dansmath.com - problem of the week - archives

Problem Archives page 27

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 . 231-240 . 241-250 . 251-260 . 261-270 . 271+ . index

261 - - Here's To Toast !
262 -- Five Running Ten
263 -- Cosine My Loan?
264 Cubes&Sqrs 2gether
265 - Four Mini-Pollies!
266- Those Factorialoots
267- Squareful Numbers
268 Nine Tangled Wires
269 AnotherBrick inWall
270-SpheresFillingSpace

problem #261 - posted wednesday, february 21, 2007
My old toaster can hold at most two slices of bread and toast them on one side at a time.
I want my bread toasted on both sides and buttered on one side. It takes 3 seconds to put
each slice into the toaster, 30 sec to toast one side of 1 or 2 slices, 3 sec to reverse a slice,
and 12 sec to butter one (that side must already be toasted). Assume only one of these
operations can be done at a time. In how short a time can I toast & butter three slices?
Show steps & reasoning. Winners judged on shortest total time, then order received.

Solution: Again, thank you for your extreme patience and concern during my summer absence!

Dan's comment: In my secret source for this problem, as some of you asked, it did take 3 sec to remove
a piece of toast; but the problem was valid (albeit unclear) as stated, such as whether you could butter
a piece while another was toasting: yes;butter while turning, no. My bad; I'll do my best scoring these!

I got a variety of solutions and times; I ranked them according to shortest time and validity. Here are some:

(Alan O'Donnell's solution, total 108 seconds) . . . Assumptions:
1 - you have 2 hands, so can hold a slice of bread in each hand. ; 2 - you can butter toast while toaster is working.
3 - since not stated, it takes 0 time to remove toast for buttering ; 4 - you want a cup of tea with your toast.
0s - pick up 2 slices to insert into toaster (simultaneously)
3s - toast loaded. put on kettle.
33s - toast pops up. reverse one slice, and replace one slice with a fresh slice of bread (simultaneously)
36s - butter 1/2 toasted slice
48s - put teabag in cup.
66s - toast pops up. replace fully toasted slice with semi-toasted/buttered slice, and turn over semi-toasted/unbuttered slice.
69s - butter fully toasted slice
81s - pour hot water onto teabag in cup.
96s - butter remaining unbuttered slice
108s - toast done. Prepare cuppa to taste and have a rest.

Then there was the question of whether putting two pieces in simultaneously meant two operations at once,
so maybe the best we can do with the restriction is 111 sec? Philippe sent in an awesome bar graph
Placing or turning (green), toasting (brown), and buttering (yellow). Yumm! (bonus point)

But after months of deliberation I applaud Nick McG, who saw no reason not to take partially toasted bread out
and do other things to save time. "What an interesting and tricky problem. Since the time required to remove a slice from the toaster
is not specified I have assumed it to be zero. - First, since it takes less time to butter than to toast it seems that we should be able to find
time for buttering in parallel with toasting. If this is true, we need to find the optimum time for just toasting.
- Second, if inserting and flipping slices took zero time we can see that 3 slices could be toasted in 90 seconds:
First 30 secs. -Toast Slice 1A (slice 1 side A) & 2A: Second 30 seconds - Toast 1B and 3A : Third 30 seconds - Toast 2B and 3B.
- However, since it takes 3 seconds to insert or flip a slice, and each slice has two sides, each slice needs a total of 66 seconds to toast.
The toaster holds 2 slices so, if we maximize the utilization of the toaster (i.e. both parts of the toaster being used at all times - either
toasting or having slices inserted or flipped) we have a minimum theoretical time of 66 * 3 / 2 = 99 sec. But we need to be careful
because for the first 3 seconds we can utilize only one part of the toaster while we insert Slice 1. So, our minimum theoretical time, MTT,
is actually given by  3 + 2*(MTT - 3) = 3 * 66  =>  MTT = 100.5 seconds. ( which we can round up to 101 seconds since all of the
possible operations take an integer number of seconds).
- So, how might we achieve this. Consider one side of the toaster: If we start by inserting slice 1 (3 seconds): toasting 1A (30 seconds) :
removing slice 1 and inserting slice 2 (3 seconds): toasting 2A (30 seconds) : flipping slice 2 ( 3 seconds): toasting 2B ( 30 seconds)
we have taken  96 seconds and we still have to butter slice 2 so we would take a minimum of 108 seconds.
- This is significantly greater than our theoretical minimum time but, at first, it seems we can do no better.
- It seems counterintuitive that we can actually do better by partially toasting one side of a slice since a) it seems to be wasting time to
remove a slice from the toaster before it is ready to be buttered  and  b) it adds 3 seconds to our insertion/flipping time each time we do it.
- For example, if we toast one side of one of the slices in two stages we are increasing the total time to toast 3 slices from 66 * 3 = 198 sec
to  66*3 + 3 = 201 seconds.  Then our MTT expression becomes 3 + 2*(MTT - 3) = 3 * 66 + 3   =>   MTT = 102 seconds.
- This is still better than what we have achieved in practice so far (108 seconds) and we find with some experimentation that a total of 102
seconds is indeed possible as shown in the tables below which show the timeline from 1 to 102 seconds both for each slice of bread and
for each half of the toaster. We can see the toaster is fully utilized ( a requirement to minimize the time)  and also that our assumption
was correct i.e. that if we minimize the toasting time then we can do the buttering in parallel. Shortest time is 102 seconds"

( No law against toasting a slice already buttered on the other side, except for the electrical damage... )
( Nick had a table with 102 rows, but I compressed it for your viewing convenience! - Dan )

Mark R. sent in solutions from 105, 120, 138, and 144 sec. depending on severity of rule enforcement.

And Yakov from Australia waxes nostalgic about these old toasters:
"You have brought on a wave of nostalgia for the old days: smoke rising from the toaster, charcoal being
scraped from charred bread, being told that charcoal is 'good for you', mum throwing smoking toast out
the window so it doesn't smoke up the kitchen any more . . . . How many of your contestants would have
had these experiences, and what have they missed??"

A classic problem to say the least!

Nick McGrath . . . . . . . 10 pts - Fabulous sol'n 102s and table, same colors as Philippe! Thanks for e-mail.
Mark Rickert . . . . . . . . 9 pts - That 105s was pretty thorough, and first received, had to go with Nick's 102.
Alan O'Donnell . . . . . . 7 pts - That was a nice easy-to-understand table for 108s, so I made it quotable!
Nats Kroy . . . . . . . . . . . 6 pts - Good idea 105s, but the toaster can't do both sides w/o human intervention
Denis Borris. . . . . . . . . 6 pts - Down to 99s if we can use a buttering brush on bread currently in toaster!
Etienne Desclin . . . . . . 5 pts - Nice early entry, and 111s was the next best answer, albeit with two hands.
David Stigant. . . . . . . . 4 pts - Nice list, total 114s; could make it 111 if place 2 slices at first second.
Philippe Fondanaiche. 4 pts - Great expl. for 111s on attached doc; great bar graph for us visual learners!
Marcello Cammarata. . 4 pts - Good to realize solutions 120s or 111s depended on the rules applied.
Yakov Macak . . . . . . . . 3 pts - 120s = 2 min. plus great images in your toast history! Got toast racks?
Claudio Baiocchi . . . . . 3 pts - 120s or 114s ; True that "no time to extract a slice" is strange. My mistake!
Tim Poe . . . . . . . . . . . . . 3 pts - 117s - Yes, old type toaster (with rust and grime) flip 2nd while 1st toasts.
K Sengupta . . . . . . . . . . 3 pts - Easy-to-read list of events, but 138s is slower bec. 3rd slice toasts alone.
Hermen Jacobs . . . . . . . 2 pts - 192s ; Extra time for those "private rooms" with one slice at a time there.
Allen Druze. . . . . . . . . . 2 pts - 144s ; some time needed to butter toast while next slice toasts; right.
Oxana Urdaneta (new). . 2 pts - Welcome (finally) Oxana! 3 min = 180s ; could be faster with overlaps.

