dan's problem archives 241-250

dan's math@home - problem of the week - archives

Problem Archives page 25

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 . 231-240 . 241-250 . 251+ index

241-- Local (loco)motion

242 Put a Hex around You

243 Altitudes with Attitude

244 - Half More Number

245 - Sphere of Influence

246 -- Beating the Cycle!

247 -- I Am the Bugman!

248- HowManyRunners?

249 - Eat a Peach (a mile)

250 - - - Pack 'Em All In!

problem #241 - posted sunday, feb 19, 2006
local (loco)motion (back to top) (note: I did this one on my podcast!)
Dan goes 49 miles in 9 hours; a mixture of running & walking.
Can you tell how fast he runs and how fast he walks in each (i)-(v)?
(i) Dan runs twice as fast as he walks . . . . . (ii) he runs 4 mph faster than he walks
(iii) he runs 7 mph and he walks 3 mph . . . (iv) he runs 5 mph and he walks 4 mph
(v) he runs 5 min, walks 1 min, repeats
a) Which of these have unique solutions? What are they?
b) Which of the above can not occur? Why not, exactly?
c) Give a range of solutions, for any that are not unique.
Show your reasoning.

Solution: (From solid contender Mark Rickert)
a) Only (iii) has a unique solution (see below). b) Only (iv) cannot occur (see below). c) See below.
For all situations, let w = walking speed ; r = running speed
(i) t = walking time ; r = 2w ; 49 = 2w(9-t) + wt
w = 49 / (18 - t) ; r = 98 / (18 - t) ; t ranges from 0 to 9
This solution is not unique ; w ranges from 49/18 to 49/9 ; r ranges from 49/9 to 98/9
(ii) t = walking time ; r = w+4 ; 49 = (w+4)(9-t) + wt
w = (13 + 4t) / 9 ; r = (49 + 4t) / 9 ; t ranges from 0 to 9
This solution is not unique ; w ranges from 13/9 to 49/9 ; r ranges from 49/9 to 85/9
(iii) t = walking time ; 49 = 7(9-t) + 3t ; t = 7/2
This solution is unique. Dan walks for 3.5 hrs (10.5 mi) and runs for 5.5 hrs (38.5 mi).
(iv) t = walking time ; 49 = 5(9-t) + 4t ; t = -4
This cannot occur because Dan would have to run for 13 hours to complete the race.
(v) Assuming Dan has to complete the race at the end of a 6 minute cycle:
Dan is running 5/6 of the time. ; 49 = (5/6)9r + (1/6)9w ; 98 = 15r + 3w
Assuming Dan runs at least as fast as he walks: 98 = 15(w+d) + 3w , where d>=0 is the
difference between his running speed and his walking speed ; 98 = 18w + 15d
This solution is not unique. w from 0 to 49/9 (when d=0), d from 0 to 98/15 (when w=0)
Therefore, r ranges from 49/9 to 98/15

WINNERS - Problem 241 . . Thanks for entering! - Dan (back to top) . leader board
Marcello Cammarata . . 10 pts - Like most, Marcello assumed I run faster than I walk
Mark Rickert . . . . . . . . . 7 pts - Nice solution, worthy of quoting (or stealing) above
Zeke Moore . . . . . . . . . . . 6 pts - Good explanations; again assumed r >= w; hmmm.
Mohamed Omar . . . . . . . 5 pts - Right; i) w = 49/(18-t) etc. underdetermined solution
Nick McGrath. . . . . . . . . 5 pts - In SF marathon I maybe walked parts faster'n I ran
Tim Poe . . . . . . . . . . . . . . 4 pts - Early entry survived a saving resubmission; good!
Zahi Teitelman. . . . . . . . 4 pts - Yes; 2xy + x(9 - y) = 49 leads to xy-dependence.
Juan Carlos Carrara . . . 4 pts - Good series of 5 cases; well-labeled eqns; r >= w
Hermen Jacobs . . . . . . . . 3 pts - It turns out ii) is possible, but with range of solutions
Denis Borris . . . . . . . . . . 3 pts - I like idea of walking backwards in iv) but even slower!
Thad . . . . . . . . . . . . . . . . 3 pts - Nice answer this wk; some algebra left out but ans ok
Jeremy Galvagni. . . . . . . 3 pts - Right; I gave "more constraints" in iii) so unique.
Al Nelson. . . . . . . . . . . . . 3 pts - Hey Al, thanks for coming back; good stab at 241.
Quasi-C . . . . . . . . . . . . . . 3 pts - Welcome back; ya been solving but not submitting!
V Balakrishnan . . . . . . . 3 pts - Actually ran 38.5 and walked 10.5 but otherw. perfekt
Phil Sayre . . . . . . . . . . . . 3 pts - You've been entering so long I thought it was me!
David Madfes. . . . . . . . . 3 pts - Yes; no walk is slowest run and no run is fastest walk.
Yakov Macak . . . . . . . . . 3 pts - High level of detail; you run clockwise in Australia?
Akifumi. . . . . . . . . . . . . . 3 pts - Right, iii unique, iv imposs, others many answers.
Luis de Brito Camacho . . . 2 pts - Hello Portugal: harmonic mean only if distances equal
Danny Roe (new) . . . . . . . 2 pts - Wanted bonus pt for having the same name as me!
Jason McConnell (new) . . 2 pts - Welcome Jason & Danny from my calculus class!

problem #242 (back to top)
posted tuesday, march 7, 2006
put a hex around you
The picture shows all possible
equiangular hexagons with integer
sides, having perimeter 12 units
(up to congruence).
How many such hexagons
are there with perimeter:
a) 18 ; b) 19 ; c) 20 ?
List each according to ordered sides
(e.g. 1,1,4,1,1,4); no need for pictures!

Show your steps and reasoning.

