Difference between revisions of "2020 AMC 10A Problems/Problem 18"

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== Solution ==
 
== Solution ==
In order for <math>a\cdot d-b\cdot c</math> to be odd, consider parity. We must have (even number)-(odd number) or (odd number)-(even number).
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In order for <math>a\cdot d-b\cdot c</math> to be odd, consider parity. We must have (even number)-(odd number) or (odd number)-(even number). Note that by
  
 
==See Also==
 
==See Also==

Revision as of 21:57, 31 January 2020

Problem

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.)

$\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192$

Solution

In order for $a\cdot d-b\cdot c$ to be odd, consider parity. We must have (even number)-(odd number) or (odd number)-(even number). Note that by

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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