problem #262 - posted tuesday, march 6, 2007
There were five runners in this year's Waterfront 10, a ten-mile "out-and-back" race, with
a turn-around at the halfway point. Each runner ran their whole race at a constant speed.
Annie reached the 3-mile mark 6 min after the winner, but before at least 1 other runner.
Ben got to the turnaround a full 7 minutes after the third-place runner did.
Carla ran each of the 10 miles of the race in 12 seconds less than 8 minutes.
Dan passed the winner 42 min into the race, then at least 1 other runner before turning around.
Emo had run 5 5/9 miles when he passed the last runner, and had met runners in both
directions already. What was the order of finish, and what was each runner's total ten-mile
time, in minutes? Show your steps and reasoning.

Solution: This is Mark's winning entry, received six hours before any others!

Hi, Dan . . . > > C's pace was 7.8 minutes per mile. (given)
> Based on the clues (numbers of runners passed, etc.):
A finished 2, 3, or 4 . . . B finished 4 or 5 . . . D finished 3, 4, or 5 . . . E finished 2 or 3
> Therefore, C finished 1. . .
Therefore, the possible orders of finish are:
C, A, E, B, D . . . C, A, E, D, B . . . C, E, A, B, D . . . C, E, A, D, B . . . C, E, D, A, B
.
Since A reached the 3-mile mark 6 minutes after C, A's pace is 2 minutes slower per mile than C's;
so A's pace is 9.8 minutes per mile. . Since D passed C 42 minutes into the race, D's pace is
42 / (10 - 42 / 7.8) = 9.1 min. per mile. This means D is faster than A; so the order is: C, E, D, A, B
Since B got to the turnaround 7 minutes after D, B's pace is 1.4 minutes slower per mile than D's;
so B's pace is 10.5 minutes per mile. . Since E ran 50/9 miles when B ran 40/9 miles, E's pace is
5/4 times faster than B's pace; so E's pace is 8.4 minutes per mile. . Therefore, we have:
1 - Carla, 78 min . . 2 - Emo, 84 min . . 3 - Dan, 91 min . . 4 - Annie, 98 min . . 5 - Ben, 105 min

Mark Rickert . . . . . . . . 10 pts - I like your 'elimination of possible orders' approach. Nice job, and quotable.
Etienne Desclin . . . . . . . 7 pts - The data do indeed 'fit together in only one way', like all good puzzles!
Tim Poe . . . . . . . . . . . . .. 6 pts - Thanks for always including the problem stmt; yes Carla 7:48mpm, Dan 9:08.
Nick McGrath . . . . . . . . 5 pts - I like your descriptive notation Ta, Sa, etc. They help 'run through' the logic!
Nats Kroy . . . . . . . . . . . 5 pts - Right: Emo fin 2, 3, or 4; Dan 3,4,5; Annie 2,3,4; Ben 4,5; means Carla won!
Denis Borris . . . . . . . . . 4 pts - Nice charts there; Carla 5/13 mi after turn, Dan 4 8/13 mi in C 5 5/13 time...
Zahi Teitelman . . . . . . . 4 pts - Conds => Carla 1, Emo 2, Ben 5. Dan 42 min for 60/13 mi; Ve = (4/5)Vb, etc.
Doug Babcock. . . . . . . . 4 pts - Carla won, pass Dan at 2520 sec (supercomposite!-D), Emo 2800 sec for 5 5/9 mi.
Yakov Macak . . . . . . . . 3 pts - Good answer as well; these are over 48 hrs after Mark's at this point ...
Philippe Fondanaiche . 3 pts - Nice notation S(n) stmt abt n; r(n) = rank; x/Vd = (10-x)/Vc for x=Dan's Dist.
Wenjun Kang . . . . . . . . 3 pts - Good logic, fine 10step solution, well-explained! Dan 42 Sd = 10 - 42 Sc.
Ken Duisenberg . . . . . . 3 pts - Right, only Carla is 'not behind' anyone so she wins; but why assume Dan is 3rd?
K Sengupta . . . . . . . . . . 3 pts - You misinterp. Carla: ran 7 min 48 sec for each mile, not all ten. Early entry!
Allen Druze. . . . . . . . . . 2 pts - Good solution except for Emo 89.6 min; shoulda been 84 min. Others ok.
Aclaira. . . . . . . . . . . . . . 2 pts - Right order, and good logical flow in your argument; earlier for more pts!
Frank Mullin . . . . . . . . 2 pts - Clues => Anne 2-4, Ben 4-5, Dan 3-5, Emo 2-3, so Carla 1 (one and won)
Claudio Baiocchi . . . . . 1 pt - Sorry about the language barrier; Google translator wasn't enough help either...

problem #263 - posted sunday, april 1, 2007
Here are a couple of cute trig problems for you; no foolin'!
a) Prove: cos(pi/7) - cos(2pi/7) + cos(3pi/7) = 1/2.
b) Find all real solutions to this transcendental equation:
cos^2 (x) + cos^2 (2x) + cos^2 (3x) = 1.

Solution: a) by Giridhar Prasannan
Consider x^7 + 1 = 0. [Letting p = pi] the roots of the equation are e^ip/7, e^i3p/7, ..., e^i13p/7
The sum of these roots should be zero since there is no x^6 term. Hence the real parts sum should also to zero.
cos p/7 + cos 3p/7 + cos 5p/7 + cos 7p/7 + cos 9p/7 + cos 11p/7 + cos 13p/7  = 0
Since cos(2p - x) = cos x, we have: cos 9p/7 = cos 5p/7 , cos 11p/7 = cos 3p/7 , and cos 13p/7 = cos p/7
Also cos7p/7 = - 1 , So we get 2(cos p/7 + cos 3p/7 + cos 5p/7 ) = 1 , i.e. cos p/7 + cos 3p/7 + cos 5p/7 = 1/2
Finally since cos(-x) = - cos x , cos 5p/7 = - cos 2p/7. Hence,
cos p/7 - cos 2p/7 + cos 3p/7 = 1/2.

b) by Hermen Jacobs: "Hi Dan, I send you my solution of problem # 263 B.  cos^2(x) = (1 + cos 2x)/2
Thus cos^2(x) + cos^2(2x) + cos^2(3x) = 1 + cos 2x + cos 4x + cos^2(3x)  The equation is
(cos 2x + cos 4x)/2 + cos^2 (3x) = 0 ;  cos 3x . cos x  + cos^2(3x)   = 0 ; cos 3x.(cos x + cos 3x ) =  0 ;
2.cos 3x. cos 2x. cos x =  0 . . . If cos p =0 then p = pi/2 + k.pi  = pi*(2k +1)/2 [same for cos 2x and cos 3x]
All the solutions are pi*(2k+1)/2  and pi*(2k+1)/4 and pi*(2k+1)/6"
[Dan's note: odd mults of pi/4 and of pi/6; the odd mults of pi/2 are included; you don't need the first ones.]