Solution: Many of you decided that a perimeter of 19 was impossible, because 19
is prime or some reason. Dare to dream, and surround your prime piece of land!
Here I stole Marcello's excellent answer, many of you had other good methods! - Dan
Opposite sides are parallel, so the sum of two consecutive sides is the same as the sum of the two opposite sides.
Take the difference of two consecutive sides, and then take the difference (in the reverse circular order) of the two
opposite sides. The sum of these differences divided by two gives the difference between the two remaining sides.
By assigning to the first two sides all possible integer pairs summing from 2 to 9, I found the following:
Sum 2: 117117 (18); 118118 (20) . . . . Sum 3: 126217 (19); 126126 (18); 127127 (20)
Sum 4: 134316 (18); 135317 (20); 135226 (19); 135135 (18) 136136 (20); 225225 (18); 226226 (20)
Sum 5: 143416 (19); 144144 (18); 145145 (20); 143325 (18) 144326 (20) 144235 (19); 234325 (19);
234234 (18); 235235 (20)
Sum 6: 151515 (18); 152516 (20); 153153 (18); 154154 (20);152425 (19);153244 (19); 152334 (18);
153335 (20); 242424 (18); 243425 (20); 243243 (18); 244244 (20); 243334 (19) 333333 (18) 334334 (20)
Sum 7: 162162 (18); 163163 (20); 161525 (20); 162253 (19); 161434 (19); 161343 (18); 162344 (20); 251524 (19);
252252 (18); 253253 (20); 251433 (18); 252434 (20); 252343 (19); 342433 (19); 342342 (18); 343343 (20)
Sum 8: 171171 (18); 172172 (20); 171262 (19); 171353 (20); 261261 (18); 262262 (20); 261352 (19); 261443 (20);
351533 (20); 351351 (18); 352352 (20); 351442 (19); 441441 (18); 442442 (20) . . . . Sum 9; 181181 (20)
Assuming reflected shapes as congruent (otherwise you would have included a 321321 hex), all sextuples
differing for cyclical permutations or inversions are the same, so, after eliminating repetitions, we find:
Perimeter 18: 117117; 126126; 134316; 135135; 225225; 144144; 143325; 234234; 151515; 242424; 333333 (total 11)
Perimeter 19: 126217; 135226; 143416; 144235; 234325; 152425; 243334 (total 7)
Perimeter 20: 118118; 127127; 135317; 136136; 226226; 145145; 144326; 235235; 152516; 153335; 243425; 244244; 334334 (total 13).

So of course the number of ways of selecting equihexes of perims 18, 19, and 20 would be 1001, a grand number! Also Jim Ferry gives a
generating func g(x) = (x^6) / [(1-x^2)(1-x^3)(1-x^4)(1-x^6)] where coeff of x^n is the number of hexagons of perimeter n. Thanks Denis! - Dan

WINNERS - Problem 242 . . Sorry I've been ssslllow! - Dan (back to top) . leader board
Nick McGrath. . . . . . . . 10 pts - Wrote a tried-and-true BASIC loop - 20th century lives!
Marcello Cammarata . . 7 pts - Holding onto first place this year, and a quotable answer.
Akifumi . . . . . . . . . . . . . 6 pts - Nice argument about the opposite pairs summing same
V Balakrishnan. . . . . . . 5 pts - Very good job with vector addition using w = e^i pi/3
Giridihar Prasannan (new) 4 pts - Welcome to my contest, good inequalities P-C-2A-B>0
Mark Rickert . . . . . . . . . 4 pts - Limited the longest to N/2 - 2, equisum to limit search
Ed Wern . . . . . . . . . . . . . 3 pts - First entry; more patterns than P = 12 ; got your #239.
Denis Borris. . . . . . . . . . 3 pts - Thanks for the link to the solution; Jim Ferry gives gen func
Jeremy Galvagni . . . . . . 3 pts - I like yr truncation of equilateral triangles to get hexes
Tim Poe. . . . . . . . . . . . . . 2 pts - Early entry but you put too much faith in symmetry
Art Morris . . . . . . . . . . . 2 pts - Welcome back, Art! We missed you & u missed a few hexs
Yakov Macak . . . . . . . . . 2 pts - You noticed the four patterns from P=12, but therere more
Phil Sayre . . . . . . . . . . . . 2 pts - It does seem primality of 19 precludes proper perimeter
Kirk Bresniker. . . . . . . . 2 pts - P=12 have 2- or 3-fold symm but all P=19 have no symm
Radu Ionescu . . . . . . . . . 2 pts - Thanks for th' attached isometric <3,3,3,3,3,3> graph paper
Charlie Acquisio (new). . 2 pts - Nice to have you aboard, good answer but wanted integers
Zahi Teitelman . . . . . . . 2 pts - Greetings, Z; good try but u fell for the symmetry/prime trap
Danny Roe . . . . . . . . . . . 2 pts - Hey Danny g'luck on your calculus final, got some hexes
Jon Stearn . . . . . . . . . . . 2 pts - Welcome back Jon, true that a+b = d+e and 2 others.
Deborah Fradelis . . . . . 2 pts - Another contestant stops by again, good reas abt parallels
Cliff Angle (new) . . . . . . 2 pts - Hi Cliff, you seem like a smart, (and acute) Angle ;-}
Claudio Baiocchi . . . . . 2 pts - That was an awesome javascript actually showing hexes!
John Smith . . . . . . . . . . 1 pt - Thanks for sending in some hexagons, "John"! Good luck.

problem #243 - posted tuesday, march 28, 2006 (back to top) Altitudes with Attitude! (thanks to Ed Wern for the title; I forgot to put one ! - Dan)

Call the altitudes of a triangle: h , k , m ;
and the radius of its inscribed circle: r .
Find the minimum value of the expression
(h + k + m) / r (over all triangles).
Which triangle achieves this minimum?
Show your steps and reasoning.

Solution: By two-time champion Nick McG...
Let the sides of the triangle be a,b,c. Let Area = K . . . 1) Then   Area , K= r(a + b + c)/2     (i)
Proof:   If the vertices of the triangle are A,B and C and the center of the incircle is at I
Then Area(ABC) = Area(ABI) + Area(BCI) + Area(CAI) = cr/2 + ar/2 + br/2 = r(a + b + c)/2.   QED
2) Area, K = 1/2 x base x altitude = ah/2 = bk/2 = cm/2 . . . So,   h = 2K/a : k = 2K/b: m = 2K/c
=>   h + k + m = 2K( 1/a + 1/b + 1/c)  (ii) . . . Now consider the given expression: (h + k + m)/r
From (i) and (ii) we have: (h+k+m)/r = 2K(1/a + 1/b + 1/c) / (2K/(a+b+c)) = (a+b+c)(1/a + 1/b + 1/c)
= 3 + (a/b + b/a) + (a/c + c/a) + (b/c + c/b)    (iii)
This expression will be minimum when the terms in paretheses are minimum.
For a general expression f(x,y) =x/y + y/x the minimum value of 2 occurs when x=y.
Proof:    Let    y = px    then  f(x,y) = x/y + y/x = 1/p + p =f(p)
Differentiating:     f '(p) = -1/p^2 + 1     so when f '(p) = 0    p^2 = 1      =>   p = 1     =>   x = y.   QED
So, (iii) is minimum when a=b, a=c, b=c =>    a = b = c . And   min. (h + k + m)/r = 3 + 2 + 2 + 2 = 9
The minimum is achieved only when a=b=c: Equilateral Triangle. . . Regards - Nick
(Dan's Note: New contestant Oliver squared (x/y - y/x) to prove min of x/y + y/x was 2. Clever!)

