Problem 9 · Recognize alternating ratios and compare prime exponents · 5 points
Made by AI and not reviewed by a real person.
Nora’s sequence of positive integers has this property: in any four consecutive terms, the product of the outer two equals the product of the inner two. If the 1st, 4th, and 9th terms are 2, 36, and 2592, what is the first position whose term is both a perfect cube and divisible by 2^100 × 3^101?
- 201
- 206
- 202
- 204
- 203
Show a hint
- Think about what the product condition tells you about ratios of consecutive terms, rather than calculating many terms.
- Find the first two ratios from the given terms. Then check the exponents of 2 and 3 at even and odd positions separately.
Show the answer and solution
Answer: D — 204
Let r and s be the first two ratios of consecutive terms. The product condition makes the ratios repeat as r, s, r, s, ... . The 9th term gives 2(rs)^4 = 2592, so rs = 6; the 4th gives 2r²s = 36, so r = 3 and s = 2. Thus the term at position 2k is 6^k, while the term at position 2k+1 is 2^(k+1)3^k. An odd-position term cannot be a cube because its two exponents differ by 1. At an even position, k must be at least 101 for divisibility and a multiple of 3 for the term to be a cube. The least such k is 102, giving position 204. The tempting choice 202 gives 6^101, which is divisible by the required number but is not a cube.