Another way using polynomials by Prachai K. "Well, this one is just plain bashing... Let cos x = a. Then
cos 2x = 2a^2 - 1, cos 3x = 4a^3 - 3a. Plug those into the equation, you get: a^2 + (2a^2 - 1)^2 + (4a^3 - 3a)^2 = 1,
which reduces to 16a^6 - 20a^4 + 6a^2 = 0, or a^2(2a^2 - 1)(4a^2 - 3) = 0. [Roots a = 0, |a| = sqrt[2]/2, sqrt[3]/2.]
The solutions are a = pi/2, 3pi/2, pi/4, 3pi/4, 5pi/4, 7pi/4, pi/6, 5pi/6, 7pi/6, 11pi/6 and + or - 2kpi, for any integer k.

Nick McGrath . . . . . . . 10 pts - Nice going, even some English spellings this time. Gunning for season win.
Mark Rickert . . . . . . . . 7 pts - Found evidence on www.mathcomp.com thanks for the source, Mark!
Prachai K. . . . . . . . . . . . 6 pts - Right, roots of z^7 + 1 = 0 are odd powers of e^ip/7, for b see above.
K Sengupta . . . . . . . . . . 5 pts - Thanks for finding deMoivre on Wikipedia, emerging as defacto source.
Etienne Desclin. . . . . . . 5 pts - I like your polynomial factoring approach, no charge for clarif, thanx 4 puzzle!
Denis Borris . . . . . . . . . 4 pts - Good remark x^7 + 1 = 0 has no x^6 term so sum of roots is 0; steps on b).
Ravi Raja . . . . . . . . . . . 4 pts - Nice to hear from you again, keep on entering. Good to get cx c2x c3x = 0.
Akifumi. . . . . . . . . . . . . 4 pts - Hard-working PhD student takes time out to do dansmathcontest; thanks!
Giridhar Prasannan . . 4 pts - Very quotable solution; nice work on part b too; see above for part a ans!
Jin Won Park . . . . . . . . 3 pts - Good c = cos x; c^2 + (2c^2-1)^2+ (4c^3 - 3c)^2 = 1 and similar for a.
Marcello Cammarata . . 3 pts - Thanks for the excel spr, monotone func on subintervals. A 14-gon too!
Philippe Fondanaiche . 3 pts - Using good ideas with Trigo; one part had pi/3 when we wanted pi/6.
Zahi Teitelman . . . . . . . 3 pts - Right; one is cos 7pi/7 = -1, others add to 1, and pair up, so ans = 1/2.
Adam Morgan. . . . . . . . 3 pts - Yes, pi/7 is root to cos x - cos 2x + cos 3x = 1/2, 8x^3 - 10x + 3 = 0.
Arijit Bhattacharyya . . 3 pts - Welcome to my contest; interesting that a) can be 1/2 or -3 from your eqn.
Phil Sayre . . . . . . . . . . . 3 pts - Invoked the name Chebyshev, and I appreciate it! T1^2 + T2^2 + T3^2 = 1.
Radu Ionescu . . . . . . . . 3 pts - Welcome back; it's been a while, good answer esp working with cos pi/14.
Claudio Baiocchi . . . . . 3 pts - Nice to see people using Derive (but not when you're Derunk;-). Good prf of b).
Garry Malashkin . . . . . 3 pts - Good proof in there; I found some of the steps hard to fill in but result ok.
Hermen Jacobs . . . . . . . 3 pts - Just part b) but an excellent method (see above) - Thanks for all these years!
Al Nelson . . . . . . . . . . . 3 pts - I like the white board attach; are you still using it in class? Good content too.
Nats Kroy . . . . . . . . . . . 2 pts - Part b only, but nice work in the polynomial realm with factoring & expands.
Haley Buxton . . . . . . . . 2 pts - Good identity in part a) to get 8c^3 - 4c^2 - 4c + 1 = 0; and that works!
Yakov Macak . . . . . . . . 2 pts - You seem to have enough "math tools" to do this job, good on ya Y-Mac!

problem #264 - posted sunday, april 22, 2007
Suppose we add a perfect square to a perfect cube, and the result is the next perfect cube.
a) Prove that the square root of the square must be the sum of two consecutive squares
b) Find the first two all-positive examples of this occurrence. Show your steps and reasoning.
[Trivial example: 0^3 +1^2 = 1^3; 1 = 0^2+1^2; find the next two.]

Solution: Several of you found this proof or a similar one online; like this:
We have (m + 1)^3 - m^3 = 3m^2 + 3m + 1 = n^2, for some integers m and n.
Hence 12m^2 + 12m + 4 = 4n^2, from which 3(2m + 1)^2 = (2n - 1)(2n + 1).
Now, 2n - 1 and 2n + 1 are relatively prime.(By Euclid's algorithm, their greatest common divisor divides
their difference, namely 2. Since both are odd, their greatest common divisor must be 1.)
Therefore we must consider two possible cases, with a and b relatively prime:
(Case 1) 2n - 1 = 3a^2, 2n + 1 = b^2 . . . . (Case 2) 2n - 1 = a^2, 2n + 1 = 3b^2
Taking the first case, we have b^2 = 3a^2 + 2. Impossible, as any square is congruent to 0 or 1, modulo 3.
In the second case, notice a must be odd. Setting a = 2k + 1, we have 2n - 1 = (2k+1)^2 = 4k^2 + 4k + 1.
Hence 2n = 4k^2 + 4k + 2 = 2[k^2 + (k + 1)2]. So
n = k^2 + (k + 1)^2.

I liked this proof from Garry, using Pell's Equation and continued fractions always gets my attention!
a) (m+1)^3 - m^3 = n^2  (1)    3m^2 + 3m + 1 = n^2    12m^2 + 12m + 3 = 4n^2 - 1
3*(2m+1)^2 = (2n - 1)*(2n + 1)  (2) integers 2n-1 and 2n+1 coprime so one must be perfect square
from (1) we have that n is odd so 2n+1 = 3(mod 4) and 2n+1 can't be a perfect square modulo 4
so 2n-1 is a perfect square and 2n-1 is odd   we have 2n-1 = (2k+1)^2 = 4*k^2 + 4k + 1 and
n = 2*k^2 + 2k + 1 = k^2 + (k+1)^2   Q.E.D.
b) Let's suppose 2n=a and 2m+1=b    From (2) => a^2 - 3*b^2 = 1  (3)
Note that if a is even and b is odd then from pair (a,b) satisfying (3) there is a pair (n,m) satisfying (1)
The equation (3) has infinite set of solution (a,b) where a,b are integer and a>0,b>0
Proof:   Suppose that a pair (a1, b1) is solution (3).    Define a pair (ak, bk) by equation :
. .(a1 + b1*sqrt(3))^k = ak + bk*sqrt(3)    (4)   obviously that (4) => (a1-b1*sqrt(3))^k=ak-bk*sqrt(3)
so (ak)^2 - 3*(bk)^2 = (ak - bk*sqrt(3))*(ak + bk*sqrt(3)) = ( (a1-b1*sqrt(3))*(a1+b1*sqrt(3)) )^k
. . . = ( (a1)^2 - 3*(b1)^2 )^k = 1^k = 1  and a pair (ak, bk) also a solution of (3). (2,1) is a solution (3) so
. . . we can to build a infinite solution set (ak,bk) from (4)    Q.E.D.
Easy to calculate first five pairs for (3)     (2,1)   and (2+sqrt(3))^2=7+4*sqrt(3) so (7,4)
and (2+sqrt(3))^3=26+15*sqrt(3) so (26,15) and (2+sqrt(3))^4=97+56*sqrt(3) so (97,56)
and (2+sqrt(3))^5=362+209*sqrt(3) so (362,209)   from first pair (2,1) 2n=2 and 2m+1=1 ==>
. . n=1, m=0 is a trivial example   from third pair (26,15)   2n=26 and 2m+1=15 => n=13, m=7
from fifth pair (362,209)   2n=362 and 2m+1=209 => n=181, m=104.

