric sequence starts at $1$ and multiplies by the same amount $r$ each time.
- So the four terms are $1$, $r$, $r^2$, $r^3$.
- Matching these to $1, p, q, z$ gives every geometric term as a power of the single unknown $r$.
- Because the terms are integers and the first is $1$, the ratio $r = p$ is itself a whole number.
💡 Multiplying by $r$ three times turns the start $1$ into $r^3$.
6.EE.B.7 Step 3 Link the shared endpoint
- Both sequences finish at the very same $z$.
- Setting the two formulas for $z$ equal ties the two unknowns together in one equation, then solving for $d$ shows exactly what the geometric side demands of the arithmetic side.
💡 Sharing the last term is the one bridge between the two rules.
4.OA.B.4 Step 4 Smallest z needs smallest r
- Since $z = r^3$ grows as $r$ grows, the smallest possible $z$ comes from the smallest allowed $r$.
- What makes an $r$ allowed?
- The formula $d = (r^3 - 1)/3$ must land on a whole number, so $r^3 - 1$ must be a multiple of $3$.
- That divisibility test is the only hurdle, so I look for the first ratio $r \ge 2$ that clears it.
💡 A bigger ratio only makes the endpoint bigger, so take the smallest ratio that works.
6.EE.A.1 Step 5 Test the ratios in order
- Check the candidates from smallest up.
- For each $r$, cube it, subtract $1$, and ask whether the result is a multiple of $3$.
- The first two fail; $r = 4$ is the first that works, giving $d = 63/3 = 21$.
💡 Try ratios in order and stop at the first one that keeps every term whole.
6.EE.A.2 Step 6 Rebuild both sequences and add
- With $r = 4$ and $d = 21$, write out both sequences and read off the five requested values.
- The arithmetic sequence is $1, 22, 43, 64$ and the geometric sequence is $1, 4, 16, 64$; both are strictly increasing integers ending at the shared $z = 64$.
- Adding the five values gives the answer.
💡 Plug the winning $d$ and $r$ back in and total the pieces.
6.EE.A.2 The arithmetic sequence starts at $1$ and adds the same amount $d$ each time. So 6.EE.A.1 The geometric sequence starts at $1$ and multiplies by the same amount $r$ each 6.EE.B.7 Both sequences finish at the very same $z$. Setting the two formulas for $z$ equ 4.OA.B.4 Since $z = r^3$ grows as $r$ grows, the smallest possible $z$ comes from the sma 6.EE.A.1 Check the candidates from smallest up. For each $r$, cube it, subtract $1$, and 6.EE.A.2 With $r = 4$ and $d = 21$, write out both sequences and read off the five reques Review
Reasonableness: Both rebuilt sequences pass every rule: $1, 22, 43, 64$ is strictly increasing with a constant step of $21$, and $1, 4, 16, 64$ is strictly increasing with a constant ratio of $4$, and they share the last term $64$. No smaller $z$ is possible because $r = 2$ and $r = 3$ both fail the multiple-of-$3$ test, and $r$ must be a whole number (it equals the integer $p$), so no fractional ratio sneaks in below $4$. The sum $149$ is choice (E); it is the largest option, which fits a problem that forces the endpoint up to $64$ before the integer rules line up.
Alternative: Tool #5 (Look for a Pattern) skips the algebra on $d$. The geometric endpoints are the cubes $8, 27, 64, 125, \dots$, while the arithmetic endpoints $1 + 3d$ are exactly the numbers one more than a multiple of $3$: $4, 7, 10, 13, \dots$ — that is, values $\equiv 1 \pmod 3$. Scan the cubes for the first one in that list: $8 \equiv 2$, $27 \equiv 0$, $64 \equiv 1$. So $z = 64$ is the smallest shared endpoint, and the rest follows the same way.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Writing the arithmetic terms $1+d, 1+2d, 1+3d$ from the common difference $d$, and substituting the winning $d, r$ to total $x+y+z+p+q$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Expressing the geometric terms as $r, r^2, r^3$ and evaluating the cubes $2^3, 3^3, 4^3$ when testing ratios.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $3d = r^3 - 1$ for the common difference $d = (r^3 - 1)/3$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Testing whether $r^3 - 1$ is a multiple of $3$ so that $d$ is a whole number, and finding $r = 4$ is the smallest ratio that passes.)
⭐ Give each hidden rule a letter, use the shared endpoint to link them in one equation, then push that endpoint down to the smallest value the whole-number rules still allow.
⭐ Give each hidden rule a letter, use the shared endpoint to link them in one equation, then push that endpoint down to the smallest value the whole-number rules still allow.
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