AMC 10 · 2025 · #11

Grade 6 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

sequences-arithmeticsequences-geometricmodular-arithmeticdivisibility-rules guess-and-check ↑ Prerequisites: sequences-arithmeticsequences-geometric

📏 Long solution 💡 3 insights

Problem

The sequence 1, x, y, z is arithmetic (add a fixed amount each step); the sequence 1, p, q, z is geometric (multiply by a fixed amount each step). They both start at 1 and end at the same value z, and 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.

Pick an answer.

(A)

(B)

(C)

103

(D)

132

(E)

149

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

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 r³ = 1 + 3d. That equation is the whole game. Tool #14 (Extreme Principle) handles the phrase "z as small as possible": since z = r³ 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, … against that whole-number condition until one passes.

1STEP 1

Name the arithmetic step

Call the added amount d: from 1 the four terms are 1, 1+d, 1+2d, and the endpoint 1+3d.

x = 1+d, y = 1+2d, z = 1+3d

2STEP 2

Name the geometric ratio

Call the ratio r: the terms are 1, r, r², and the endpoint r³, and since r equals the integer p, r is whole.

p = r, q = r², z = r³

3STEP 3

Link the shared endpoint

Both finish at the very same z, so set 1+3d = r³ and solve for the step: d = (r³ - 1)/3.

r³ = 1 + 3d ⟹ 3d = r³ - 1 ⟹ d = (r³ - 1)/3

4STEP 4

Smallest z needs smallest r

Since z = r³ grows with r, take the smallest r that keeps d whole — that is, the first one where r³ - 1 is a multiple of 3.

d = (r³ - 1)/3 ∈ Z ⇔ 3 ∣ (r³ - 1)

5STEP 5

Test the ratios in order

Cube each candidate and subtract 1: 7 and 26 both miss, but 63 is a multiple of 3, so r = 4 is the first to pass, with d = 21.

2³ - 1 = 7, 3³ - 1 = 26, 4³ - 1 = 63 = 3 · 21

6STEP 6

Rebuild both sequences and add

The sequences are 1, 22, 43, 64 and 1, 4, 16, 64, sharing z = 64, so the five requested values total 149.

x+y+z+p+q = 22+43+64+4+16 = 149 → (E)

Answer

149

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.

💡Key takeaway

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.

Name the arithmetic step
Name the geometric ratio
Link the shared endpoint
Smallest z needs smallest r
Test the ratios in order
Rebuild both sequences and add