The actual numbers, the next few solutions, and a pattern are in this table from Etienne of Belgium:
7^3 + 13^2 = 343 + 169 = 512 = 8^3 (and 13= 2^2 + 3^2 )
104^3 + 181^2 = 1124864 + 32761 = 1157625 = 105^3 (and 181=9^2 + 10^2 )
1455^3 + 2521^2 = 1456^3 (and 2521= 35^2 + 36^2 )
20272^3 + 35113^2 = 20273^3 (and 35113= 132^2 + 133^2 )
282359^3 + 489061^2 = 282360^3 (and 489061= 494^2 + 495^2 )
3932760^3 + 6811741^2 = 3932761^3 (and 6811741= 1845^2 + 1846^2 )
n |  cube       square
0 |  0          1
1 |  7          13       c1 = 8*s0 - 1
2 |  104        181      c2 = 8*s1
3 |  1455       2521     c3 = 8*(s2 + s0) - 1
4 |  20272      35113    c4 = 8*(s3 + s1)
5 |  282359     489061   c5 = 8*(s4 + s2 + s0) - 1
6 |  3932760    6811741  c6 = 8*(s5 + s3 + s1)

Nick McGrath . . . . . . . 10 pts - First and formost again, Nick; good proof of cube-different squares!
Mark Rickert . . . . . . . . 6 pts - Similar proof, valid algebra; second example ok first is 8^3 - 7^3 not 9^3 - 8^3
K Sengupta . . . . . . . . . . 5 pts - I like the references to MathWorld and Pell's Equation (Mark too)
Garry Malashkin . . . . . 5 pts - Good use of recursions: (2 + \/ 3)^3 = 26 + 15 \/ 3 ; 26/2 = 13; 13^2 = 169
Denis Borris . . . . . . . . . 4 pts - I couldn't reconcile your integer sequences links but the followup helped
Claudio Baiocchi . . . . . 4 pts - The two 'Lemmata' were more than enough to prove sum-of-squares property.
Philippe Fondanaiche . 4 pts - Nice proof using rel prime factors and Pell's Equation. Keep on entering!
Etienne Desclin. . . . . . . 3 pts - Good table of examples, and intuition (not proof) about patterns that emerge
Ravi Raja . . . . . . . . . . . 3 pts - I like mult by 4 then factoring out a 3. Very clever use of quadratic residues
Nats Kroy . . . . . . . . . . . 2 pts - Part b answer was great; good to list \/(3x^2+3x+1) to see if it's an integer
Jean Moreau de Saint-Martin (new) 2 pts Bienvenue a mon concours; good proof! See examples of first sol's above.
Zahi Teitelman. . . . . . . 2 pts - Right that 105^3 - 104^3 = 181^2 and 181 = 10^2 + 9^2, same with 8, 7, 13.
Radu Ionescu . . . . . . . . 1 pt - You found the one solution; right that 3x(x+1) = (y+1)(y-1) this yields more
Hermen Jacobs . . . . . . . 1 pt - Used trusty Basic to find solution with 13^2 = 8^3 - 7^3; let it run longer.
Phil Sayre . . . . . . . . . . . 1 pt - "7 and 2. 104 and 9." Somewhat right; I sense there's a man in a hurry here ;-}

problem #265 - posted wednesday, may 23, 2007
The 'minimal polynomial' of a number z is the f(x) of smallest degree with integer coefficients
having z as a root. For example the min poly for z = \/3 is f(x) = x^2 - 3.
Find min polys for these "algebraic" numbers: a) z = \/ 7 + \/ 11 . .b) z = 2 \/ 3 + 6 i
c) z = sqrt[5] + cuberoot[2] . d) z = tan(18 deg). Show steps & reasoning.

Solution: Thanks to K Sengupta [my comments in brown]

(a) By conditions of the problem: x = V7 + V 11 . . . Or, x^2 ­ 18 = 2V 77 [sq both sides; ok x not z.]
Or, x^4 ­ 36*x^2 + 324 = 308 . . Or, x^4 ­ 36*x^2 + 16 = 0. Consequently, f(x) = x^4 ­ 36*x^2 + 16

(b)  By the problem: (x- 2V3)^2 = (6i)^2 . . . Or, x^2 ­ 4*V3*x + 12 = -36
Or, 4*V3*x = x^2 + 48 . . . Or, (4*V3*x)^2  = (x^2 + 48)^2 . . . Or, 48*x^2 = x^4+ 96*x^2 + 2304
Or, x^4 + 48*x^2 + 2304 = 0 . . . . Consequently,  f(x) =  x^4 + 48*x^2 + 2304

(c) By the problem: x ­ V5 = cuberoot(2) . . . Or, (x - V5)^3 = 2 . . . Or, (x^3 + 15x) ­ V5(3x^2 + 5) = 2
Or, x^3 + 15x ­ 2 = V5(3*x^2 + 5) . . . Or, (x^3 + 15x ­ 2)^2 = 5*(3*x^2 +5)^2
Or, x^6  + 30*x^4 - 4*x^3 [+ 225x^2] ­ 60x + 4 = 45*x^4 + 150*x^2 + 125 [oops]
x^6 ­ 15x^4 ­ 4x^3 + 75x^2 ­ 60x ­ 121 = 0 ; f(x) = x^6 ­ 15*x^4 ­ 4*x^3 + 75*x^2 ­ 60x ­ 121

(d) Put   x = tan h, where h = 18 degrees ; Accordingly, 5h = 90 degrees, giving: 2h = (90 ­ 3h)
Or, tan 2h = tan (90-3h) = cot 3h . . . Or, tan 2h = 1/(tan 3h) [these are like Chebyshev quotients]
Or, 2*tan h/(1 - tan^2 h) = (1 - 3*tan^2 h)/(3*tanh - tan^3 h) (resubstituting the value x = tan h)
Or, 2x/(1-x^2) = (1- 3*x^2)/(3x- x^3) . . . Or, 6*x^2 ­ 2*x^4 = 1 - 4*x^2 + 3*x^4
Or, 5* x^4  - 10*x^2 + 1 = 0 . . . Consequently, f(x) =  5* x^4  - 10*x^2 + 1

