Logic Problem Solution:
The Skating Elvises
Paul (clue 1), Wayne (4), and Art (6) finished 3rd-5th in some order in
the figure skating judging. If Paul in clue 1 had placed 3rd with the
Toronto Elvis 2nd and the "Heartbreak Hotel" skater 1st, the Baltimore
Elvis in clue 4 would have to be Paul, which he isn't (11), or the
skater to "Heartbreak Hotel," which he isn't (9). So, Paul was 4th
or 5th. If Art in clue 6 had placed 3rd with Burnsides 2nd and the
Elvis from Glasgow 1st, since the Baltimore skater isn't Burnsides
(11), the Baltimore Elvis would have been 3rd followed by the skater
to "Jailhouse Rock" 4th and Wayne 5th (4). Paul then would have skated
to "Jailhouse Rock"--no (8). So, Art wasn't 3rd in the event; Wayne
was. By clue 4, therefore, the Baltimore Elvis skated to 1st place
and the performer to "Jailhouse Rock" was 2nd. Since the Baltimore
skater's song choice wasn't "Heartbreak Hotel" (9), in clue 1, Wayne
in 3rd place performed to "Heartbreak Hotel," with the Toronto Elvis
then 4th and Paul 5th in the "Elvises on Ice" competition. Then Art
in clue 6 placed 4th, with the Glasgow skater 2nd and Burnsides 3rd.
Wayne Burnsides is from Memphis and Paul from Paris (10). By clue 2,
Stan won the figure skating and Gold finished 2nd. Gold is Lou. The
one who skated to "Can't Help Falling in Love" isn't Art or Paul from
Paris (3) and is Stan. By clue 7, Art performed to the Presley hit
"The Wonder of You," and Paul is Soleil. Paul chose "Return to
Sender" for his background song. Art is Cadillac and Stan Parker
(5). The 1st-5th place finishers in "Elvises on Ice" are
- 1st - Stan Parker, Baltimore, "Can't Help Falling in Love"
- 2nd - Lou Gold, Glasgow, "Jailhouse Rock"
- 3rd - Wayne Burnsides, Memphis, "Heartbreak Hotel"
- 4th - Art Cadillac, Toronto, "The Wonder of You"
- 5th - Paul Soleil, Paris, "Return to Sender"
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