2020 AMC 10A Problem 18

Below is the video solution and professionally curated solution for Problem 18 of the 2020 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:paritydeterminantmultiplication principle

Difficulty rating: 1540

18.

Let (a,b,c,d)(a,b,c,d) be an ordered quadruple of not necessarily distinct integers, each one of them in the set {0,1,2,3}.\{0,1,2,3\}. For how many such quadruples is it true that adbca\cdot d-b\cdot c is odd? (For example, (0,3,1,1)(0,3,1,1) is one such quadruple, because 0131=30\cdot 1-3\cdot 1 = -3 is odd.)

4848

6464

9696

128128

192192

Video solution:
Solution video thumbnail
Play video

Click to load, then click again to play

Written solution:

Only parity matters. Modulo 22, the condition is that adbcad-bc is 11, meaning the matrix (abcd)\begin{pmatrix}a&b\\ c&d\end{pmatrix} is invertible over F2\mathbb F_2.

There are (41)(42)=6(4-1)(4-2)=6 invertible 2×22\times2 matrices over F2\mathbb F_2. Each parity pattern lifts to 24=162^4=16 choices from {0,1,2,3}\{0,1,2,3\}, so there are 616=966\cdot16=96 quadruples. Thus, C is the correct answer.

Problem 18 in Other Years