K Sengupta . . . . . . . . . . 9 pts - Very good answer except the one coefficient; thanks for entries!
Claudio Baiocchi . . . . . 7 pts - I like yr correction to precision of stmt: real rel prime integer coeffs.
Nick McGrath . . . . . . . 6 pts - Good that tan(36+54) = infinity, so denom = 0; tan36*tan54 = 1.
Ed Wern . . . . . . . . . . . . 5 pts - Welcome back Ed; MathWorld says tan(pi/10) = \/(25 - 10\/5) / 5.
Mark Rickert . . . . . . . . 5 pts - Good idea, expand powers of \/7 + \/11 etc and elim radical terms
Hermen Jacobs. . . . . . . 4 pts - Early entry Hermen! Nice work on a, b, c; small coeff error on b.
Al Nelson . . . . . . . . . . . 4 pts - Nice hand-writ pdf; d) is just 4th deg, u had extra factor x^4 - 10x + 5.
Phil Sayre . . . . . . . . . . . 4 pts - First 3 parts were good and basic, a surd came along and helped on d.
Radu Ionescu . . . . . . . . 3 pts - First entry; right polys no steps on a,b,d; resub nicely solved part c.
Denis Borris . . . . . . . . . 3 pts - I was ok on a, b, c but tan(2*9) = 2tan9/(1-tan29) = \/2 (\/5 - 1) etc ?
Zahi Teitelman. . . . . . . 3 pts - Yes the factor theorem works with a, b, c. And d) extra 5x^4 + 10x^2 + 1.
Philippe Fondanaiche . 3 pts - Very nice work, mais you missed an x on yr 12i term x4 - 96x2 + 2448
Nats Kroy . . . . . . . . . . . 3 pts - Wow, a vacation with no internet is like three weeks off, in the 1960s!
James Laverty (new) . . . 2 pts - Welcome to my contest; nice answers on a,b; good intuition on degrees c, d.
Bengt Blomster (new). . . 2 pts - Hello Bengt; I liked that cos72 = 1/(2phi) = (\/5 - 1)/4. Keep on entering!
Wenjun Kang . . . . . . . . 2 pts - Nice answer; right amount of detail, well done (received before 266 up)
Ajit Athle . . . . . . . . . . . 2 pts - Your a, b, c are ok; in d you had tan36 so got its poly z^4 - 10z^2 + 5.

problem #266 - posted friday, august 31, 2007
a) Which number is larger, S or T ? (Both a numerical comparison and calculator-free proof)
b) Which is larger, S/99 or T/100 ? (Give the numerical comparison, and a proof if you can)
c) Figure out the limit, as n --> oo, of
(oo is infinity . . . Again, try for a numerical value and an exact proof)

Solution: Welcome new contestant and dansmath.com website fan Joe Fendel, who writes:
A) I'd say T is larger. If we compare S^(100) and T^(100), we get 99!*S and 100!. Divide by 99! and we
are comparing S and 100. Clearly S < 100, since 100^99 is term-wise more than 99!. Thus T > S.
Excel bears me out: S is about 37.6231 and T is about 37.9927.
B) Here I'd say S/99 is bigger than T/100. Let's start with 99! / 99^99, just because that's such a fun number.
If we multiply this by (99/100)^99, we get a smaller number, which is equal to 99! / 100^99, and this is the
same as 100! / 100^100. The 99th root of the larger number is S/99, and the 100th root of the smaller
number is T/100. And so S/99 must be smaller than T/100 (since they're each larger than 1).
Again, Excel confirms: S/99 is about 0.380031 and T/100 is about 0.379927.
C) This sure looks like it's headed toward 1/e, which is about 0.3679. Makes sense in the context of limits
involving exponents, too. How can we prove it? Well, 1/e is the limit as n -> oo of (n / n+1)^n.
That looks a lot like that "if we multiply this by (99/100)^99" step above! So for example, let x be the limit
of [(n!)^(1/n)]/n as n -> oo. Then since for all n, we have x = x^(n+1)/x^n, it's true in the limit, and thus
x is the limit of the ratio of (n+1)!/(n+1)^(n+1) and n!/n^n, which is (n / n+1)^n, and that leads to 1/e!

Dan's note: Joe was one of the few people not to use Stirling's formula, which as you may know, says:
n! ~ [\/(2 pi n)] * [(n/e)^n] ; but did you know that there's a whole series making this more accurate:
n! ~ [\/(2 pi n)] * [(n/e)^n] * [1 + 1/(12n) + 1/(288n^2) + . . . ] : Cool!

Philippe Fondanaiche 10 pts - Nice attachmt; is this your first weekly win? Fe'licitations Philippe!
Nick McGrath . . . . . . . 7 pts - Shows logT - logS > 0; good proof of monotonicity (long word!)
K Sengupta . . . . . . . . . . 5 pts - Very interesting & original steps; a bit tricky, good to Wikipedia Stirling!
Moreau deSaint-Martin 4 pts - I liked the second proof you sent in, and your use of Stirling's f!
Joe Fendel (new) . . . . . . . 4 pts - Thanks for finding my site (and liking it!) Quotable answer too. (above)
Mark Rickert . . . . . . . . 4 pts - Not all proofs included but first entry received and numericals were ok!
Radu Ionescu . . . . . . . . 3 pts - Early entry gains point; major steps left out of b, c but right idea Radu.
Art Morris . . . . . . . . . . 3 pts - Welcome back Art! Good numericals and nice prf on a) and Stirling too!
Marcello Cammarata. . 3 pts - Nice answer but I think logT = log1+log2+...+log100)/100 not times...
Etienne Desclin. . . . . . . 3 pts - Very pretty PDF layout 8-} Yes the sequence in c) does indeed decr.
Zahi Teitelman. . . . . . . 3 pts - Stirling yes but in the limit, not alw. exact. I liked Excel LN, FACT, EXP.
Claudio Baiocchi . . . . . 3 pts - Thanks for the support and the great proof of part c). Radius of conv!
Nats Kroy . . . . . . . . . . . 3 pts - In c) limit is 0.37 to 2 places, cool use gamma function to smooth out n!
Steve Tonnesen (new) . . 3 pts - Welcome to my contest; nice answers on a,b; good intuition on degrees c, d.
Hermen Jacobs. . . . . . . 2 pts - Good to program it but I think you used a z=n! in place of n in expon.
Denis Borris . . . . . . . . . 2 pts - Thanks for the resub to suggest Stirling; not sure abt ratio argt in b)...
James Laverty . . . . . . . 2 pts - Interesting approach in a); thanks for trying for prf of b); Stirling rules c)
Phil Sayre . . . . . . . . . . . 2 pts - A bit after the Labor Day rush, good Stirling; pretty close call on b).

problem #267 - posted thursday, september 27, 2007
Next-to-last problem of this season! 11th season starts with Problem 269.
A natural number n is 'squareful' if, whenever a prime goes into n,
the square of that prime also goes into n.
a) Find all squareful numbers less than or = to 300.
b) Prove that n is squareful if and only if n is the product of a square and a cube.
c) Prove there are an infinite number of pairs of consecutive squareful numbers, like 8'n'9.

