1990 AMC 8 Problem 21

Attempt Problem 21 of the 1990 AMC 8 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 1990 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

A list of 88 numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three are 16,16, 64,64, 10241024:
?,x‾,x‾,x‾,x‾,16,64,1024?, \underline{\phantom{x}}, \underline{\phantom{x}}, \underline{\phantom{x}}, \underline{\phantom{x}}, 16, 64, 1024
164\dfrac{1}{64}
14\dfrac{1}{4}
11
22
44
Answer: B
Concepts:recursionwork backwards
Small hint:
If a term is the product of the two before it, then dividing a term by the one just before it recovers the term two positions back
Big hint:
Work backwards from the end, filling one earlier term at a time by dividing
Solution:
Since each term is the product of the two before it, dividing any term by the term just before it recovers the term two positions earlier. Working backwards from the last three terms 16,16, 64,64, 10241024:
64÷16=4,16÷4=4,4÷4=1,4÷1=4,1÷4=14. \begin{gathered} 64 \div 16 = 4, \quad 16 \div 4 = 4, \\ 4 \div 4 = 1, \quad 4 \div 1 = 4, \\ 1 \div 4 = \dfrac14. \end{gathered}
These fill in the earlier terms, giving the full list 14,\dfrac14, 4,4, 1,1, 4,4, 4,4, 16,16, 64,64, 10241024. The first number is 14\dfrac14.
Thus, the correct answer is B .
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.