dan's math@home - problem of the week - archives
 
 
Problem Archives page 17
with Answers and Winners. For Problems Only, click here.
 
1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+ . prob index
 
161- Dimples All Over
162 Your Fav Subject?
163 - Riding The Train
164 - Three More Sqrs
165 -- Here's To 2003!
166 -- Happy And Sad
167 ExerciseEfficiently
168 Arranged by Hight
169 - Pay For Grades?
170 - Area's The Same
 
 
Problem #161 - Posted Friday, December 27, 2002
Dimples All Over ! (back to top)
A regulation golf ball is spherical and has 384 dimples, arranged in a triangular pattern.
Most of the dimples are surrounded by six other dimples, but some are surrounded by
only five. How many dimples have only five neighbors? Give a mathematical proof if possible.
Explain steps and reasoning; resubmissions carry a 1-pt penalty.
 
Solution: (From Nick McGrath, persistent and industrious entrant; sorry about my e-mail!)

"Since we are told the dimples are surrounded by either 5 or 6 neighbors we can consider the problem as one
of tiling a sphere with a combination of hexagons and pentagons. In the case of the golf ball each of the "faces"
(hexagonal or pentagonal) has a dimple at its center as well as one at each of the vertices. Since it is important
that the ball has a symmetrical pattern for consistency of flight the "faces" must be symmetrically oriented
relative to the center of the ball. In addition, we can assume regular pentagons and hexagons.
[Dan's note: In practice, actual regular hexagons won't provide the proper gradual curvature necessary for a spherical shape.]
 
First question: Why not all hexagons? [ ... Nick provides a proof by contradiction.]
Second question: If we have a combination of hexagons and pentagons, how many pentagons are there?
Counting ends of edges we have 3V=2E. Also, F=P+H (P is the # of pentagons and H is the # of hexagons).
Counting sides of edges we have 2E=5P+6H so 3V=5P+6H. Substituting ito Euler's formula: V+F-E=2
gives us (5P+6H)/3 + (P+H) - (5P+6H)/2 = 2 => 10P+12H+6P+6H-15P-18H = 12 => P=12.
We must have12 pentagons; so of the 384 dimples on the golf ball, exactly 12 have only 5 neighbors.
(In fact the planes of the 12 pentagons, if extended, would form a regular dodecahedron. See my geometry page for more
info and pictures of tessellations and Polyhedra. - Dan)
 
 
WINNERS - Problem 161 . (back to top) . leader board . . . .
Nick McGrath . . . . . . 10 pts - Nice job, clear analogy to polyhedra ; the dual of dimples!
Jayavel Sounderpandian . . 7 pts - Right; use X + Y = V and Euler's equation V-E+F=2.
Mohamed Omar . . . . . 5 pts - Welcome back; good to notice V=384 and E=1146, giving X=12.
Phil Sayre. . . . . . . . . . . 3 pts - Start with equator & 60-degree planes, nice, but some v have 4 edges.
Tim Poe . . . . . . . . . . . . 3 pts - Your icosahedral intuition was right; do all real g balls have 392?
Arthur Morris . . . . . . 3 pts - Finished off M.Omar's idea with F=764. Your eqns almost solved it.
Quasi-C. . . . . . . . . . . . 2 pts - Good start, 2E = 5X + 6(384-X); then you noticed X is even.
Nikita Kuznetsov . . . . 2 pts - I got your #164 entry; sorry about the bounced e-mails!
Hermen Jacobs . . . . . . 1 pt - Not sure about the geometry of 3 * 2^7 , maybe 7 dimensions?
 
 
Problem #162 - Posted Monday, January 6, 2003
Your Favorite Subject ? (back to top)
Last year, near the end of my "Survey of Math" course, we had time for one more topic:
Polyhedra, Divisors, or Curves. I decided to let my fifteen students vote their preferences :
Each student submitted their 1st, 2nd, and 3rd choice topics. There were 3 ways to count
the ballots, defined as follows: (1) Plurality: The topic with the most first-choice votes ("prefs") wins.
(2) Instant Runoff: Eliminate the third-place topic (in number of prefs), and add the 2nd-choice votes of
that topic to prefs of the first two. (3) Point system: Two pts for each 1st choice, one for each 2nd choice.
If I counted the votes by plurality, Polyhedra won, but by the instant runoff method,
Divisors was the choice. And under the point system, Curves was the winner!
How was this possible?! Give a scenario with 15 actual preferences.
 