Solution: a) Etienne (de Belgique) nous donne les 28 solutions des chiffres "pleins de pouvoir":
 01) 1 02) 4   = 2^2 03) 8   = 2^3 04) 9   = 3^2 05) 16  = 2^4 06) 25  = 5^2 07) 27  = 3^3 08) 32  = 2^5 09) 36  = 2^2 * 3^2 10) 49  = 7^2 11) 64  = 2^6 12) 72  = 2^3 * 3^2 13) 81  = 3^4 14) 100  = 2^2 * 5^2 15) 108  = 2^2 * 3^3 16) 121  = 11^2 17) 125 =  5^3 18) 128  = 2^7 19) 144  = 2^4 * 3^2 20) 169  = 13^2 21) 196 =  2^2 * 7^2 22) 200 = 2^3 * 5^2 23) 216 = 2^3 * 3^3 24) 225 = 3^2 * 5^2 25) 243 = 3^4 26) 256 = 2^8 27) 288 = 2^5 * 3^2 28) 289 = 17^2
b) Et Jean Moreau de Saint-Martin (de Paris) m'a envoye':
<== Suppose n = (s^2)(c^3), product of a square and a cube. If a prime p goes into n, it goes into
at least one of the nulmbers s and c; its square p^2 goes into s^2 and/or c^3, so this square goes into n.
==> Suppose n a squareful number; write it as the product of its prime divisors p, q, r, ... with exponents a, b, c, ...
n = (p^a)(q^b)(r^c)... Every exponent is at least 2 and every odd exponent is at least 3.
Let c be the product of all primes with an odd exponent in n (if no such prime, take c=1). Then c^3 goes into n.
Writing n/c^3 as product of primes, every prime has an even exponent, so n/c^3 is a square s^2 (possibly 1).
Then n  = (s^2)(c^3), product of a square and a cube. QED.

c) Consider the (Pell) equation . . . x^2 = 8y^2 + 1. . . . (x,y) = (3,1) is a solution. There is no maximum solution:
if (x,y) is a solution, (3x + 8y, x+3y) is another (and bigger) solution. So this equation has infinitely many solutions.
Each solution gives a pair (8y^2, x^2) of consecutive integers which are squareful integers: x^2 is a square and 8y^2
is the product of a square y^2 and a cube 8=2^3. The first pair is (8, 9). This proves that there are infinitely many
pairs of consecutive integers which are squareful integers. Remark. The Pell equation x^2 = 27y^2 + 1 gives other
such pairs, beginning by (675, 676). And so on... By this method we have pairs of squareful numbers, one of them
being a square. There are pairs without squares, for example: (12167 = 23^3, 12168 = (2^3)(39^2) ).

Joe Fendel . . . . . . . . . . . 10 pts - Wow, a weekly winner in the second entry. Fast track to partner ;-}
Tim Poe . . . . . . . . . . . . . 7 pts - Nice how you elim a certain amount by each condition imposed.
Nats Kroy. . . . . . . . . . . . 6 pts - Good mult table pitting squares against cubes - clever Nats!
Nick McGrath . . . . . . . . 5 pts - True, these are widely-er known (Sloane) as powerful numbers.
Philippe Fondanaiche . 5 pts - Nice recursion for Pell's eqn x2 - 8y2 = 1: x' = 3x + 8y; y' = x + 3y
K Sengupta . . . . . . . . . . 4 pts - Small typo double expon for triple, good proof (1 is ok) and Pell's.
Garry Malashkin . . . . . 4 pts - That's a pretty complete proof; good that ak + bk \/2 = (1 + \/2)^k.
Claudio Baiocchi . . . . . 3 pts - Right; you prove a stronger result, the infin. consec. sqfl and sqs!
David Stigant . . . . . . . . 3 pts - Welcome back; thanks for showing the nice 'Haskell' program!
Moreau deSaint-Martin 3 pts - Good recursion, nice 12167=23^3, 12168=2^3 39^2 both non-sq.
Ed Wern . . . . . . . . . . . . 3 pts - List > 1 with 27 elts and 27 as an elt! I like any symmetry I can get.
Marcello Cammarata . . 3 pts - Powers of primes 2..17 ; new recursion (a,b)->(4a(a+1),(2a+1)^2)
Mark Rickert . . . . . . . . 3 pts - Movin up that all-time list; yet another recursion an+1 = 6 an - an-1.
Phil Sayre . . . . . . . . . . . 3 pts - Last entry but full of interest; (1+\/2)^k, (2+\/3)^k, (5+2\/6)^k, fund units!
Denis Borris . . . . . . . . . 2 pts - That was the right link, rather short answer; next prob more interesting!
James Laverty . . . . . . . 2 pts - Numbers like 12 aren't squareful/powerful bec 3 goes in but 9 doesn't.
Etienne Desclin. . . . . . . 2 pts - Nice proof especially in part b) read others' ans here for info part c.
Steve Tonnesen . . . . . . . 2 pts - Keep on entering, good thinking and correct on those first 4 pairs.
Zahi Teitelman. . . . . . . 1 pt - Same as James; 63 has the same expon structure as 12 and isn't sqfl.

problem #268 - posted wednesday, october 10, 2007
Last problem of this season! Eleventh annual contest starts with Problem 269.
I was working on the circuits in my house, and found a bundle of nine identical wires,
going into the wall and up to the attic. In the attic were the nine wires going down to
the main floor. Suppose I can twist any number of wires together at either end, and I
have a circuit tester that lights when a loop is complete. How can I match up all nine
pairs of wire ends, using the smallest number of trips up to (and down from) the attic?
Assume the wires are insulated except for a few bare cm at the ends. Wires cannot be pulled thru
the wall or pipe. Show your steps and reasoning.

Solution: Here's a detailed solution by longtime solver Tim Poe; his first weekly win since Problem 233:

"First, I note that the problem asks for the minimum number of round trips to the attic, not the number of tests.
I assume that your circuit tester is portable (more or less a standard multimeter) and you can take it with you
and test from either end. I also assume you begin downstairs.

Downstairs: connect a pair of wires, a trio of wires, and a quartet of wires.

Attic (.5 round trips): Using the loop tester, find the ends of each group. (A wire that completes a loop with
only one other is one of the pair; one that makes a loop with two others is one of the trio, and one that makes
a loop with three others is part of the quartet.) Once this is done, you*ll have the same bundles as you have
downstairs, but within each bundle you won*t know which wire is which. Label the pair (in any order) A and B;
the trio (again in any order) C, D and E; and the quartet (still in any order) F, G, H and I. Connect them in the
following groups: {A,C,F}, {B,G}, {D,H}, {E} and {I}

Downstairs (1 round trip): Disconnect the bundles, labeling those that were paired as *2* and *2*, those from
the trio as *3*, *3* and *3*; and those from the quartet as *4*, *4*, *4* and *4*. Start checking for circuits:
a *2* that makes a loop with both a *3* and a *4* is wire *A* -- a *2* that makes a loop with a *4* but not a *3* is wire *B*
a *3* that makes a loop with a *2* and a *4* is wire *C* -- a *3* that makes a loop with a *4* but not a *2* is wire *D*
a *3* that makes a loop with nothing is wire *E* -- a *4* that makes a loop with a *2* and a *3* is wire *F*
a *4* that makes a loop with a *2* but not a *3* is wire *G* -- a *4* that makes a loop with a *3* but not a *2* is wire *H*
a *4* that makes a loop with nothing is wire *I*

So you have all wires identified with only one round trip.