WINNERS - Problem 243 . . (back to top) . leader board
Nick McGrath. . . . . . . . 10 pts - You can read and admire your solution above. Go Nick!
Ed Wern . . . . . . . . . . . . . 7 pts - Your push-pull argument made sense but lost a couple points
Marcello Cammarata . . 6 pts - Got 9(a+b+c)/3 *3(1/a+1/b+1/c) = 9*Arith /Harm >= 9
Mark Rickert . . . . . . . . 5 pts - Your eqns were messy enough to pass muster, also 4 pts #239
Art Morris . . . . . . . . . . . 4 pts - Relat. betw inscr rad and alt is mysterious harmonic mean
Radu Ionescu . . . . . . . . 4 pts - Good prf uses x/y + y/x >= 2 and unnamed area formula
Oliver Schebanek (new) . 4 pts - Nice to hear from you, good Law of Sines & squaring
V Balakrishnan . . . . . . 4 pts - You used that arith > harmonic mean and made it work
Jeremy Galvagni . . . . . .3 pts - Right, equi has short alt & lg rad, among all rt tri isosc wins
Claudio Baiocchi . . . . . 3 pts - Right; 2A = r(a+b+c) will event. do the trick. +1 pt for 239
Tim Poe . . . . . . . . . . . . . 3 pts - I like your combo of Heron's and quick fix making >= 9
Akifumi . . . . . . . . . . . . . 3 pts - Nice use of area formula and arith mean > geom mean
Hermen Jacobs. . . . . . . . 3 pts - Good intuit about equilateral, interesting loop proof of min
Phil Sayre . . . . . . . . . . . . 3 pts - Good proof; minor typo fixed fast, only 1/2 pt deduc!
Denis Borris . . . . . . . . . . 3 pts - I guess you figured it w/o my clarif; yr ans seemed fine!
Patrik Petersson . . . . . . 3 pts - Welcome back after a long break! Good ans arith > geom.
Quasi-C . . . . . . . . . . . . . 2 pts - That's a nice fudging sharp corner argt; could get unflimsed
Kirk Bresniker. . . . . . . . 2 pts - I'm not doubting your Burrington's 1940 handbook...
Thad . . . . . . . . . . . . . . . . 1 pt - Your belief was right about 9 for equitri - late entry scores pt

problem #244 - posted saturday, april 8, 2006
Half More Number! (back to top)
Find the smallest positive integer such that if the ones digit is moved (from the right) all
the way to the left, the resulting number is exactly 50% more than the original number.
Show your steps and reasoning.

Solution: From 3-time winner (2002/03, 03/04, 04/05) Nick McGrath :
Let the original number be of the form 10A + x where x is the ones digit.
Then we seek the smallest A such that 3/2 * (10A + x) = x*10^k + A   (k is the number of digits of A.)
Rearranging we have 30A + 3x = 2*x*10^k + 2A => 28A = x(2*10^k - 3)
The number in parentheses is 199...97 where the dots represent an unknown number of 9's
Since 28A == 0 mod 4 the RHS is also 0 mod 4 and since (2*10^k - 3) is odd we conclude x == 0 mod 4.
Thus the only candidates for x are 4 or 8. Try x=4. We have 28A = 4 * 199...97 => 7A = 199...97
Now divide 7 into 199...97 til we find the smallest soln which turns out to be 28571 => 10A + x = 285714.
Now x=8 ; 28A = 8 * 199..97 => 7A = 2 * 199...97 clearly smallest A is double the value we got for x=4.
Thus the smallest number that meets the given criterion is 285714.

Ken D. had a shorter way to say it: We want: (10^z)*y + x = 1.5(10x+y) ; or x = (y/28) * (2*(10^z) - 3)
Since (2*(10^z) - 3) is always odd, and y has to be a single digit, y must be 4, so that the remaining 7 in the
denom. can divide evenly into (2*(10^z) - 3). The first such number divisib. by 7 is when z = 5 ; x = 28571.

And a bonus point to newcomer Joe Schank: "The problem did not specify the base of the numbers. If base 4
is allowed, then the smallest such integer is 12 (base 4). If decimal integers, the smallest such integer is 285714.
Base    Answer
2       No such number exists
3       1201023 (41610) --> 2120103 (62410)
4       124 (610) --> 214 (910)
5       12325 (19210) --> 21235 (28810)
6       No such number exists
7       1327 (7210) --> 2137 (10810)
8       13505642728 (195,225,78610) --> 21350564278 (292,838,67910)
9       13856750329 (557,885,50410) --> 21385675039 (836,828,25610)
10      285,714 --> 428,571

WINNERS - Problem 244 . . Sorry I've been so ssslllow! - Dan (back to top) . leader board
Nick McGrath. . . . . . . . 10 pts - First full answer i was happy with, and happy to quote!
Tim Poe. . . . . . . . . . . . . . 7 pts - Proved N even and a mult of 3 (three-ven); good search
Patrik Petersson . . . . . . 6 pts - Nice notation, 3N/2 = 10^n ao + ... a1; thanx 4 web complimt

V Balakrishnan. . . . . . . 5 pts - Good; (10a + b)*1.5 = b*10^(k-1) + a; involves 28.

Marcello Cammarata . . 5 pts - First with correct answer but need more expl of method
Radu Ionescu . . . . . . . . . 4 pts - Your first entry was 12 digits; cut in half by the second!
Ed Wern . . . . . . . . . . . . . 4 pts - So you're getting 199997 / 7 = 285714 is how we get it
Akifumi . . . . . . . . . . . . . 4 pts - Yes, 14N = (10^(k-1) - 1)*ao needs ao = 4, then good n.
Mark Rickert . . . . . . . . . 3 pts - I like the way you look at things mod 14, and test cases
Kirk Bresniker. . . . . . . . 3 pts - Good C# code; no I didn't get your 239, can u send it?
Denis Borris. . . . . . . . . . 3 pts - This is a cute problem, huh? Yes, 571428 too & tack-ons.
b basic (new) . . . . . . . . . . 3 pts - Nice to have you with us, good way to say 199997/7
Hermen Jacobs. . . . . . . . 3 pts - Hello my good man Hermen (with his trusty Basic code)
Jeremy Galvagni . . . . . . 3 pts - yes; n = 10a + b ==> 14a = (10^n - 1.5)*b; use decimals
Oliver Schebanek . . . . . 3 pts - Welcome back; like the rem 0 for (2*10^k - 3)/7 quot.
Giridhar Prasannan. . . 3 pts - Nice use of fancy summation gifs (and correct algebra!)
Michael Callahan (new) 3 pts - Welcome to the dansmath community, good answer!
Al Nelson. . . . . . . . . . . . 3 pts - Hello Al; good solution (there were several ahead of you)
Ajit Athle . . . . . . . . . . . 3 pts - Gets from 197z = 28(10x + y) to n = 285714 w/proof!
Quasi-C . . . . . . . . . . . . . 3 pts - Yes; first 99...9 that's a mult of 7 is 999999; go from there!
Joe Schank (new) . . . . . . 3 pts - Great get, Joe; since I didn't specify a base the best is 124.
Claudio Baiocchi . . . . 2 pts - Using a decimal for 1/17 gets 1176470588235294, hmm
Florian Fischer (new) . . 2 pts - Hello, FF. First ans posed good eqn, resub solved it!
Phil Sayre. . . . . . . . . . . . 2 pts - Hi, Phil; what's so funny looking about [2/7 * 10^6] ?
Zahi Teitelman . . . . . . . 2 pts - I'm with you 150%; the number is 285714 = 199997/7
Ken Duisenberg . . . . . . 2 pts - Good to have you back this year ; see the quote above!
Deborah Fradelis . . . . . 2 pts - I liked the .ppt attchmt; I was hoping for multimedia show!
Mario Roederer . . . . . . 2 pts - Used a .ssh on formula 1.5(10x + y) vs y*10^n + x
Dan Greene (new) . . . . . . 2 pts - A new guy; hope u come back! Good ans; glad u liked prob
Cliff Angle . . . . . . . . . . 2 pts - Excel-lent approach; thanks for support of site & podcast
Thad . . . . . . . . . . . . . . . 2 pts - I admire a person who knows how to handle a TI-89!
Juan Carlos Carrara . 2 pts - Good use of equations; that's pretty much a proof!
Ravi Raja . . . . . . . . . . 1 pt - Thanks for sending in; it is possible to less than double n.

