Week of February 4
Problem XVI  Super Bowl Problem
Grades eligible: All
In NFL Football, quarterbacks are evaluated statistically by a passer rating. This rating is obtained by a rather lengthy formula.
(Definitions: A completion is a successful pass; an interception is a pass completed to the opposing team, and does not count as a completion.)
1. Take the number of completions, and divide that by the number of pass attempts, and multiply this number by 100; from that number subtract 30, and divide the difference by 20.
2. Take the number of touchdown passes, and divide that by the number of pass attempts, and multiply this number by 100; and divide this product by 5.
3. From the number 9.5 take away the number of interceptions divided by the number of pass attempts multiplied by 100, and divide this difference by 4.
4. Take the total number of passing yards, and divide that by the number of pass attempts; from that number subtract 3, and divide the difference by 4.
5. Sum all four of these numbers, and divide by six hundredths (.06) for the passer rating.
Example: The following is the statistical line for Kansas City Chiefs’ QB Len Dawson from Super Bowl IV.
Quarterback 
CompletionsAttempts 
Passing Yards 
Touchdowns 
Interceptions 
Len Dawson (KC) 
1217 
142 
1 
1 
1. (12/17) x 100 = (1,200/17)  30 = (690/17)/20 = (690/340) = (69/34)
2. (1/17) x 100 = (100/17)/5 = (100/85) = (40/34)
3. [9.5 – ((1/17) x 100) = [9.5 – (100/17)] = (61.5/17)/4 = (61.5/68) = (30.75/34)
4. (142/17) – 3 = (91/17)/4 = (91/68) = (45.5/34)
5. (69/34) + (40/34) + (30.75/34) + (45.5/34) = (185.5/34)/.06 = (18,550/204), or about 90.9
Problem: Rank the five passer ratings of the following quarterbacks from highest to lowest, and show your work to confirm your order.
Quarterback 
CompletionsAttempts 
Passing Yards 
Touchdowns 
Interceptions 
Daryle Lamonica (Oakland, Super Bowl II) 
1534 
208 
2 
1 
Terry Bradshaw (Pittsburgh, Super Bowl XIV) 
1421 
309 
2 
3 
Joe Theismann (Washington, Super Bowl XVIII) 
1635 
243 
0 
2 
Joe Montana (San Francisco, Super Bowl XXIV) 
2229 
297 
5 
0 
Rex Grossman (Chicago Bears, Super Bowl XLI) 
2028 
165 
1 
2 
Submit solutions to mike@Hinsdale60521.com; all work must be shown to receive consideration. The submission deadline is Friday at 10 p.m., and the solution will be provided on Saturday morning.
Week of January 7
Problem XV  Roman Numerals
The Romans made use of seven numeral letters to express their numbers—i.e.,
I = 1
V = 5
X = 10
L = 50
C = 100
D = 500
M = 1,000
Rule I. A less numeral before a greater decreases its value; e.g., IV = 4, or 5 – 1.
Rule II. A less numeral after a greater increases its value; e.g., VI = 6, or 5 + 1.
Rule III. When expressing numbers 1,000 or less, no numeral is repeated more than four consecutive times; e.g., IX = 9, not VIIII; XL = 40, not XXXX; XC = 90, not LXXXX.
Problem: Express the following numbers as Roman numerals.
a. 3
b. 25
c. 109
Now express the following Roman numerals as numbers.
a. VIII
b. XXI
c. LXII
Extra Challenge: express the year 2013 as a Roman numeral.
Submit solutions to mike@Hinsdale60521.com; all work must be shown to receive consideration. The submission deadline is Friday at 10 p.m., and the solution will be provided on Saturday morning.
Week of December 17
Problem XIV  Common Measures
A measure of a quantity is some number by which it can be divided without leaving a remainder. For example, 4 is a measure of 8, and 5 is a measure of 20.
That measure is a common measure which measures more than one quantity; e.g., 7 is a common measure of 14 and 28, and 8 is a common measure of 48, 64, and 72.
The greatest number that measures several quantities is called the greatest common measure (or also the greatest common factor).
Find the greatest common measure of:
1) 35 and 77
2) 48 and 244
3) 25, 75, and 125
4) 42a and 90a
Grades eligible: All
Submit solutions to mike@Hinsdale60521.com; all work must be shown to receive consideration. The submission deadline is Friday at 10 p.m., and the solution will be provided on Saturday morning.