Tim Poe. . . . . . . . . . . . . 10 pts - I like the permutation cycle notation. First weekly win since 2005!
Claudio Baiocchi . . . . . 7 pts - Adapted a 10-wire solution with and w/o a "ground wire" (cheating)
Nick McGrath . . . . . . . . 5 pts - Another yearly win to go along with 2004 and 2005; jolly good!
Marcello Cammarata . . 4 pts - I wonder if 'Marcello Cammarata' is Italian for Martin Gardner?
Ken Duisenberg . . . . . . 4 pts - Nicely worded solution; great to have a fellow puzzler along!
Joe Fendel . . . . . . . . . . . 4 pts - I'm proud to show my face in your google search results! ;-}
David Stigant . . . . . . . . 4 pts - Nice thorough explan; bonus pt for proof it can't be done in one.
Denis Borris . . . . . . . . . 3 pts - I managed to patch yr correct 15-wire sol'n to your resub instruc!
Etienne Desclin. . . . . . . 3 pts - Cool notation 1111,222,33 as twisty groups, then go upstairs...
Philippe Fondanaiche . 3 pts - I see you are part of the "four pairs plus one" camp. Ca marchera!
Mark Moyer (new) . . . . . 3 pts - Great to have you along for the dansmath ride... keep entering!
Steve Tonnesen. . . . . . . 3 pts - Reduced the number of trips in resub, solved while on a driving trip!
Ed Wern . . . . . . . . . . . . 3 pts - I like yr fake sol'ns: call electrician, buy a really long wire, yell w/partner.
Mark Rickert . . . . . . . . 2 pts - Early entry with update; not sure why you corrected 3 trips with 4.
James Laverty . . . . . . . 2 pts - You answered yr own ques abt interior wires; fewer trips needed
Nats Kroy. . . . . . . . . . . . 2 pts - A loop has to involve more than one wire, still solved in 3 trips.
Yakov Macek . . . . . . . . 2 pts - I dug your series of pictures; a how-to manual! Still too many trips.
Mary Noggin (new) . . . . 1 pt - "I'd hold a lightbulb between my wet feet, climb a ladder, wet my fingers..."

.
problem #269 - posted tuesday, october 23, 2007
First problem of the season! My eleventh annual contest! %;-}
 [ Another Brick In The Wall ] Figure out the values of the 15 bricks marked A, B, ..., N, O. Each value is the sum of the two below it, like F = A + B. Some values are given, L is 5 more than J. (Time's up on this one.)

Solution: Wow, there were a lot of you with entries this time! Yes, it was a particularly easy one to start off YearEleven,
and it did draw in some new contestants, so I have done my job %;-} The numerical answers for the bricks were:
. . . . . . . . . . O: 165
. . .. . . . . M: 80 , N: 85
. . . . .J: 35 , K: 45 , L: 40
. . F: 14 , G: 21 , H: 24 , I: 16
A: 2 , B: 12 , C: 9 , D: 15 , E: 1

Here are a couple of GIFs from JinWon's PDF (I couldn't grab the text)
And now Garry's solution! . . .
"We have the system : O = M + 85 ; M = J + K ; 85 = K + L ; J = 14 + G ; K = G + 24 ; L = 24 + I ; G = B + C ;
I = 25 - C and  L = J + 5 . . . nine equations, nine variables ( B,C,G,I,J,K,L,M,O )
I use the program which can solve a system of linear equations with exactly nine equations.
It get O=165, M=80, J=35, K=45, L=40, G=21, I=16, B=12, C=9. Easy to calculate A=14-B so A=2, and D=24-C so D=15."

Many of the rest of you showed complete equation-solving details, and I thank you (and usually reward you) for that! - Dan

Nick McGrath . . . . . . . .10 pts - Last year's winner, & off to a fast start again! Go Nick McG!
Tim Poe . . . . . . . . . . . . . 7 pts - Thanks, it was nice to be back, or bach in my case, back then.
Mark Rickert . . . . . . . . 5 pts - Hey Mark; good to notice the key bricks were G and I (not Joe)
Denis Borris . . . . . . . . . 5 pts - This "one of my top cuties" was photographed in a bookstore.
Claudio Baiocchi . . . . . 4 pts - Good method, solve for variables "floor-by-floor"-- it worked!
Etienne Desclin. . . . . . . 4 pts - Liked yr method of gray variables, filled in by black numbers!
Art Morris . . . . . . . . . . . 3 pts - Art "one-full-decade-of-entries" Morris 'Excel-s' yet again ;-}
Ondrej Calda (new) . . . . 3 pts - A new math person 'Czechs in' from a new country for my contest!
Garry Malashkin . . . . . 3 pts - Nine equations in 9 variables sounds do-able, by hand or 'solver'.
Alan O'Donnell. . . . . . . 3 pts - Yes it's true this one was easier, but tougher challenges lie ahead!
 Phil Sayre. . . . . . . . . . 3 pts - Picked the top-down method, eh? Joe Fendel . . . . . . . . . . 3 pts - B=14-A, yes, etc. You got em all! Jin Won Park . . . . . . . 3 pts - Nice quoted ans (no attachmt needed) Nats Kroy. . . . . . . . . . . 3 pts - Brick pic didn't arrive but ans ok! Steve Tonnesen . . . . . . 3 pts - K+L=85, subst in, lots of eqns!! Hermen Jacobs . . . . . . 3 pts - Good sys of 11 eqns, sorry abt 268 Yakov Macek . . . . . . . . 3 pts - Yer welcome 4 ez starter problem* *James Laverty . . . . . . . 3 pts - I liked "pyramid of values" & sys. Marcello Cammarata . 3 pts - Nice mix of eqns, thoughts, sol'ns. Allen Druze . . . . . . . . . 3 pts - Very clear expl'n, thnks! All good. David Stigant . . . . . . . 3 pts - Good eqns to sort things out, ok! Dawn Haught (new) . . . 3 pts - 85=K+L, L=G+19, G=K-24, go Dawn! Ravi Raja . . . . . . . . . . . 3 pts - J=x=F+G etc. Ok, who needs x? Mark Moyer . . . . . . . . . 3 pts - From diagram K+L=N & so on!<

problem #270 - posted sunday, november 11, 2007 - Second problem of my eleventh contest!
 Spheres Filling Space (and vice-versa) (back to top) There are three common unit-sphere packings in 3D space, listed in increasing density: i) Stacked square-grid layers, ii) Stacked hexagonal layers, iii) Indented hexagonal layers. a) Find the exact density of each packing, then round to 0.1%. b) Fill in the main spaces for each packing i), ii), iii), with small spheres that are twice as dense as the unit spheres.Find the size of the small spheres in each case, and then rerank the total mass densities from lowest to highest, giving both the exact values and the approximations to 0.1%. Show steps and reasoning;

Solution: Dan's note: Wow, I was ready to post this a couple of days ago but I realized there is more than one size hole
in the third packing, and the "main spaces" might either be the largest, or the largest and the more numerous next-largest.
None of this problem's responses took this into account, so I made the max score 8, then 6, then 4, 3, 2. Details below.