Solution: (From Tim Poe):
"Pleurality is easy. Polyhedra must have more than five votes, while the other two have less than
Polyhedra. So let's start with 6 prefs for Polyhedra, 5 prefs for Divisors, and 4 prefs for Curves.
Voila: by Pleurality, Polyhedra wins.
Moving on to Instant Runoff. The third place finisher in our scenario above is Curves. If 3 of those
voting for Curves had a second choice of Divisors, while the 4th had a second choice of Polyhedra,
then the runoff votes are 6 + 1 = 7 for Polyhedra and 5 + 3 = 8 for Divisors.
Voila: by Instant Runoff, Divisors wins the race.
Finally, we have the Point system. In our scenario, we already know all 1st place votes, and the 2nd
place votes for those whose 1st choice was Curves. Let us say that all those with 1st preferences of
Polyhedra and Divisors all chose Curves for their 2nd choice. By the Point system, you'd have:
For Polyhedra: 6 1st pref + 1 2nd Pref = (6*2)+1 = 13
For Divisors: 5 1st pref + 3 2nd Pref = (5*2) + 3 = 13
For Curves, 4 1st pref + 11 2nd Pref = (4*2) + 11 = 19
Voila: by the Points system, Curves wins."
 
[Dan's note: This is a "strange occurence," isn't it? I wonder what the outcome woulda been if
they'd used other methods (besides plurality) in Florida in 2000?
Still not sure exactly how many correct solutions there are for this one... Sorry no bonus points.]
 
WINNERS - Problem 162 . (back to top) . leader board . . . .
I'll use the notation 3PDC to mean three people had order of choices 'Polyhedra, Divisors, Curves.'
Nick McGrath . . . . . . 10 pts - Simplest and first solution: 6PCD+4CDP+5DCP
Tim Poe . . . . . . . . . . . . 7 pts - That's it; three ways to win! See solution above.
Jon Stearn . . . . . . . . . . 5 pts - Welcome back! 3CDP+1CPD+5DCP+6PCD=OK
Nic Hargreaves . . . . . . 4 pts - Yes, a spreadheet does it again! Same distrib as NickMcG.
Les Billig . . . . . . . . . . .4 pts - I like the table: PDC vs.123. 1PDC+6PCD+1DPC+4DCP+3CDP
Ed Wern . . . . . . . . . . . 4 pts - "I would vote to have the final topic be voting theory."
A. Teitelman . . . . . . . 3 pts - Great answer; all ok except C gets 19 pts not 20.
Hermen Jacobs . . . . . 2 pts - I agree with your inequalities; need to specify the ballots
Jack Dostal . . . . . . . . 2 pts - Another .xls! Good try but your D can't win the runoff.
Quasi-C. . . . . . . . . . . . 2 pts - Your D only had 3 prefs so it's out of the runoff.
Allen Druze. . . . . . . . 2 pts - D can win but you have to specify indiv ballots
Arthur Morris . . . . . . 1 pt - P won't be the inst winner if 2nd choice gets more from 3rd
Phil Sayre. . . . . . . . . . 1 pt - Elim the 3rd and give its 2nd place votes to those others.
 
 
Problem #163 - Posted Saturday, January 25, 2003
Riding The Train (back to top)
Two students were on a commuter train from New York to Boston. One said, "The trains
coming from Boston pass us every five minutes. I wonder, how many Boston trains arrive
in New York per hour, given equal speeds in both directions?" "Twelve, I figure," said the
second, "sixty minutes divided by five." The first disagreed. Who was right?
 
Solution: (From Mohamed Omar):
"The first student is correct. Consider the system with respect to the train that is viewing the other trains.
He/she sees a train pass every 5 minutes, but the train is moving at twice his speed in this frame of reference.
Consequently, trains leave twice as slow as the second guy thinks, which means they arrive only 6 per hour."
 
WINNERS - Problem 163 . (back to top) . leader board . . . .
Nick McGrath . . . . . . 10 pts - Right; the time is doubled if you're not moving.
Hermen Jacobs . . . . . . 7 pts - I like your scenario with the emergency stop
Arthur Morris . . . . . . 5 pts - Good use of symmetry; stopped train = station
Mohamed Omar . . . . . 4 pts - Yes, if train passes every 5 min going twice as fast
Jayavel Sounderpandian . . 4 pts - Nice use of a 2-hour span although not given 1 hr trip
Crystal Lai (new) . . . . 3 pts - Welcome to my contest! Can you decide how many trains?
Tim Poe . . . . . . . . . . . . 3 pts - Right ans per hour but trip not nec 1 hour
Quasi-C. . . . . . . . . . . . 3 pts - One of these relativistic arguments, eh? Cool.
Ed Wern . . . . . . . . . . . 3 pts - Good reasoning; second student is correct all right.
Jack Dostal . . . . . . . . . 2 pts - Right; 10 min between consecutive trains. No Excel?
Nikita Kuznetsov . . . . 3 pts - Yes it was fairly easy but you said 10 trains per hour...
Zahi Teitelman . . . . . . 3 pts - Correct; in a period of 1 hour only 6 trains each direc
AZ Runner. . . . . . . . . . 3 pts - Welcome back, AZ. They do see 'double' the trains.
Nic Hargreaves . . . . . . 3 pts - Yes, the first was right to disagree. It's everyone's right!
Yonlawan Chirawatcharadej . 2 pts - Right; it's not 12, but not 13 either; I think it's 6.
Paul Robinson (new) . . 2 pts - Condition of exactly same speed not required to disagree
Joe Alvord. . . . . . . . . . 2 pts - Right; they take twice as long; which student got it?
Allen Druze. . . . . . . . . 2 pts - Yes the second is 10 min behind the first; who's right?
 
