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,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

Answer: E
Concepts:arithmetic sequencegeometric sequencemodular arithmetic
Difficulty rating: 1500
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+qx + y + z + p + q =22+43+64+4+16= 22 + 43 + 64 + 4 + 16 =149.= 149. Thus, E is the correct answer.

← Problem 10#10
Full Exam

Problem 11 in Other Years