problem #245 - posted friday, april 28, 2006.
Sphere of Influence (back to top)
Let P1 , P2 , . . . , Pn be points on a sphere of radius 1 .Prove that the sum of the squares of
the distances (in R3) between all pairs of points is at most n^2, and state conditions for this sum
to be equal to n^2. Show your steps and reasoning.

Solution: Some examples from Kirk B; used Platonic solids to find sets of 4, 6, 8, 12, 20 points!

For n=2, the sum of squares of distances = 2^2 = 2^2 ; For n=3, the ssd = 3 * (sqrt(3)^2) = 9 = 3^2

For n=4, the ssd = 6 * (4/sqrt(6))^2 = 16 = 4^2 (see Phil's tetrahedron formula below)

For n=6, the ssd = 12 * (sqrt(2)/2)^2 + 3 * 2^2 = 24 + 12 = 36 = 6^2 (octahedron)

For n=8, the ssd = 12 * (2/sqrt(3))^2 + 12 * (2*sqrt(2)/sqrt(3))^2 + 4 * 2^2) = 64 = 8^2 (cube)

For n=12, the calculating gets tough. One really neat fact I learned in researching this is that the 12

vertices of the icosahedron can be found as the twelve vertices of three orthogonal golden rectangles!

http://documents.wolfram.com/mathematica/Demos/Notebooks/BuckyballConstruction.html

That's pretty cool. To inscribe the icosahedron in the unit radius sphere, let the diagonal of the golden
mean rectangle = diameter = 2. The short length is then equal to 2 /sqrt(4+4 phi^2) and the long length
is equal to 2phi/sqrt(4+4 phi^2). Tote up the (n)(n-1)/2 = 66 lengths, square, and sum, and you find
the ssd = 30*(16/(4+4 phi^2)) + 30*(16 phi^2/(4+4 phi^2)) + 6 * 2^2 = 144 = 12^2
(He also shows for the dodecahedron's 20 vertices, the ssd = 20^2 using coords from MathWorld)

Marcello showed that for n general points (xi, yi, zi) that sum of squared distances...
... then used Lagrange Multipliers to prove the n^2 is indeed a maximum. Go Italia!
Let x^2 + y^2 + z^2 = 1 be the equation of the sphere and (xi, yi, zi) (i = 1,, n) the n points on it.
Then d2 = sum of square distances = 1/2 Sum(i,j) ((xi-xj)^2 + (yi-yj)^2 + (zi-zj)^2) (i,j = 1, ., n).
The factor 1/2 appears because both i, j go from 1 to n, so each term is duplicated; terms like xi - xi
are also accounted for, but they equal 0 and don't affect the sum. Developing the squares:
d2 = 1/2 Sum(i,j) (xi^2 + yi^2 + zi^2 + xj^2 + yj^2 + zj^2 - xixj - yiyj - zizj)
= 1/2 Sum(i,j) (1 + 1) - 1/2 Sum(i,j) (xixj + yiyj + zizj) = n^2 - f, where
f = Sum(i,j) (xixj + yiyj + zizj). So d2 <= n^2

In fact the formula will also show that d2 = n^2 - (x1+x2+...+xn)^2 where v^2 means v.v dot prod.
so the sum is the max n^2 whenever the centroid (proportional to x1+x2+...+xn) is the zero vector.

WINNERS - Problem 245 . . Listen to my podcasts! - Dan %;-} (back to top) . leader board
Nick McGrath. . . . . . . . 10 pts - Nice coordinate-laden proof (using algebra, of all things!)
Marcello Cammarata . . 7 pts - I like the analogy of like charges constrained to a sphere.
Patrik Petersson . . . . . . 5 pts - Staging a strong comeback after years of dormancy... good!
Mohamed Omar . . . . . . 4 pts - Good use of coordinates, and some nice typsetting on the pdf!
Kirk Bresniker . . . . . . . 4 pts - Cool to bring in actual coords of the Platonic solids, I like!
Joe Schank. . . . . . . . . . . 4 pts - Fine attchmt, and correct that S = n^2 when Pavg = (0, 0, 0).
Oliver Schebanek . . . . . 3 pts - I see your proof, seems you're raising a vector to a power?
b basic . . . . . . . . . . . . . . 3 pts - Good use of centroid x1+...+xn; notation problems w/vec^2
Jeremy Galvagni . . . . . . 3 pts - I liked your examples; read the various proofs above!
Denis Borris. . . . . . . . . . 3 pts - Proof by omission: "a bit of algebra shows that..." %;-}
Claudio Baiocchi. . . . . . 3 pts - Yes, Sum |Pi - Pj|^2 = Sum 2[1 - (Pi ,Pj)] dot products
Radu Ionescu . . . . . . . . . 2 pts - True that if the pts are in diam opp pairs then sum is n^2

V Balakrishnan. . . . . . . 2 pts - Proved that for n pts in reg polygon on equator, S = n^2.

Hermen Jacobs. . . . . . . . 2 pts - Good expansion of sum of (xi - xj)^2 but not all pos terms
Phil Sayre. . . . . . . . . . . . 2 pts - Yes for n=4 tetrahedron of side s has Rsph = (s \/6)/4.

problem #246 - posted saturday, may 20, 2006.
Beating the Cycle! (back to top)
Suppose we have nine cards, numbered 1-9 (or Ace, 2, . . . , 9) separated into three piles
X, Y, and Z, of three cards each,such that : pile X beats pile Y, Y beats Z, and Z beats X.
We say "pile X beats pile Y" if a random card from X has a bigger number than a random
card from Y, most of the time. How is it possible to arrange the cards into piles like this?
Show your steps and reasoning.