 
Problem #164 - Posted Thursday, February 6, 2003
Three More Squares . . . (back to top)
"I found three perfect squares in arithmetic progression!" said Fred. "That's the smallest
solution. But I have three that are even," Steven countered, "and also a formula for an
infinite family of progressions! For each triple, I see the first two are the squares of
p^2 - q^2 - 2pq and p^2 + q^2." a) What were Fred's small and Steven's even solutions?
b) What's the square root of that third square in each p, q progression?
 
Solution: (From Quasi-Conscious):
b) We are given that the first in the series is the square of P^2 - Q^2 - 2PQ ,
That is P^4+Q^4 + 2P^2Q^2 - 4P^3Q + 4PQ^3 - 2P^2Q^2 = (P^2+Q^2)^2 - 4PQ*(P^2-Q^2)
The second in the series is (P^2+Q^2)^2, so the difference is 4PQ*(P^2-Q^2), so the third in the
sequence is (P^2+Q^2)^2 + 4PQ*(P^2-Q^2) = P^4 + Q^4 + 2P^2Q^2 + 4P^3Q - 4 PQ^3
= (P^2 - Q^2 +2PQ) ^2 . . . So the root of the third square is P^2 - Q^2 + 2PQ .
a) Using the above it is a simple matter to parameterize P & Q, and generate the sequences of three.
The first is 1; 25; 49 using P=2, Q=1. The smallest even trio is 4; 100; 196 for P=3, Q=1
 
WINNERS - Problem 164 . (back to top) . leader board . . . .
Hermen Jacobs . . . . . . 9 pts - The winner ; (some algebra missing; judge's deduction)
Tim Poe . . . . . . . . . . . . 7 pts - Wait till I bust some geometric progressions on you!
Nick McGrath . . . . . . 5 pts - I guess the equations helped; and you're welcome!
Nic Hargreaves . . . . . . 4 pts - Couldn't read your @#&* Word equations but ans. ok
Quasi-C. . . . . . . . . . . . 4 pts-Well-put answer, so I put it! Are there inf many primit.triples?
Arthur Morris . . . . . . 3 pts - Second answer came in behind those above, but good job!
Jack Dostal . . . . . . . . . 3 pts - Squares are ok but your q^2 - p^2 - 2pq needs q > p
Mohamed Omar . . . . . 3 pts - Good answer, not too far behind time-wise, keep it up!
Jayavel Sounderpandian . . 3 pts - You had all bases covered with 0,0,0 then 1,1,1 and 4,4,4.
Phil Sayre . . . . . . . . . . 3 pts - Right; good answer; it's ok to include a little algebra too
Zahi Teitelman . . . . . . 2 pts - 3rd number can't be 1; (are you the same as A.Yeitelman)?
Nikita Kuznetsov . . . . 2 pts - Partial credit for partial answer; college keeps you busy huh?
Allen Druze . . . . . . . . 1 pt - First two parts ok and under the wire... thanks!
 
 
Problem #165 - Posted Sunday, February 16, 2003
Here's To a peaceful 2003 ! (back to top)
a) In the sequence : 12345678910111213 . . . what position does 2003 first appear?
b) How many different ways can 2003 be written as the sum of three squares?
c) Find the smallest number that's the sum of three squares in this many ways.
Rearrangement does not count as different.
 
