1994 AIME Problem 1

Attempt Problem 1 of the 1994 AIME 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 1994 AIME solutions, or check the answer key.

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1.

The increasing sequence 3,3, 15,15, 24,24, 48,48, \ldots consists of those positive multiples of 33 that are one less than a perfect square. What is the remainder when the 19941994th term of the sequence is divided by 1000?1000?

Answer: 63
Concepts:modular arithmeticcounting integers in a rangeperfect square
Difficulty rating: 1640
Small Hint:

A number k21k^2-1 is divisible by 33 exactly when kk is not divisible by 33

Big Hint:

Index the eligible values of kk in pairs, then reduce the required square modulo 10001000

Solution:

The terms are k21k^2-1 for integers k2k\geq2 not divisible by 3.3. In each block of three consecutive kk’s there are two eligible values. The 1994=29971994=2\cdot997th corresponds to k=3(997)+1=2992.k=3(997)+1=2992. Since 29929928(mod1000),2992\equiv992\equiv-8\pmod {1000}, k21(8)2163(mod1000).\begin{aligned}k^2-1&\equiv(-8)^2-1\\&\equiv63\pmod {1000}.\end{aligned}

Full Exam

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