On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$
children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the $15$
children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who
answered yes.
How many pieces of candy in all did the principal give to the children who always
tell the truth?
$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$
Note that:

Truth-tellers would answer yes-no-no to the three questions in this order.
Liars would answer yes-yes-no to the three questions in this order.
Alternaters who responded truth-lie-truth would answer no-no-no to the three questions in this order.
Alternaters who responded lie-truth-lie would answer yes-yes-yes to the three questions in this order.
Suppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.
The conditions of the first two questions imply that
\begin{align*} T+L+A&=22, \\ L+A&=15. \end{align*}
Subtracting the second equation from the first, we have $T=22-15=\boxed{\textbf{(A) } 7}.$
Remark
The condition of the third question is extraneous. However, we know that there are $9$ alternaters who responded lie-truth-lie, $6$ liars, and $9$ alternaters who responded truth-lie-truth from this condition.
~sigma ~[orenbad](https://artofproblemsolving.com/wiki/index.php/OrenSH) ~MRENTHUSIASM