AMC 10A 2020 Problem 18: Parity

Let $( \textbf a, \textbf b, \textbf c, \textbf d)$ 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 $ \textbf a\cdot \textbf d - \textbf b\cdot \textbf c$ is odd?

2020 AMC 10A Problem-18

Parity

4 out of 10

Mathematics Circle

Knowledge Graph

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Use some hints

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We need exactly one term to be odd, one term to be even. Because of symmetry,let us set $\textbf a \textbf d$ to be odd and $\textbf b \textbf c$ to be even,then multiple by $2$.

Now can you complete the sum using odd and even property?

See If  $\textbf a \textbf d$ is odd, then both $\textbf a$ and $\textbf  d$ must be odd, therefore there are $2$.$2$=$4$ possibilities for $\textbf a \textbf d$.

now consider $\textbf b \textbf c$, we can say that $\textbf b \textbf c$ is even,then there are $2$.$4$=$8$ possibilities for $\textbf b \textbf c$ . However, $\textbf b$ can be odd.in that case $2$.$2$=$4$ more possibilities for $\textbf b \textbf c$. Thus there are $8$+$4$=$12$ ways for us to choose $\textbf b \textbf c$ and also $4$ ways are there to choose $\textbf a \textbf d$.

Considering symmetry, to $\textbf a \textbf d $- $\textbf b \textbf c$ be odd,there are $12$.$4$.$2$ = $96$ quadruples .So, the answer is $96$.