2025 AMC 10A Problem 11

Below is the professionally curated solution for Problem 11 of the 2025 AMC 10A, 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.

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Concepts:arithmetic sequencegeometric sequencemodular arithmetic

Difficulty rating: 1500

11.

The sequence 1,x,y,z1, x, y, z is arithmetic. The sequence 1,p,q,z1, p, q, z 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

Solution:

From the arithmetic sequence, z=1+3d,z = 1 + 3d, so z1(mod3).z \equiv 1 \pmod 3. From the geometric one, z=p3z = p^3 for some integer ratio p2.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=641(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+q=22+43+64+4+16=149.x + y + z + p + q = 22 + 43 + 64 + 4 + 16 = 149. Thus, E is the correct answer.

Problem 11 in Other Years