2025 AMC 10A Problem 11

Attempt Problem 11 of the 2025 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

The sequence 1,1, x,x, y,y, zz is arithmetic. The sequence 1,1, p,p, q,q, zz is geometric. Both sequences are strictly increasing and contain only integers, and zz is as small as possible. What is the value of x+y+z+p+q?x + y + z + p + q?
6666
9191
103103
132132
149149
Answer: E
Concepts:arithmetic sequencegeometric sequencemodular arithmetic
Small hint:
Write z=1+3dz = 1 + 3d from the arithmetic sequence and z=p3z = p^3 from the geometric one
Big hint:
Since z≡1(mod3),z \equiv 1 \pmod 3, test increasing integer ratios pp until p3≡1(mod3)p^3 \equiv 1 \pmod 3
Solution:
From the arithmetic sequence, z=1+3d,z = 1 + 3d, so z≡1(mod3).z \equiv 1 \pmod 3. From the geometric one, z=p3z = p^3 for some integer ratio p≥2.p \ge 2. We want the smallest such z,z, so test p=2,3,4.p = 2, 3, 4. Only p=4p = 4 works, since p3=64≡1(mod3).p^3 = 64 \equiv 1 \pmod 3. That forces d=21,d = 21, and the sequences are 1,22,43,641, 22, 43, 64 and 1,4,16,64.1, 4, 16, 64. So x+y+z+p+qx + y + z + p + q =22+43+64+4+16= 22 + 43 + 64 + 4 + 16 =149.= 149. Thus, E is the correct answer.
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