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2025 AMC 10A Problem 11

Problem 11 of 25IntermediateAlgebraNumber Theory

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?

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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.

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Concepts: arithmetic sequence · geometric sequence · modular arithmetic

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.