Original note: Ok, not so many contestants this time, but you begged me to make up more interesting problems!
There was a split on this one: a few of you didn't take into account the relative numbers of types of small spheres.
Here's Jean Moreau de Saint-Martin with a great detailed report, followed by my calculation (which I believe to be correct).

i) Stacked square-grid layers - Each sphere can be packed in a cubic box, with side 2.
Volume of a unit sphere 4pi/3, volume of the cube 8. Density (4pi/3)/8 = pi/6 = 0.5235987... rounded 52.4%
ii) Stacked hexagonal layers - The package of a sphere is now a prism with height 2, hexagonal base circumscribed to a unit circle
The hexagon has side 2/sqrt(3)., perimeter 6x2/sqrt(3) = 4sqrt(3), area (1/2)4sqrt(3) = 2sqrt(3).
The prism has volume 2x2sqrt(3) = 4sqrt(3). Density (4pi/3)/(4sqrt(3)) = pi/(3sqrt(3)) = 0.604599...  rounded 60.5%
iii) Indented hexagonal layers - Let A, B the centers of 2 spheres in contact in 2 different layers. AB = 2.
Consider the prism of case ii), centered in A. B is located on a vertical side of this prism, so AB has a horizontal component of 2/sqrt(3).
The vertical distance between layers is the vertical component of AB, which is, by Pythagoras' theorem, sqrt(8/3).
Each sphere can be associated with a hexagonal prism having height sqrt(8/3) and (as in the stacked case) base area 2sqrt(3).
The volume of this prism is 4sqrt(2). Density (4pi/3)/(4sqrt(2))=pi/sqrt(18) = 0.740480... rounded 74.0%

*** that was part a) *** now part b) ***

i) The small spheres have their center at the apices of the cubes packaging the unit spheres, with distance sqrt(3) from the centers of the cubes.
They have radius (sqrt(3)-1), their volume is (sqrt(3)-1)^3 = 6sqrt(3) - 10 times the volume of a unit sphere.
The square grid of small spheres has the same mesh width as the square grid of unit spheres. So there is one small sphere for every
unit sphere, but they are twice as dense as the unit spheres, so the mass density is magnified by the factor
1 + 2(6sqrt(3)-10) = 12sqrt(3) - 19. Finally the density is (12sqrt(3) - 19)pi/6 = 0.934419... rounded 93.4%

ii)  The small spheres have their center at the apices of the prisms packaging the unit spheres. From the center A of a prism, the small
sphere in contact has center B, and AB has projections 1 vertical, 2/sqrt(3) horizontal, so AB = sqrt(7/3) by Pythagoras' theorem.
So the small spheres have radius (sqrt(7/3)-1), their volume is (sqrt(7/3)-1)^3 = (16/9)sqrt(21) - 8 times the volume of a unit sphere.
Each small sphere lies on 3 unit spheres. The centers of these 3 spheres form a triangle with area half of the prism base area.
So  there are 2 small spheres for every unit sphere; moreover they are twice as dense as the unit spheres, so the mass density is magnified
by the factor 1 + 2x2((16/9)sqrt(21) - 8) = (64/9)sqrt(21) - 31.
Finally the density is (  (64/9)sqrt(21) - 31)pi/sqrt(27) = (64sqrt(7) - 31sqrt(27))pi/27 = 0.959623... rounded 96.0%

iii) Each unit sphere lies on 3 other unit spheres. Their centers form a regular tetrahedron with side 2.
Each tetrahedron center is the center of a small sphere. The distance from this center to the tetrahedron apices is sqrt(3/2).
So the small spheres have radius (sqrt(3/2)-1), their volume is (sqrt(3/2)-1)^3 = (9sqrt(6) - 22)/4 times the volume of a unit sphere.
Let A the center of a unit sphere, in contact with (in the upper layer) spheres with centers B,C,D. A is an apex for 4 tetrahedra:
3 determined by A, one of points B,C,D, and 2 neighbours of A in its layer; and one tetrahedon ABCD. A is an apex for 4 other tetrahedra
with spheres in the lower layer. As each center A belongs to 8 tetrahedra, and each tetrahedron has 4 apices, the tetrahedra are twice as
many as the unit spheres, and so are the small spheres with density 2.
Thus the mass density is magnified by the factor 1 + 2x2(9sqrt(6)-22)/4 = 9sqrt(6)-21.
Finally the density is ( 9sqrt(6)-21)pi/sqrt(18) = (sqrt(27) - sqrt(24.5))pi = 0.774103... rounded 77.4%

From lowest density to highest density, the new ranking is
iii) Indented hexagonal layers, 77.4% i) Stacked square-grid layers, 93.4% ii) Stacked hexagonal layers, 96.0%.

 Dan's note: There was a variety of opinion on the number of small spheres (radius \/(3/2) - 1 = 0.2247) per unit sphere: 1, 2, or 3. But as I notice, there are larger (octahedral) holes between the spaces of layers, with small spheres of radius \/ 2 - 1 = 0.414, which have volume (4 pi/3)(\/2 - 1)^3 = (4 pi/3)(5\/2 - 7) and these are in a 1:1 ratio with the unit spheres, so the new density is (pi/6)(\/2) + (pi/6)(5\/2 - 7) = 0.7405 + 0.0372 = 0.7777 = 77.8%. If the small tetrahedral holes (2 to a unit sphere) are also filled ('main' spaces but not the mainest), the density becomes (pi/6)(6\/2 - 7) + (pi/3)(9\/6 - 22) = 0.7777 + 0.0119 = 0.7896 = 79.0% Either way this formerly densest packing finishes last, and the winner is the stacked hexagonal at 96.0% total (mass) density.

Another French contestant Philippe F attached a beautiful word document from which I stole these pictures:
cubical
stacked hexagonal (i'm leaving out all his nice equations!)
indented square / hex

Moreau deSaint-Martin . . . 8 pts - This early answer came the closest; see quoted solution above!
Nick McGrath . . . . . . . . . . . 6 pts - I believed you until I found some of those spheres were bigger.
Marcello Cammarata . . . . . 4 pts - First answer, claimimg the reverse order, good expl of sphere sizes
Philippe Fondanaiche . . . . 4 pts - That was an awesome document you produced, with embedded picts!
Claudio Baiocchi . . . . . . . . 3 pts - Correct original densities, smaller spheres are twice as heavy per unit vol
Etienne Desclin. . . . . . . . . . 3 pts - Got i and ii right; last was pi\/2 / 5 good expl but off a bit on density
James Laverty. . . . . . . . . . . 3 pts - Excellent answer on first part; I do come through with partial credit
Nats Kroy . . . . . . . . . . . . . . 3 pts - Great argmt with boxes LMN and limit of number of spheres inside
Hermen Jacobs . . . . . . . . . . 2 pts - I believe you on the first cubical part, the tetrahedral is good locally
K Sengupta . . . . . . . . . . . . . 2 pts - Very good job on part a) all three parts, enjoy the space ball debate.
 THANKS to all of you who have entered, or even just clicked and looked. My site is in its 11th season - OVER 99,000 HITS so far! (Not factorial.) Help it grow by telling your friends, teachers, and family about it. YOU CAN ALWAYS FIND ME AT dansmath.com - Dan the Man Bach - 2007 A.D.

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