Solution: (I picked parts from a few of you; thanks for all the great ideas!)
a) First we need to show that 2003 doesn't occur before its 'natural position'. Gaurav states,
"As both these numbers will have the first digit non-zero, the first number must have 200 as its
suffix. Also as the next number starts with 3, the smallest such numbers will be 3200 & 3201.
But as 2003 itself is smaller then 3200, the first occurrance of 2003 will be at its numerical place
(in sequence 1 2 3...2001 2002 2003...) only." As for the position, Nikita says, "There are
9 1-digit numbers, 90 2-digit, 900 3-digit and 1003 4-digit numbers (before 2003), we have
9+2*90+3*900+4*1003 = 6901; so the first digit of 2003 appears in the 6902th position."
b) Quasi-C points out: "For any n, n^2 is of the form 4m or 4m+1. So for a number of the
form 4k+3 to be the sum of three squares, it must be the sum of three odd squares." (Cool; this
limits the search by a factor of 8. - Dan) Also Nic says "45^2 > 2003 and 3*26^2 > 2003, so if the
largest number is 'a' we have: a^2 + b^2 + c^2 = 2003, a>=b>=c, 25<=a<=44, and a, b, c, odd.
A search among these yields (a,b,c) = (31,31,9), (33,25,17), (35,27,7), (39,19,11), (37,25,3)"
So there are 5 ways to make 2003 as the sum of three (non-zero, possibly equal) squares.
If you (mis)interpreted 'three squares' as three different squares, then there are four ways.
c) The smallest number that's the sum of three squares in 5 ways is 194 = a^2 + b^2 + c^2
where (a,b,c) = (9,8,7), (11,8,3), (12,5,5), (12,7,1), (13,4,3); i.e. 9^2 + 8^2 + 7^2 = 194, etc.
If you thought 0^2 counted then 146 is smallest with five; if you wanted 4 it was 161 (or 101).
 
WINNERS - Problem 165 . (back to top) . leader board . . . .
Nick McGrath . . . . . . 10 pts - Yes, 9*1+90*2+900*3+1003*4 = 6901 so 6902nd.
Gaurav Agrawal . . . . . 7 pts - Welcome back; good answer, and '3200, 3201' is right too.
Nic Hargreaves . . . . . . 5 pts - Good reasons why 2003 after 2002, nice use of sprsht sorting.
Hermen Jacobs . . . . . . 4 pts - Got it on 3rd try, after other ans. were in. Good persistence!
Jack Dostal . . . . . . . . . 4 pts - Thorough answer; "if 0 allowed then 146, else 194."
Nikita Kuznetsov . . . . 3 pts - I never said the squares be different; yes 161 smallest for 4 ways
Phil Sayre . . . . . . . . . . 3 pts - 146 smallest 5-way-3-square if 0 allowed; 'zat what you meant?
Tim Poe . . . . . . . . . . . . 3 pts - Good list of squares ; 6902-05 is correct; but why?
Jayavel Sounderpandian . 3 pts - Computer searches are great. Why can't they find my car keys?
Arthur Morris . . . . . . 2 pts - Woulda been more pts without the addition error - oops!
Quasi-C. . . . . . . . . . . . 2 pts - Nice proof of 2003 after 2002 but same addition error...
Jon Stearn . . . . . . . . . 2 pts - Right abt 3200-3201 but what positions? 101 for 4ways-0-ok.
Mohamed Omar . . . . 2 pts - Good answer, not too far behind time-wise, keep it up!
Ed Wern . . . . . . . . . . . 2 pts - Two points for two parts, did you find the smallest for c?
Zahi Teitelman . . . . . 1 pt - There were multi-digits in a lot of those numbers...
Allen Druze . . . . . . . . 1 pt - Was there a miscalc on the 6983? No logic given, Got c yet?
 
 
Problem #166 - Posted Friday, February 28, 2003
: ) Happy And Sad : ( . . . (back to top)
Each day, Stevie is either happy or sad.
If he is happy one day, then four times out of five he is happy the next day.
If he is sad one day, then he is sad the next day one time out of three.
In the long run, what are the chances that Stevie is happy on any given day?
Solution: (From Nick McG, stretching out his lead...)
"Let p(x) be the probability that Stevie is happy on a given day, x, and p(x+1) the probability that he is
happy on day x+1. Then p(x+1)=4/5*p(x)+2/3*q(x) where q(x) is the probability he is sad on day x.
q(x)=1-p(x) so p(x+1)=4/5*p(x)+2/3*(1-p(x))=2/15*p(x)+2/3.
In the long run we must have that p(x+1)=p(x) so: p(x)=2/15*p(x)+2/3 => p(x)= 15/13*2/3=30/39=10/13.
The probability that Stevie is happy on a given day is thus 10/13. (This is 0.769230 ~ 76.9% of the days.)
 
Bonus point for the first to clearly explain to me Allen's answer below (Allen eligible).
 