Solution: From Zeke Moore :
The sum of the 9 cards is 45, so the parity of the situation suggests we want 3 piles with sums
of 15 each. That suggests the 3x3 magic square:
6 7 2
1 5 9
8 3 4 . . . . Let's make each column into a pile... (note: the rows would work too - Dan)
Pile X: 6, 1, 8
Pile Y: 7, 5, 3
Pile Z: 2, 9, 4
If a random card is drawn from each of 2 piles, there are 3x3 or 9 different ways to make a pair.
So one pile will "beat" the other if it has the high card in at least 5 of the 9 pairings.
Pile X vs Pile Y: Pile X wins 5 of the 9 matchups (6-5, 6-3, 8-7, 8-5, and 8-3).
Pile Y vs Pile Z: Pile Y wins 5 out of 9 (7-2, 7-4, 5-2, 5-4, and 3-2).
Pile Z vs Pile X: Pile Z wins the best-of-9 series (2-1, 9-1, 9-6, 9-8, and 4-1).
------------
Denis Borris and others pointed out there are 5 solutions; some piles win 6-3, some 5-4:
"In case you want them all, there are 5 solutions (X,Y,Z order):
159,348,267 ; 168,357,249 ; 178,356,249 ; 168,457,239 ; 178,456,239
I guess we can say that's 15 solutions, since each can be rearranged (cycled) 3 ways."

WINNERS - Problem 246 . . Subscribe to my podcast! %;-} (back to top) . leader board

Art Morris . . . . . . . . . . . 10 pts - Art's strategy was to put a small, med, and lg in each grp
Claudio Baiocchi . . . . . . 6 pts - Good answer and two piles win by 6-3; sh. be Z vs X
Marcello Cammarata. . . 5 pts - Intuition told that 1 and 9 go together; went well from there
Zeke Moore . . . . . . . . . . . 5 pts - Sum of 1 to 9 is 45 so sum 15 each grp; used magic sq!
Oliver Schebanek. . . . . . 4 pts - Got 932,871,654 after solving some clever equations
Nick McGrath. . . . . . . . . 4 pts - Also mentioned magic sq 168,357,249; 9 poss pairs ea.
Denis Borris . . . . . . . . . . 3 pts - Showed each wins 5-4 then sent (all 5) possible solutions
Mark Rickert . . . . . . . . . 3 pts - Five minutes of fooling around" led to (*); then program
Tim Poe. . . . . . . . . . . . . . 3 pts - 9! orders = 362880; pile compar only 3000; then 5 diff
V Balakrishnan. . . . . . . 3 pts - Gave several working solutions; (a couple of repeats)
Yakov Macak. . . . . . . . . 3 pts - Good on ya mate, nice to list what beats what in pairs.
Ed Wern . . . . . . . . . . . . . 3 pts - Trial gave error, then fixed up solution to 942,861,753
Kirk Bresniker. . . . . . . . 3 pts - Gave a source with a cool graph (combinatoric style)
b basic. . . . . . . . . . . . . . . 3 pts - Nice solution with visible evidence; should i use yr name?
Jeremy Galvagni . . . . . . 3 pts - This was a well known paradox to you (and now yr students!)
Patrik Petersson . . . . . . 3 pts - You gave and proved solutions; your 6 were 2 recycled.
Adam Panagos (new) . . . . 3 pts - Welcome to my contest, nice detail & multiple solutions!
Deborah Fradelis . . . . . . 3 pts - Got the popular 348,267,159 and proved it pair-by-pair
Zahi Teitelman. . . . . . . . 3 pts - I checked your numbers; they were fine as you showed
Felix Chan (new) . . . . . . . 3 pts - Glad you came here thru my podcast, welcome & good ans
Sean Allaband (new) . . . . 2 pts - Your criterion didn't check the 9 pairs (3*3) but good try!
Rewon C (new) . . . . . . . . . 2 pts - Solved abc,def,ghi; a>d, b>e, d>g; that forced it to work!
Thad . . . . . . . . . . . . . . . . 2 pts - Your "more aesthetic" solution didn't work after switch
Hermen Jacobs. . . . . . . . 1 pt - Thanks for sending in the doubt, but it really is possible!

Al Nelson. . . . . . . . . . . . 1 pt - Good answer (but after 247 was up so just 1 pt for this one)

problem #247 - posted monday, may 29, 2006
I Am the Bugman! (back to top)
"Bugs" can have six legs (insects), eight legs (spiders), or ten legs (decipedes).
(a) Is it possible to find a set of bugs that has any even number of legs > 4?
(b) How many sets of bugs have 100 legs total?
(c) How many bug collections are possible with 100 bugs and 800 legs?
List or describe b) and c), giving the number of insects, spiders, and decipedes.
Show your steps and reasoning. (Time's up on this one.)

Solution: a) From Nick McG: " Let 2N = 6a + 8b + 10c ; we can divide by 2 and get
N + 1 = 3a + 4b + 5c +1 = 3(a -1) + 4(b + 1) + 5c = 3a + 4(b - 1) + 5(c + 1) = 3(a + 2) + 4b + 5(c - 1)
So the next N is acheivable, can show that 6, 8, 10 are too (I'm paraphrasing a bit - Dan)

b) From Patrik P : The problem is to find the number of solutions to the equation 6x + 8y +10z = 100.
or the equivalent equation 3x + 4y + 5z = 50. It is easy to see that 3x + 4y must be divisible by 5.
After some testing i found the following solutions; there are a total of 26 sets of bugs that have 100 legs:
(x, y, z) = (0, 0, 10) (0, 5, 6) (0, 10, 2) (1, 3, 7) (1, 8, 3) (2,1, 8) (2, 6, 4) (2, 11, 0)
(3, 4, 5) (3, 9, 1) (4, 2, 6) (4, 7, 2) (5, 0, 7) (5, 5, 3) (6, 3, 4) (6, 8, 0) (7, 1, 5)
(7, 6, 1) (8, 4, 2) (9, 2, 3) (10, 0, 4) (10, 5, 0) (11, 3, 1) (12, 1, 2) (14, 2, 0) (15, 0, 1)

c) From Akifumi I : It is # of (a,b,c) in N^3 s.t. 3a + 4b + 5c = 400 and a + b + c = 100.
Then c = a. (subtract quadruple of the 2nd equation from the 1st). So the set of equation becomes:
8a + 4b = 400 and 2a + b = 100 --> 2a + b = 100. A possible value of a is exactly all the integers
between 0 and 50. Hence the number of solutions is 51. From the argument above, it is easy to list :

(0, 100, 0), (1, 98, 1), (2, 96, 2), . . . , (49, 2, 49), (50, 0, 50)

Dan's notes: I think I made part c) too easy; you guys could have handled 768 legs or something.
Also Akifumi noted the number of ways of doing b) is the same as the number of ways of solving the Dioph.
3a + 4b + 5c = 50, which is the coefficient of x^50 in the power series for 1 / [(1 - x^3)(1 - x^4)(1 - x^5)].
Also see my old problem 81 - Change 4A Dollar. His N^3 is triples of natural nos (might need "whole nos")

WINNERS - Problem 247 . . Subscribe to my podcast! %;-} (back to top) . leader board

Claudio Baiocchi . . . . . . 10 pts - Nice job; any method to finding the 26? Good eqns on c).