Week of December 3
Problem XIII  Fellowship
Three sailors, A, B, and C, freighted a ship with 340 tons of tea. A loaded 120 tons, B loaded 96, and C the remainder. In a storm, the sailors were compelled to throw 85 tons overboard; how much must each sailor throw overboard?
Grades eligible: All
Submit solutions to mike@Hinsdale60521.com; all work must be shown to receive consideration. The submission deadline is Friday at 10 p.m., and the solution will be provided on Saturday morning.
Week of November 19
Problem XII  De Morgan's Problem
For the following problem, refer to the diagram attached in this Word Document.
A and B are objects viewed under a microscope, moving towards the point Q within the line CD. A and B move in such a way that AQ is ¼ of an inch when BQ is ½ of an inch; and AQ is 1/9 of an inch when BQ is ⅓ of an inch; and so on, AQ always being xsquared inches when BQ is x inches. The microscope’s magnifying power is adjusted so that AQ always appears to be the same length. Does B appear to move towards or away from Q? Explain your answer.
Grades eligible: All
Submit solutions to mike@Hinsdale60521.com; all work must be shown to receive consideration. The submission deadline is Friday at 10 p.m., and the solution will be provided on Saturday morning.
Week of November 12
Problem XI  Solar System
Winner: Kayleigh Hester (Hinsdale Middle School)
Honorable Mention: Alec Hill, Will Lopez, Allegra Dugan, Jenny Witt, Avalon CharltonPerrin, Lindsay Peters, and Annie Krillenberger
Of the following theories, determine which of the three you think is the correct one, and explain why you believe that to be the case. The more correct and thoughtful reasons provided, the better chance you will have of winning this week’s contest.
1. Ptolemaic (or Geocentric) Theory—The Earth is stationary at the center of the system, and the Sun, Moon, and all of the planets revolve around it.
2. Copernican (or Heliocentric) Theory—The Sun is at the center of the system, and the Earth and all of the other planets revolve around it.
3. Tychonian Theory—All of the planets revolve around the Sun, which revolves around the Earth.
Solution
The most correct of the three theories given above is the second, called the Copernican Theory after its founder, Nicolaus Copernicus, an eminent clergyman, mathematician, and astronomer who arrived at this conclusion around 1500. This theory is also sometimes called heliocentric, which means suncentered.[*]
I think most of those who attempted this problem were aware that the Earth and planets revolve around the Sun, but take this fact for granted which, in reality, is not at all obvious. Even the illustrious Lord Bacon did not accept the Copernican Theory (although a number of the Ancients believed in it). After all, the Ptolemaic Theory, i.e., that the Sun and planets revolve around the Earth, is the most natural conclusion to be drawn at first sight. The language we use to describe the Sun in relation to the Earth is really Ptolemaic in nature; e.g., the Sun rises in the East and sets in the West, or the Sun passes from East to West in the sky—even though it is well known that this motion is really apparent.
But there are a number of reasons that can be given to confirm the truth of the Copernican Theory, and I will list a few of them in this place.
1. Using the telescope, Galileo found that the phases of Venus resemble the phases of the Moon. And since we know that the Moon revolves around the Earth, Galileo inferred that Venus likewise revolves around the Sun.
2. Galileo also observed that Jupiter’s four moons revolve around Jupiter in a system similar to the one proposed by Copernicus.
3. Sir Isaac Newton discovered the Law of Universal Gravitation, or the rule that all bodies are attracted to each other by the force of Gravity. The Sun being far and away the largest body in our system, it consequently has the strongest gravity, and acts upon all of the planets much more than the planets severally act upon it.
4. There is a phenomenon astronomers call Aberration, or the apparent change in position of stars in the sky.[†] This phenomenon can only be explained by the combination of the motion of light and the motion of the Earth.
M.E.
[*] Actually, the Sun is not at the center of any of the planets’ orbits, but rather at a focus within their elliptical orbits. Kepler discovered that a planet describes an ellipse, not a circle or series of circles, in revolving around the Sun. An ellipse contains two points called foci, whose combined distances from any given point of the circumference are equal to the length of the ellipse’s transverse diameter, or length across an ellipse. In the case of our Solar System, within each planetary orbit, the Sun resides at one of these two foci.
[†] For those at all acquainted with Astronomy, the Aberration is neither the same thing as nor connected at all to the Precession of the Equinoxes.
Problem X  Mortality
Mortality rates suggest the length of life we expect someone to reach, according to data already compiled.
Use the table below to answer the following questions. Express all answers as fractions reduced to lowest terms or as decimals to the nearest thousandth. No calculators are allowed.