WINNERS - Problem 166 . (back to top) . leader board . . . .
Nick McGrath . . . . . . . 10 pts - On top again with a good'n'concise, yet complete, answer!
Steve Carlon (new) . . . . . 7 pts - Glad you decided to enter my contest. Good answer, Steve.
Jayavel Sounderpandian. 5 pts - Good use of 'long run' where tomorrow = today.
Arthur Morris. . . . . . . . 4 pts - Came up with a rep cycle of 13 days: HHHHSHHHSSHHH
Nikita Kuznetsov . . . . . 4 pts - Converging sequence rather than equation, largely ok.
Gaurav Agrawal . . . . . . 3 pts - Good to call it a Markov Chain process with transition prob
Zahi Teitelman . . . . . . . 3 pts - Yes, p(x) = x = x*4/5 + y*2/3; y = x*1/5 + y*1/3, x+y = 1; x=10/13.
Quasi-C. . . . . . . . . . . . . 3 pts - Looked at it as a large population of largely happy people.
Ed Wern . . . . . . . . . . . . 3 pts - Total happy days = 4/5 hap + 2/3 sad ; this leads to solution.
Hermen Jacobs . . . . . . . 3 pts - Good idea; 2x2 matrix raised to powers, rows become same!
Phil Sayre . . . . . . . . . . . 3 pts - Ok, with help from Python; thanks for providing fraction ans!
Jon Stearn . . . . . . . . . . . 3 pts - Nice concise eqn. x = (4/5)x + (2/3)(1-x) --> x = 10/13.
Anirban Bhattacharyya . 2 pts - Welcome back A.B. Somewhere you assumed equal likely.
Tim Poe . . . . . . . . . . . . . 2 pts - Not sure what you meant by 30 pairs; all possible?
Marcello Cammarata (new) 2 pts - Benvenutti! Did I get that right? See reasoning above for answer.
Joe Alvord . . . . . . . . . . . 2 pts - Good idea to use a 3-col sprsht; your recursion coeffs were off.
Allen Druze . . . . . . . . . . 2 pts - Said only "2/3 ( 1 + 2/15 + 4/225 + . . . ) = 10/13." Mysterious!
Les Billig . . . . . . . . . . . . 2 pts - No proof but ran ten 1000-day experiments; avg 77.3% happy!
Wanda Cahill (new) . . . . 1 pt - E-mail fan becomes contest entrant. Welcome and keep it up!
 
 
Problem #167 - Posted Sunday, March 16, 2003
Exercising Efficiently. . . (back to top)
Excess Exercise Ernie can do 20 pullups, or 30 situps, or 40 pushups, in one minute.
Doing one pullup burns 1.5 calories; one situp burns 1.8 cal; a pushup burns 1.2 cal.
Ernie is too 'pressed' for time; you need to design him two Efficient Exercise Efforts:
(i) Panic Workout: Lasts seven minutes, burns exactly 300 calories, with a total of 200 reps,
(ii) Full Session: At least 30 min, burning at least 1200 calories, with as few reps as possible.
Each workout must contain at least one rep(etition) of each exercise. State how many reps of
each type, and total reps, for each workout.
 
Solution:
a) This part is a system of three linear equations: L=# of pullups, S=# of situps, H=# of pushups.
Then total time : L/20 + S/30 + H/40 = 7 minutes , or 3L + 2S + 1.5H = 420 seconds,
Calories: 1.5 L + 1.8 S + 1.2 H = 300 ; Reps : L + S + H = 200. Solve the last 3 eqns and we get
L = 56 pullups, S = 72 situps, H = 72 pushups.
 
b) The second part is a linear programming system of inequalities: 30 minutes = 1800 seconds
T = 3L + 2S + 1.5H >= 1800 sec ; C = 1.5 L + 1.8 S + 1.2 H >= 1200 cal ; Minimize R = L+S+H.
As you found, the planes intersect at H=350, S=375, L=0 , 725 reps, but you need at least 1 pushup:
Several of you sent in more than one optimal answer of 726 reps . The most was 10, aggregate was 11.
 
 L pullups  S situps  H pushups  T time (sec) C calories  R reps
 350 375 0 (illegal) 1800 1200 725 (illegal)
 349 376 1 1800.5 1201.5 726
 350 375 1 1801.5 1201.2 726
 351 374 1 1802.5 1200.9 726
352 373 1 1803.5 1200.6 726
353 372 1 1804.5 1200.3 726
354 371 1 1805.5 1200 726
 349 375 2 1800 1200.9 726
 350 374 2 1801 1200.6 726
 351 373 2 1802 1200.3 726
 352 372 2 1803 1200 726
 350 373 3 1800.5 1200 726
 
Quasi-C decided, rightly so, that the most productive one is the 'max burn' workout of 1201.5 cal.
Another approach would be cal/sec : 1201.5/1800.5 = 0.66731 beats 1200.9/1800 = 0.66717.
 