Marcello Cammarata . . . 7 pts - Good descrip. on a), nice param sol'n (x, 100-2x, x) for c).

Akifumi . . . . . . . . . . . . . . 5 pts - I liked the use of generating function; see above comment.

Nick McGrath . . . . . . . . . 5 pts - Cool induction proof on part a) leading to N + 2 legs.

Patrik Petersson . . . . . . . 4 pts - Congruence literacy, we number theorists appreciate it.

Kirk Bresniker . . . . . . . . 4 pts - Recursive C programs; where would geeks be without them!

Oliver Schebanek . . . . . . 4 pts - Says legs appear in pairs in nature. Is that always true? Ok!

Tim Poe . . . . . . . . . . . . . . 3 pts - Pulled out the old "easily seen" card from the proof deck!

b basic . . . . . . . . . . . . . . . 3 pts - Nice link to MathWorld; should I use your real name?

Mark Rickert . . . . . . . . . 3 pts - True; for c) avg # of legs is 8 so I = D. I shoulda made it harder.

Adam Panagos . . . . . . . . 3 pts - I like your reasoning style; thanks for becoming a regular!

Al Nelson . . . . . . . . . . . . 3 pts - Slick idea for b) start with z = 10 then reduce by 1 and look.

Ed Wern . . . . . . . . . . . . . 3 pts - Good method as usual; I updated your (and Brady's) score(s)

V Balakrishnan . . . . . . . 3 pts - Accurate use of "representable" as heard on my podcast...

Jeremy Galvagni . . . . . . 3 pts - Right to use the mn - m - n for smallest non-rep of rel prime

Mario Roederer . . . . . . . 2 pts - a) and b) good; 1343 sol'ns for c) might have some repeats?

Denis Borris . . . . . . . . . . 2 pts - Are boxes of chicken legs less grody than bugs? Nice mod 10

Ravi Raja . . . . . . . . . . . . 2 pts - Welcome back after a restful break; 4 short on part b) ...

Hermen Jacobs . . . . . . . . 2 pts - Got most of them for b) and c); prove all evens reachable.

Zahi Teitelman . . . . . . . 2 pts - Your attached table had 26 ways but you said there were 6

Phil Sayre . . . . . . . . . . . . 2 pts - Diophantus would like this, yes (but he may've found more ways)

John Stearn . . . . . . . . . . 2 pts - Good to see you back at bach; good expl for a), how many in b)?

Doug Babcock . . . . . . . . 2 pts - Right;100 - 10d - 8s must be divis by 6, nice steps Doug.

Allen Druze . . . . . . . . . . 1 pt - You must have an extra one in b); not sure abt 5151 and 321201.

Thad . . . . . . . . . . . . . . . . 1 pt - Later entry good job on c); and just 12 ways for b); keep entering!

Etienne Desclin (new) . . . 1 pt - Good answer (after 248 up); resub was free this time; bienvenue!

problem #248 - posted tuesday, july 4, 2006
How Many Runners? (back to top)
A 'number' of runners entered a recent road race.
These are two true statements regarding the 'number':
a) It has 3 distinct digits. . . b) If you add 99 the number reverses.
Here are four other statements regarding the 'number':
1) It is divisible by the sum of its digits . . . 2) It is not prime. . .
3) It has only one common digit with the product of its digits. . .
4) The sum of the first and last digit is one more than the middle digit
If I told you which of these 'other' statement(s) were false, you'd be
able to determine the number. How many runners entered the race?
Show your steps and reasoning.

that's me (middle)

Solution: From Jeremy G : (Note: JG didn't consider middle digit zero, eight other no-good numbers)
First the numbers that satisfy a) and b): the three digit number abc can be written as 100a+10b+c
and its reversal as 100c+10b+a. Setting the reversal 99 more and the original gives c=a+1.
From this it is easy to constuct the 56 (Dan's note: 64) numbers that the number in question could be.
a, b, c are the first three columns in the chart at the end. The next four columns are the truth table.
1 denotes true, 0 is false. For example 132 IS divisible by the sum of its digits and IS not prime
but DOES NOT have only one digit in common with the product of its digits and DOES not have
a middle digit that is one less than the sum of its first and last digits.
Finally to find determinable value. I considered numbers in s1 to s4 to be binary digits and
converted these to base 10 - the last column. Then I scanned for unique digits. There is only one 5.

It corresponds to the number 485. Statements 1 and 3 are false. Here is the chart, pasted from excel:

a b c s1 s2 s3 s4

1 3 2 1 1 0 0 12
1 4 2 0 1 0 0 4
. . . . . . etc . . . . . etc . . . . .
4 7 5 0 1 1 0 6
4 8 5 0 1 0 1 5 <-----
4 9 5 0 1 0 0 4

. . . . . . etc . . . . . etc . . . . .