Table I  Kersseboom’s Table
Age 
Persons living 
Age 
Persons living 
0 
1,400 
56 
434 
4 
993 
60 
382 
8 
913 
64 
329 
12 
878 
68 
273 
16 
849 
72 
217 
20 
817 
76 
160 
24 
783 
80 
100 
28 
735 
84 
55 
32 
687 
88 
21 
36 
645 
92 
5 
40 
605 
96 
0.6 
44 
569 
100 
0 
48 
530 


52 
482 


1. What is the probability (i.e., chance) of someone being alive at 20 years of age?
2. What is the probability of someone being alive at 84 years of age?
3. Given a person 20 years old, what are his chances of attaining 40 years? 60 years?
4. Does a 20yearold have a better chance of attaining 60 years, or a 40yearold of attaining 80 years? Explain your answer.
Extra Challenge: Examine the table below with the one from above, making the comparisons required.
Table II  M. de Parcieux’s Table
Age 
Persons living 
20 
814 
30 
734 
40 
657 
50 
581 
60 
463 
70 
310 
80 
118 
90 
11 
95 
0 
Starting from 40 years of age, which table will have more persons living at age 60? at age 80? Explain your answer.
Grades eligible: All
Submit solutions to mike@Hinsdale60521.com; all work must be shown to receive consideration. The submission deadline is Friday at 10 p.m., and the solution will be provided on Saturday morning.
Week 9 Winner: Alec Hill, 6th Grade (Hinsdale Middle School)
Honorable Mention: Will Lopez, Charlotte Novy, Avalon CharltonPerrin, Cam Garber, Kayleigh Hester, Allegra Dugan, Jenny Witt, Regan Hultquist, Ellie Geier, Kyle Ayres, Kate Gabriel, Elizabeth Floersch, Joshua Terry, Emma Wang, Lindsay Peters, D.J. Walsh, Hrishi Deshmukh, Mahir Hussain, Tyler Knapp, J.P. Gaffigan, Casey Schrader
Week of October 22
Problem VIII  Alligation
No one has yet solved this week's problem, so students will be allowed another week to attempt to solve it. The ninth Problem of the Week will appear on Monday as well.
This problem calls for some rather advanced arithmetic, so I will offer a hint. When trying to solve a problem of this kind, it is important to understand that the quantities mixed together must be balanced; that is, in this case, the quantities 15, 17, 18, and 22 combined together in some way must average 20. Now, 15, 17, and 18 are less than 20, while 22 is greater than 20; so my hint is to first separate the lesser quantities from the greater. (It should occur to the student that the amount of 22carat gold required must be greater than any of the other amounts, as this is the only quantity of the four given that is more than 20.)
And, one more thing, since the compound must be 40 oz., the sum of the amounts of 15, 17, 18, and 22carat gold required must be 40 oz.
Problem: A goldsmith must make a compound of 40 oz. of gold at 20 carats fine, but he has no 20carat gold at his disposal. How many ounces of 15, 17, 18, and 22carat gold must he mix together to form this compound?
Grades eligible: All (Recommended for higher grades, but may be attempted by all)
Submit solutions to mike@Hinsdale60521.com; all work must be shown to receive consideration. The submission deadline is Friday at 10 p.m., and the solution will be provided on Saturday morning.
Week 7 Winner: Will Lopez, 6th Grade (Hinsdale Middle School)
Week 6 Winner: Mehlum Anjarwala, 2nd Grade (Madison School)
Honorable Mention: Casey Schrader, 5th Grade (Monroe School), Kendall Schrader, 5th Grade (Monroe School)
Week 5 Winner: Kendall Schrader, 5th Grade (Monroe School)
Honorable Mention: Melina Slone (Prospect School), Dylan Tang (Madison School), Kevin Jin (Prospect School), Nolan Sharer
Week 4 Winner: Kevin Jin (Prospect School)
Week 3 Winner: Carson Boggs, 3rd Grade (Monroe School)
Honorable Mention: Ryan Halpin (Madison School), Mary Heneghan (Oak School)
Week 2 Winner: Jack Kroll, 5^{th} Grade (Monroe School)
Honorable Mention: Luke Speziale (3^{rd} Grade), Audrey Gallwas (3^{rd} Grade), Owen Bots (4^{th} Grade), Quinten Lane (3^{rd} Grade), Colin Lane (5^{th} Grade), Elizabeth Alcala (3^{rd} Grade), Trevor Schmitz (3rd Grade)
Week 1 Winner: Joe Boggs, 2nd Grade (Monroe School)
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