WINNERS - Problem 167 . (back to top) . leader board . . . .
Gaurav Agrawal . . . . . . 9 pts - The winner with 726 reps but no proof that your method gives min.
Nick McGrath . . . . . . . . 7 pts - Good answer even 'resorting to a program.' no pt deduc 4 2nd ans.
Arthur Morris. . . . . . . . 5 pts - Wow, you're still entering my humble contest! I like 'cal/min' idea.
Tim Poe . . . . . . . . . . . . . 4 pts - Your solution had exactly 1200 cal but a bit more was ok too.
Steve Carlon . . . . . . . . . 4 pts - I don't know if we count half reps; I've done them myself though
Jack Dostal . . . . . . . . . . 4 pts - I like your discussion and full list of ten answers; I think I have 11th
Quasi-C. . . . . . . . . . . . . 3 pts - Good use of min not reps; yes max cal in min min is best for effic.
Phil Sayre . . . . . . . . . . . 3 pts - Glad to provide you-all with a 'Nice Mental Workout'. Thanks.
Jayavel Sounderpandian. 3 pts - 350, 375, 0 --> 350, 375, 1 turns out to be minimal; do you know that?
Hermen Jacobs . . . . . . . 3 pts - I know, too bad you can't draw the 3D coords. You could scan...
Wanda Cahill . . . . . . . . 2 pts - Got the 3x3 sys right but 798,1,1 takes 40 min and isn't minimal.
Zahi Teitelman . . . . . . . 2 pts - First part ok, but b) 730 not min doesn't have to be a mult of 5 reps.
Allen Druze . . . . . . . . . . 2 pts - Got the first part ok, then got 350, 375, 0 but you need a pushup.
Marcello Cammarata. . . . 2 pts - Yes, 56,72,72, and solving sys gives (349.875, 374.437, 1)
Yonlawan Chirawatcharadej 1 pt - 750 total reps is pretty close but 726 is min; no ans for 1st part
 
 
Problem #168 - Posted Monday, March 31, 2003
Arranged by Hight ?. . . (back to top)
No, that's not a misspelled word. The hight h(p) of a polynomial p(x) is defined as the sum of
the degree n and the absolute values of the coefficients: h(x^2 - 3x + 5) = 2+1+3+5 = 11.
a) If n is allowed to be 0, how many polys p(x) are there with integer coeffs and hight
not exceeding 1, 2, 3, 4, 5 ? (That's five answers. For the first 3 give an actual list of the p(x).)
b) What's the hight of a monic cubic poly with roots a, b, and 3 ? (Ans. in terms of a and b.)
"Monic polynomial" means leading coeff. is 1. Explain reasoning.
 
Solution: First we settle the issue of the degree of the zero poly p(x) = 0 : Gaurav defines degree (order)
of a poly as the max power on a term with non-zero coeff; hence deg(0x^2 + 3x) = 1. deg(0) is undefined.
 
a) Let's list the polys p = p(x) of each hight h = h(p), sorted by degree: . . h = 1 (2 polys) : p = 1, -1. . . .
h = 2 (4) : p = x, -x, 2, -2 . . . . h = 3 (10) : p = x^2, -x^2, 2x, -2x, x+1, x-1, -x+1, -x-1, 3, -3.
h = 4 (24) : p = x^3, 2x^2, x^2+x, x^2+1, 3x, 2x+1, x+2, 4, and associates with some or all negative signs ;
h = 5 (58) : p = x^4, 2x^3, x^3+x^2, x^3+x, x^3+1, 3x^2, 2x^2+x, 2x^2+1, x^2+2x, x^2+2, x^2+x+1,
. . . . 4x, 3x+1, 2x+2, x+3, 5, and associates.
Total numbers of polys having hight h up to : 1 (2) , 2 (6), 3 (16) , 4 (40) , 5 (98).
New contestant Alan gives the total polys of hight up to n as V(n) with a Fibonacci-like recursion
V(n+1) = 2V(n) + V(n-1) ; it seems to need an artificial start of V(0) = 1 and V(-1) = 0.
 
b) p(x) = (x-a)(x-b)(x-3) = x^3 - (a+b+3)x^2 + (ab+3a+3b)x - 3ab ; use "abs(coeff)" = | coeff | ;
those answers without abs vals got h(p) = 3 + (a+b+3) + (ab+3a+3b) + 3ab = 7 + 4(a+b+ab),
but the correct hight was really h(p) = 3 + abs(a+b+3) + abs(ab+3a+3b) + abs(3ab).
 