WINNERS - Problem 248 . . Vote for my podcast! %;-} (back to top) . leader board
Marcello Cammarata . . 10 pts - Did the truth table idea too and used 64 candidates. Buon lavoro!
Tim Poe . . . . . . . . . . . . . . 7 pts - Very logically sound, not ruling out leading digit zero. 72 cand's!
Ed Wern . . . . . . . . . . . . . 5 pts - I agree with the 64 cand's; not sure about 'A or not A' ever b false.
Nick McGrath . . . . . . . . 5 pts - Tested all approp N from 102 to 879; 4 matched none, etc. !:485
Mark Rickert. . . . . . . . . 4 pts - Raised question of 849; product of dig 288 is this 1 dig common?
Ravi Raja . . . . . . . . . . . . 4 pts - Liked the idea of narrowing it down by cond 1, 2, 3, 4; FTF works.
Al Nelson . . . . . . . . . . . . 3 pts - Good work making a list (of 56, not 64); one of the "dot-x-l-s" folks!
Denis Borris . . . . . . . . . 3 pts - Only 3 of those 56 nos had cond 4 true; only 485 uniquely cond'd
Claudio Baiocchi. . . . . . 3 pts - If all false then 839; if some false some true then 485; allez Pascal!
Hemingway (new) . . . . . . 3 pts - Thanks for joining the dansmathclub; cond 4 leaves least true; true!
Mario Roederer . . . . . . . 3 pts - I like a man who admits to using brute force (.xls) when nec/poss.
Allen Druze . . . . . . . . . . 3 pts - By "Hell Dan" I think you meant Hello, or was it a devil of a problem?
Phil Sayre. . . . . . . . . . . . 3 pts - Phil used Python to write his 'favorite line of code' for the truth table!
Zahi Teitelman . . . . . . . 3 pts - Got 56 cases and of those assume cond 3 is false; leads to 485. Hmm..
Jeremy Galvagni. . . . . . 3 pts - Allow me to quote you up top (and to round you & Zahi up 1/2 point)
Etienne Desclin . . . . . . . 3 pts - Nice color cells for diff cond's true of the 64 candidates. Bien fait!
Oliver Schebanek . . . . . 3 pts - Ok to assume no leading zero I think, for real 3-digit nos, eh?
Adam Panagos. . . . . . . . 3 pts - Assume d1 > 0 then 485 wins the truth table; glad you had fun!
Art Morris . . . . . . . . . . . 2 pts - One of the first entries, but your 263 had the mid 1 more not less
Radu Ionescu. . . . . . . . . 2 pts - Your 243 was true for cond 4, but shared a truth-string with 364.
Kirk Bresniker . . . . . . . 2 pts - Concluded that 263 was the chosen one, not sure which cond's...
Kirk Bresniker (prob 239) 5 pts - Got your re-send; another good one slipped by; sorry!
Patrik Petersson . . . . . . 2 pts - True that if 1, 2, and 4 are false then N might be prime but not nec.
Hermen Jacobs . . . . . . . 2 pts - Also thought it's 243 assuming 1, 2 false and (then) 3, 4 true...
Nats Kroy . . . . . . . . . . . 2 pts - 485 was the one; yr answer was good (just after most others)
K Sengupta (new) . . . . . . 2 pts - Welcome; interesting you also proved that b = 2a was nec.
Jin Won Park (new) . . . 2 pts - Glad to have you enter; good answer (my Korean screen font works!)
Hendrik van Eijsden (new) 1 pt - Welcome to dansmathworld; the 263 club rulez! Keep entering!
Danny Roe . . . . . . . . . . 1 pt - Good to hear from you; hope you're still listening to my podcast!
Mahul Bhattacharya (new)1 pt - True that z = x+1; but your answer of 152 has some problems.
Ajit Athle . . . . . . . . . . . 1 pt - Good argument "if 2nd cond.were false..." After 249 up so just 1 pt.

problem #249 - posted saturday, july 22, 2006

Eat a Peach (a Mile) (back to top)
A truck driver has 3000 peaches in Ashland and wants to move them to
Blythe, 1000 miles away by freeway. The truck can hold up to 1000 peaches,
but the driver has an addiction: if there are any peaches in the truck, the
driver will eat them at a rate of one peach per mile. Give a plan to deliver
the most peaches possible to Blythe. (Hint: There is safe fruit storage at all
points along the way.) Show your steps and reasoning.

Solution: A few of you may have thought that the driver needs peaches in order to travel; this is a
different problem; the driver only eats them if they are available. So the answer of 833 1/3 is the
largest I have found, and seems maximal. (The other way has a proof that 533 1/3 is maximum.)

From Akifumi: Answer = 833.3... peaches. If the truck can carry up to 3000 peaches at once, there is

nothing to think. To keep peaches as much as possible, we have to reduce the number of times the

truck carries peaches on each subintervals between A and B. I split the interval into 3 sections.

The first section is where the truck has to carry peaches 3 times. This section starts A. We want to

minimize the length of this section. The second section is where the truck has to carry peaches twice.
The last section is where the truck can carry all the peaches in 1 time.

The amount of peaches decreases to 2000 at 333.3... miles from A. (rate of decrease = 3 peaches/mile)
The amount of peaches decreases from 2000 to 1000 at 333.3... + 500 = 833.3... miles from A.
The amount of peaches decreases from 1000 to 833.3... after the truck goes the last section. (because

there are 166.6.. miles in this section.)

New contestant Garry Malashkin's Steps: P1 at 333 1/3 mi, P2 at 833 1/3 mi. start: (Ashland= 3000 peaches, others=0)

Ashland (load 1000) -> P1(unload 1000*(2/3) ) after: ( Ashland=2000,P1=1000*(2/3),P2=0,Blythe=0) ; then P1 -> Ashland

Ashland (load 1000) -> P1(unload 1000*(2/3) )      after: ( Ashland=1000,P1=1000*(4/3),P2=0,Blythe=0) ; then P1 -> Ashland
Ashland (load 1000) -> P1(unload 1000*(2/3) )      after: ( Ashland=0,P1=2000,P2=0,Blythe=0)
P1(load 1000) -> P2(unload 500)                  after: ( Ashland=0,P1=1000,P2=500,Blythe=0) ; then P2 -> P1
P1(load 1000) -> P2(unload 500)                 after: ( Ashland=0,P1=0,P2=1000,Blythe=0)
P2(load 1000) -> Blythe(unload 2500 / 3 )           after: ( Ashland=0,P1=0,P2=0,Blythe=2500 / 3)
Total 2500 / 3 (= 833 1/3) peaches in Blythe.

WINNERS - Problem 249 . . Vote for dansmathcast on podcast alley! %;-} (back to top) . leader board

Akifumi. . . . . . . . . . . . . 10 pts - Good discussion, leading up to 833 (and 1/3) peaches being delivered!
Tim Poe. . . . . . . . . . . . . . 7 pts - Your method was for whole peaches; 832 or 833 should B enough!
Art Morris . . . . . . . . . . . 5 pts - Yes, it's an issue when during the mile that the peach is eaten!
Garry Malashkin (new) . 5 pts - Thanks for joining the contest and getting quoted on your first try!
Oliver Schebanek . . . . . 4 pts - Good thinking;; each mile to the right uses 1 peach; minimize these!
Patrik Petersson . . . . . . 4 pts - Yep, that's the right scheme, go to 333 1/3 then to 833 1/3. Good!
Ed Wern. . . . . . . . . . . . . 4 pts - 15th entry but 7th correct one! Be sure to tell your son he's on the board!
Subba Ready (new) . . . . 4 pts - Welcome; nice that you got 2001 peaches to the 333 mile marker.
Nick McGrath . . . . . . . . 4 pts - First entry and the good proof of 533 1/3 merited an extra point!
Al Nelson . . . . . . . . . . . . 3 pts - I like the inclusion of an algebraic proof along with the 833 1/3 peaches.
Zahi Teitelman . . . . . . . 3 pts - That's cool that you looked for a calculus proof of maximality...
Kirk Bresniker. . . . . . . . 3 pts - 1000 to halfway was 500; using 1/4, 1/2, 3/4 was 750, also got 833!
Claudio Baiocchi. . . . . . 3 pts - Got 833 (or 834) depending on when the peach was eaten each mile.
Adam Panagos. . . . . . . . 3 pts - Nice notation, L(i) peaches at location P(i), used miles 334 and 833.
Jeremy Galvagni. . . . . . 3 pts - Two answers; 833 and 833 1/3 for continuous vs discrete solutions.
Mark Rickert. . . . . . . . . 3 pts - Same amt as Nick's 533 1/3; but Blythe is very dry and needs fruit.
K Sengupta . . . . . . . . . . 3 pts - Your answer was 533 1/3, made by eating peaches on the way back
Denis Borris . . . . . . . . . 3 pts - Incredibly persistent and good thinking (score was neg 2 but I relented ;-)
Etienne Desclin. . . . . . . 3 pts - Thanks for entry; 750 was the best for the 250, 500, 750 mile marks.
Marcello Cammarata . . 2 pts - That 667 was my first improvement over 500; you got 750 in resub.
Ravi Raja. . . . . . . . . . . . 2 pts - Well explained, double R, and 533 1/3 is the best for your assumption
Hermen Jacobs . . . . . . . 2 pts - Right, you can get 3000 - 3x peaches at x mile mark; nice response
David B (new) . . . . . . . . . 2 pts - Welcome to the site; 700 is a good answer, and well-explained.
Ajit Athle. . . . . . . . . . . . 2 pts - Got 533 1/3 by overeating and storing stuff at 466 2/3 mile mark
Allen Druze . . . . . . . . . . 2 pts - 666.66 was a good answer; you started right by 2000 at 1/3 mark
Nats Kroy . . . . . . . . . . . . 2 pts - You did get the 833 1/3 to B, but a few days later hence reduced score
Phil Sayre. . . . . . . . . . . . 2 pts - Interesting 829 peaches to B using n=6 segments; 833 using 1000
Abhijit Parashar (new) . 1 pt - Thank you for entering my contest. good idea storing after one mile!

