2020 AMC 10A Problem 18

P16-20~4 minPrint

Number Properties Basic Counting Use an invariant, parity, or coloring argument Exploit symmetry to reduce work or pair up objects

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

Show solution without answering

Key insight

The difference is odd when exactly one of the two products is odd. A product of two numbers from 0 to 3 is odd for only 4 of 16 pairs.

Solution

Only odd and even matter here, so first figure out when a difference is odd.

Odd minus even is odd, and even minus odd is odd. But odd minus odd and even minus even are even. So exactly one of $ad$ and $bc$ must be odd.
When is $ad$ odd? A product is odd only when both numbers are odd. In $\{0,1,2,3\}$ the odd numbers are $1$ and $3$ . So $a$ has $2$ choices and $d$ has $2$ choices: $2 \times 2 = 4$ pairs give an odd $ad$ .
There are $4 \times 4 = 16$ pairs $(a, d)$ in all, so $16 - 4 = 12$ pairs give an even $ad$ .
The same counts hold for $(b, c)$ : $4$ odd, $12$ even.
Case 1: $ad$ odd and $bc$ even: $4 \times 12 = 48$ quadruples.
Case 2: $ad$ even and $bc$ odd: $12 \times 4 = 48$ quadruples.
Total: $48 + 48 = 96$ .

Check with the example: $(0,3,1,1)$ has $ad = 0$ (even) and $bc = 3$ (odd), so it is in Case 2.

The answer is $\boxed{\textbf{(C)}\ 96}$ .

Why this works

For questions about odd and even, replace each number by just "odd" or "even" and count choices. A product is odd only if every factor is odd, and a difference is odd only if the two parts differ in type. The two cases here mirror each other, so you can count one and double it.

Alternative approach

Think in chances. Each of $ad$ and $bc$ is odd with chance $\frac{4}{16} = \frac14$ , and they do not affect each other. Exactly one is odd with chance $\frac14 \cdot \frac34 + \frac34 \cdot \frac14 = \frac38$ . There are $4^4 = 256$ quadruples, and $\frac38 \times 256 = 96$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

A difference is odd exactly when its two terms have opposite parity, so we need one of $ad$ , $bc$ odd and the other even.

Count ordered pairs $(a, d)$ from $\{0,1,2,3\}^2$ ( $16$ pairs) by the parity of $ad$ . The product is odd only when both factors are odd, and each has $2$ odd choices: $2 \cdot 2 = 4$ pairs. The remaining $16 - 4 = 12$ pairs give an even product. The same counts hold for $(b, c)$ .

Quadruples with $ad$ odd and $bc$ even: $4 \cdot 12 = 48$ . Quadruples with $ad$ even and $bc$ odd: $12 \cdot 4 = 48$ . Total
$48 + 48 = 96.$

The answer is $\boxed{\textbf{(C)}\ 96}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Algebra, Chapter 20: Special Functions (§20.3)