WINNERS - Problem 168 . (back to top) . leader board . . . .
Gaurav Agrawal . . . . . . 9 pts - Good lists but you forgot x+3 and associates. Nice def of poly.
Phil Sayre . . . . . . . . . . . 6 pts - I like your deg. vs. ht. table; it looks like Pascal's (Chu's) Triangle.
Tim Poe . . . . . . . . . . . . . 4 pts - Correct numbers of p(x) for each ht; 2nd try was closer on part b.
Alan O'Donnell (new) . . 4 pts - Yes, neg coeffs are fine; great recursion, welcome to the contest!
Marcello Cammarata. . 4 pts - Thanks for the lists even for h=4 & 5; you&others needed abs vals.
Quasi-C. . . . . . . . . . . . . 3 pts - Second part fine; left off a few 3-termers on part a); deg(0) = ?
Nick McGrath . . . . . . . 3 pts - Neg coeffs are ok; can get more p's. Got the 2nd part right.
Steve Carlon . . . . . . . . . 3 pts - Totals ok except for h(0) undef; needed the abs vals too.
Hermen Jacobs . . . . . . . 3 pts - Note that h(x^n) = n+1; throws off your totals. Ok on b).
Arthur Morris. . . . . . . . 2 pts - Need some more p(x)'s; try not to assume pattern too early.
Zahi Teitelman . . . . . . . 2 pts - There are a few for each hight; need abs vals but ok if a,b>0.
Nikita Kuznetsov . . . . . 1 pt - It wasn't as easy as it looked at first; semi-Pascal's tri. at best.
 
 
Problem #169 - Posted Friday, April 11, 2003
Pay For Grades?! . . (back to top)
Sarah's parents decide to reward her (with money!) for her school grades. Sarah has five
classes (English, History, Japanese, Math, and Science). She will get $15.00 for each A,
$10.50 for a B, $7.50 for a C, $1.50 for a D. Of course, she gets nothing for an F.
The last three semesters Sarah has avoided F's, and received the same amount of money
(a whole number of dollars) each semester, although the grade distribution of A's, B's, C's, D's
was different each time. . a) How much money did Sarah receive after each semester?
b) What were her grades and semester G.P.A.* each time? (Don't worry about which grades
she got in which classes.) * GPA is avg grade pts per class: 4 pts for A, 3 for B, 2 for C, 1 for D, 0 for F.
(I'm not condoning or condemning this pay idea.) (One point max while this sentence is up.)
 
Solution: As Nick and others noted, if a, b, c, d are the numbers of A's, B's, etc., Sarah's total payoff was
T = 15a + 10.5b + 6.5c + 1.5d = 15a + 10b + 6c + d + 0.5(b+c+d) = whole number ==> b+c+d even,
so a has to be odd since a+b+c+d = 5 classes. Putting a = 3 or a = 1 limits the possibilities quite a bit,
and a = 5 is ruled out since there's only 1 way to get 5 A's. This gives us a table of 22 solutions (a,b,c,d)
with 16 different values of T. The only T to appear 3 times is T = 48. So Sarah got $48 each semester.
The three ways are : (a,b,c,d) = (3,0,0,2) GPA=2.8 , (1,3,0,1) GPA = 2.8 , (1,1,3,0) GPA = 2.6 ;
i.e. Sarah got 3A's, 2D's , or 1A, 3B's, 1D , or 1A, 1B, 3C's. Steve and Nikita found the other possible
(non-integer) values of T, listed below, and Quasi-C tried factoring out 1.50 and looking mod 2, 3, or 5.
 
WINNERS - Problem 169 . (back to top) . leader board . . . .
Steve Carlon . . . . . . . . 10 pts - Nice job; lists other amts with 3 ways: 52.50, 43.50, 40.50, 34.50.
Wanda Cahill . . . . . . . . 7 pts - Good algebra noting 2c=a, c+2d=b, etc to get equal sums.
Gaurav Agrawal . . . . . . 5 pts - Saw the clue: A can't be even or she wouldn't get whole # of $.
Nick McGrath . . . . . . . . 5 pts - Woulda been first but one GPA was miscomputed -- picky!
Hermen Jacobs . . . . . . . 4 pts - Good old BASIC does it again! T = 21, 27, . . . , 75; three 48's.
Tim Poe . . . . . . . . . . . . . 4 pts - Yep, I used a spreadsheet too. They can be useful, can't they?
Quasi-C. . . . . . . . . . . . . 4 pts - Gee they are all multiples of $1.50 ; $48 = 36*$1.50 = total.
Jayavel Sounderpandian. 4 pts - Early entry; good reasonong on odd A's; GPA's incorrectly done.
Arthur Morris. . . . . . . . 3 pts - You got the answers ok but did total grade pts instead of GPA.
Jon Stearn . . . . . . . . . . . 3 pts - Some grades were "too rich," meaning Sarah was too smart?
Phil Sayre . . . . . . . . . . . 3 pts - Listing all 56 ways from AAAAA to DDDDD; is this a shoe store?
Marcello Cammarata. . 3 pts - Benvenuto again, and nice table supplied (by attachmt, tsk, tsk.)
Zahi Teitelman . . . . . . 3 pts - Number of A's can't be 5, 4, or 2. That's right, limit the search.
Ed Wern . . . . . . . . . . . . 3 pts - Step-by-step Excel saga; I agree, too much $ for a 2.6 GPA.
Jack Dostal . . . . . . . . . . 2 pts - You found three ways for Sarah to make $43.50; need a whole #.
Nikita Kuznetsov . . . . . 2 pts - Did you misread 'different grade distribs' as 'different GPA's'?
Allen Druze . . . . . . . . . . 2 pts - Correct grades given, but GPA's weren't all 2.8. Close though!
 
