1988 AIME Problem 6

Attempt Problem 6 of the 1988 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 1988 AIME solutions, or check the answer key.

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

6.

It is possible to place positive integers into the vacant twenty-one squares of the 5×55\times5 square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (∗).(*).
Answer: 142
Concepts:arithmetic sequencepattern recognitionsystem of equations
Small hint:
Index the rows and columns from 00 through 44
Big hint:
A grid whose rows and columns are arithmetic has the form A+Bi+Cj+DijA+Bi+Cj+Dij
Solution:
Index rows and columns from 00 to 4.4. The row and column conditions give the general entry A+Bi+Cj+Dij.A+Bi+Cj+Dij. The four shown values yield A+4B=0,A+B+C+D=74,A+2B+4C+8D=186,A+3B+2C+6D=103.\begin{aligned}A+4B&=0,\\A+B+C+D&=74,\\A+2B+4C+8D&=186,\\A+3B+2C+6D&=103.\end{aligned} Solving gives A=52,A=52, B=−13,B=-13, C=30,C=30, and D=5.D=5. The asterisk is at (i,j)=(0,3),(i,j)=(0,3), so its value is A+3C=52+90=142.A+3C=52+90=142.
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.