Exactly six different essays will appear in a coming issue of a magazine. Three of the essays- G, I, and Jare by the writer K, and the other three essays-M, N, and P-are by the writer Y. Each essay will appear exactly once in the issue, and an essay must appear on each of the pages 3, 6, 9, 12, 15, and 18. The order in which the essays appear in the issue will be governed by the following conditions: The essays on pages 3, 9, and 15 must all be by the same writer. I must precede P. M must precede J. If an essay by K appears on page 3, which of the following is a pair of essays either of which could appear on page 18?

Explanation: The six pages are 3, 6, 9, 12, 15, 18. Essays by K: G, I, J. Essays by Y: M, N, P. Rules: (1) Pages 3, 9, 15 have the same writer. (2) I < P (I precedes P). (3) M < J (M precedes J). Given: K's essay (G, I, or J) is on page 3. By Rule (1), the essays on pages 3, 9, and 15 must all be by **K**. The K essays are G, I, J. Thus, pages {3, 9, 15} must contain {G, I, J}. The remaining pages are {6, 12, 18}, which must contain the Y essays {M, N, P}. We need to find the possibilities for page 18 from {M, N, P}. Rule (3) is M < J. Since M is in {6, 12, 18} and J is in {3, 9, 15}, M must be on page 6 or 12, or J must be on page 15. If J is on page 15, M can be on 6, 12, or 18. If M is on page 18, it violates M < J because J is on page 3, 9, or 15 (all before 18). Since J is in {3, 9, 15}, M must be on 6 or 12. Therefore, M **cannot** be on page 18. The possible essays for page 18 are N or P. (The explanation confirms this logic: 'Y's Essay will appear on page 18 as New Rule, and as Rule III, M < J, so M \(\neq\) page 18, then only N & P = page 18').