 
Problem #170 - Posted Monday, April 21, 2003
Area's The Same ! . (back to top)
Two things come to mind when looking at right triangles:
The Pythagorean Theorem relating the three sides, a, b, c ;
and the formula for the area A. Sometimes, all four of
these quantities are integers: a 3-4-5 triangle has area 6.
What is the smallest natural number A that's the area of two
different integral right triangles? (Switching a and b isn't different.)
What's the smallest natural number A that's the area of three i.r.t.'s?
What's the smallest A (if there is one) that's the area of four i.r.t.'s?
Show your steps, reasoning, or search method. Give sides and areas.
Partial answers get partial credit. One point penalty for resubmissions.
Sum of squares of legs
= square of hypotenuse
Solution: From Nick McG, first entrant (moves into the all-time top ten-- congrats Nick!)
"It is well known and easy to show that the sides of a right triangle are given by
a = 2mn, b = m^2-n^2, c = m^2+n^2 where m and n are arbitrary integers with m >= n.
The area, A, of such a triangle is 1/2*base*height=1/2*2mn*(m^2-n^2)=mn(m^2-n^2).

Writing a small program looping through values of m and n readily finds the first two solutions. We have:
For 2 equiarea triangles : A=210 from triangles (12,25,37) and (20,21,29)
For 3 equiarea triangles : A=840 from (15,112,113) ; (24,70,74) ; (40,42,58)
For 4 equiarea triangles: My program ran out of steam for this one, so I resorted to a Google search to discover
that one of my favorite sites, MathWorld, has the solution. http://mathworld.wolfram.com
A=341,880 from (111, 6160, 6161), (231, 2960, 2969), (518, 1320, 1418), (280, 2442, 2458)
 
Our good friend Monsieur Fermat showed us how to generate an arbitrary number of equiarea right
triangles." http://mathworld.wolfram.com/RightTriangle.html
(Dan's Note: I used spreadsheet columns for A = m n (m+n) (m-n) ; A needs to be a significantly composite number.)
WINNERS - Problem 170 . (back to top) . leader board . . . .
Nick McGrath . . . . . . . . 9 pts - Nice answer; one side off (35 not 25); too close to cost you 1st place
Tim Poe . . . . . . . . . . . . . 7 pts - Visual Basic strikes again! up to a million was a safe bet
Steve Carlon . . . . . . . . . 5 pts - Good ; try m and n up to 100 and see if you get a set of four.
Jon Stearn . . . . . . . . . . . 4 pts - So that's what a .cpp file is ; C++ sufficized! Brute force worked.
Marcello Cammarata. . 4 pts - Excellent proof of m^2 - n^2 method; nice use of partial derivs.
Arthur Morris. . . . . . . . 3 pts - Took you'n'MATLAB two tries but the 341880 was worth it.
Phil Sayre . . . . . . . . . . . 3 pts - Thanks for dusting off the Number Theory book. (Got supercomposites?)
Hermen Jacobs . . . . . . . 2 pts - Yes there are four equal-area triangles, or indeed any number!
Gaurav Agrawal . . . . . .2 pts - You inspired me to look for all the (m, n) pairs by hand!
Jack Dostal . . . . . . . . . . 2 pts - Combo of C++ & Excel to compute and sort. (Try Mathematica! ;-)
Vince LoCascio . . . . . . 2 pts - Correct : a^2 + b^2 = c^2 so a^2 = (c+b)(c-b) limits choices
Quasi-C. . . . . . . . . . . . . 2 pts - Are you trying some sort of Chinese Remainder maneouewvre?
Nikita Kuznetsov . . . . 2 pts - Theory of A = 105*4^(2n-3) would be growth factors of 16
Allen Druze . . . . . . . . . 2 pts - You got the first two ; ever find a set of four? Here you go.
 
 
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