New Calculator Rule Calculators
with algebraic manipulation capability (such as the TI-89 and
TI-92) are not permitted. This is same as the American
HS Mathematics Examination policy.
New Calculator Rule Calculators
with algebraic manipulation capability (such as the TI-89 and
TI-92) are not permitted. This is same as the American
HS Mathematics Examination policy.
Contest Dates HS Contest
dates are Dec. 1, Jan. 5, Feb. 2, Mar. 9, and Apr. 13. A registration
form for our April Algebra Course I Contest is enclosed.
Do you have a schedule conflict with our contest dates? Our rules
say that, in case of vacations, special testing days, or other
disruptions of the normal school day, you may give the
contest on an earlier day-but you must still mail scores by
Friday of the official contest week. If scores are late for due
cause, please attach a brief explanation. We reserve the right
to consider as unofficial late scores lacking such an explanation.
Received Your HS Contest Package
Late? If you have not yet received the contests,
phone 1-201-5686328 so we can ship another set. If you just
recently got the contests, please take Contest #1 as soon as
possible, even if it's late!
The Score Report and the Cumulative
Column Students on your score report must take the
contest at the exact same time. Do not include students taking
the contest during any later class period. Below is part of a
score report. The Total column is for Contest 2 totals
only. The Indiv. Cumulative is for student totals for the
first 2 contests. This column is optional; but high scoring students
not tallied here cannot be named in our newsletter. Chris Lewis
got 5's on the first 2 contests and had a cumulative total of
10. Pat Harris got a 5 and had a cumulative total of 9. Team members
may vary each contest-use your school's 5 best scores each time,
and submit additional sheets if needed. Be sure to check
the "Contest Number" box, and enter your Team Score
at the top of the form.
Please PRINT Last Name, First Name | ||||||||
Cumulative | ||||||||
| 1. Lewis, Chris | ||||||||
| 2. Harris, Pat | ||||||||
| 3. Smith, Lee | ||||||||
| 4. Nelson, Jan | ||||||||
| 5. Sun, Ronnie | ||||||||
| TEAM TOTALS | ||||||||
Completion of the "Cumulative" column is optional, but must be completed for any student who might be listed as a League high scorer.
Authentication of Scores To
give credibility to our results, we authenticate scores high enough
to win recognition. Awards indicate compliance with our rules.
Please ask students to read the Selected Math League Rules
on the back of this newsletter and sign a sheet to confirm knowledge
of the rules. Keep the signed copies. Do not send
them to us unless we request authentication from you.
General Comments About the
Contest 675 students in Tom Armbruster's school took Contest
#1. Nancy Tooker had 11 students get a 4 or 5. Pam Coryell "thought
this was very good for openers. It let younger students get correct
answers. They'll now take further contests. P.S. I love your cartoons."
Sandia Davenport liked the first 4 problems of contest 1, but
not the 5th (too susceptible to calculator solution) or #6 (too
guessable). Ted Heavenrich has "some math students who were
disappointed by the lack of mathematical challenge." Bryan
Knight thought it "tougher than in prior years, but teachers
and students enjoyed it." Susy Wong called it "challenging."
Bob Smith said "Great contest to kick off the year."
Problem 1-2: A Comment From
Us This idea is from The USSR Olympiad Problem Book
(written before calculators).
Problem 1-3: Appeal (Denied)
Two appeals for (3,0) were denied, since (3,0) is not
the sought-after value of k.
Problem 1-4: Alternate Solution
Dave Farber and Dennis Gwirtz said that 15+8 = 23, so an 8-15-17
Pythagorean triple works, and 15-8 = 7 is the answer. Not even
algebra is needed!
Problem 1-5: Comments, Appeals
(Denied), Alt. Sol'n Bryan Knight and Bob Smith sent alternate
solutions. Here's Bob's: As in our solution, a3+b3+c3
= -1998-78(a+b+c). Since x = a
is a root, x3+78x+666 = (x-a)´
(x2+ax-666/a) = (x-a)(x-b)(x-c).
Thus, (x-b)(x-c) = x2-(b+c)x+
bc = (x2+ax-666/a). Thus,
a = -b-c and a3+b3+c3
= -1998. Susy Wong, Robert Morewood, and Ted Heavenrich each had
³ 1 student use a Sharp EL-9600,
an HP-48, or a TI-85 to handle complex numbers. Morewood's "students
agree that the algebraic solution is much easier then the direct
calculator approach, once you understand the algebra." Tim
Butler used a TI-83 to locate the real root at »
-5.9. The cubic became (x+5.9)(x2-5.9x+112.8-)
= 0. After solving the quadratic for imaginary roots, he cubed
and added the three cubes together. Some students got the real
and imaginary roots and wrote their answer as an indicated
sum of three cubes, rather than an actual sum. Credit was
denied since all reasonable simplifications must be made where
failure to make them might indicate a possible lack of knowledge.
An appeal for (-1998.0000,1.074 ´
105) was denied. Finally, Dave Farber got -1980 as
an answer, but he just knew it had to be -1998. He went
back and found his error!
Problem 1-6: Comments and
an Alt Solution Robert Morewood "particularly liked the
Friday the 13th question. It was perfect for Halloween."
One student said she "wasted time trying to figure out what
the guy with the dark cloud overhead had to do with 1-6."
Dave Farber and Ben Wearn (who asked if there were any other all-female
teams) used (mod 7) to keep track of the days in each month. The
sequence of days in each month of a normal year is {31,28,31,30,31,30,31,31,30,31,30,31}.
In mod 7, this becomes {3,0,3,2,3,2,3,3,2,3,2,3}. Sum the sequence
to form a mod 7 shift sequence that tells you how many days of
the week the 13th of the month is shifted, compared to January.
For example, February is shifted 3 days. March is shifted 3+0
= 3 days. April is shifted 3+0+3 = 6 days. May is shifted 3+0+3+2
= 8 days = 1 day, etc. This sequence is {0,3,3,6,1,4,6,2,5,0,3,5}.
A quick look shows that the 13th must fall on a Friday at least
once (since all 7 digits appear) and at most 3 times (Feb., Mar.,
and Nov. are all shifted 3 days compared to Jan.). A similar analysis
holds for leap years.
Prob #, % Correct (top 5 each school)
| 1-1 97% | 1-4 84% |
| 1-2 90% | 1-5 10% |
| 1-3 72% | 1-6 50% |