problem #250 - posted thursday, august 10, 2006
Pack 'Em All In ! (back to top)
Shirley needs to ship off 250 small rectangular boxes that each measure 1" by 1" by 4" (inches).
(a) What are the dimensions and surface area of the rectangular packing box of smallest
surface area, if there can be no empty space in the (big) box? (b) Does this answer change
if we allow empty space in the box? If so, give the dimensions and area of this smallest-area box.
Show your steps and reasoning.

Solution: Some of you claimed this means there must be at least one dimension that's a mult of 4, but
I never said all the small boxes needed to be oriented the same direction. Indeed the 10 x 10 x 10 solution
is impossible, ruling out the minimum surface area of 600 sq in. Here's Mark Rickert:
a) Since the volume is 1000, we are constrained to edges of the form 2^m * 5^n.
A cube has the smallest surface area. In general the surface area is kept small if the lengths of the edges
are nearly equal. The rectangular prisms with volume 1000 with the smallest surface areas are:
10x10x10 (600) , 5x10x20 (700) , 5x8x25 (730) , 4x10x25 (780)
If there's a 10x10x10 packing, I don't know it (I suspect there wouldn't be a part B if there were).
However, there is a 5x10x20 packing (5x10 base, stand the boxes on end: 50 per layer, 5 layers)

b) The volume must be greater than 1000. Let's look at candidate volumes and their factors:
Volume    Factors    Minimum Surface Area
1001    7, 11, 13        622
1002    2, 3, 167        1682
1003    17, 59           2158
1004    2, 2, 251        2016
1005    3, 5, 67         1102
1006    2, 503           3022
1007    19, 53           2158
1008    2^4, 3^2, 7      620 (8x9x14) very close to theoretical minimum of 600
After this, there are some that come close, but aren't less than 620. Since we've gotten as low as 620,
we only need to check up to 1050 because a cube with volume 1051 has a surface area greater than 620.
Here are some close ones:
1014    2, 3, 13^2       650 ; 1020    2^2, 3, 5, 17    664
1024    2^10             640 ; 1026    2, 3^3, 19       678
1040    2^4, 5, 13       628 ; 1050    2, 3, 5^2, 7     650
Can we actually pack the boxes in the container? The answer is yes. Make the base 9x14, stand the
boxes on end (126 per layer, two layers).

And this fills the need for a proof that 10x10x10 is impossible; from new contestant Garry M:
Fill the 10x10x10 box with 2x2x2 cubes; let's suppose that the corner 2x2x2 box is black and
the next box is white and so on (as chess-board only 3D). Easy to find that total quantity of black
2x2x2 boxes is greater than total quantity of white 2x2x2 boxes so total quantity of black 1x1x1
boxes also is greater than total quantity of white 1x1x1 boxes.
Suppose we assemble the big box from 1x1x4 boxes. For each 1x1x4 box there are four cases of colors:
(BBWW,WWBB,WBBW,BWWB) B-black, W-white. So each 1x1x4 box has equal quantity black
and white 1x1x1 boxes and therefore the big box has also equal quantity black and white 1x1x1 boxes.
We come to the contradiction. QED

WINNERS - Problem 250 . . Vote for dansmathcast on podcast alley! (back to top) . leader board

Nick McGrath . . . . . . . 10 pts - I granted you first place due to your proof that 10x10x10 won't work.
Marcello Cammarata . . 9 pts - First correct entry but no proof that 10x10x10 is impossible, so...
Tim Poe . . . . . . . . . . . . . 7 pts - Glad you experimented to try for the 10x10x10; not sure if mult of 4 nec
Mark Rickert. . . . . . . . . 6 pts - Clever of u to suspect there's no 10x10x10 because part b existed!
Kirk Bresniker . . . . . . . 5 pts - Nice; 9x10x12 (270 box max) gives A=636, 8x9x14 (252) A=620
Oliver Schebanek . . . . . 5 pts - Nice 4-color 1000 cubes 10x10x10 by i^(x+y+z); 7x11x13: A=622
Al Nelson . . . . . . . . . . . . 4 pts - Right, the small boxes must be unaltered; not confident abt dim mult of 4
Ed Wern . . . . . . . . . . . . . 3 pts - First entry; yes10x10x10 no good; your 8x8x16 was 640 sq in; close!
Garry Malashkin . . . . . 3 pts - Nice proof with 2x2x2 boxes colored b/w; diff # of these in 10x10x10.
Claudio Baiocchi. . . . . . 3 pts - Vol 1008 for 8x9x14; good, best A=620. Not sure what the 2020 was.
Adam Panagos. . . . . . . . 3 pts - Assumed all boxes lined up lengths; still got 700 and 620; yes!
Adam Ramage (new) . . . 2 pts - Nice early entry; welcome to my contest! The 6x6x7 was 252 boxes?
Ravi Raja. . . . . . . . . . . . 2 pts - Your 700 for the 5x10x20 was good; nice try with 9x9x12 and 8x10x12
Hermen Jacobs . . . . . . . 2 pts - Got your answer and resub for part a); b) was interesting question
Etienne Desclin. . . . . . . 2 pts - True 10 not mult of 4; resub 640 better than 660 (but min was 620)
Nats Kroy. . . . . . . . . . . . 2 pts - I liked the idea of layers 10x10; fit into 11 long (A=640); resub 620
Denis Borris . . . . . . . . . 2 pts - Congrats Tyler! 7x12x12 close (A=624) resub got 8x9x14 (A=620)
Justin Bradley (new) . . . 2 pts - Hi; glad you like my podcast! Good try 5x8x25: A = 730; 8x11x12: A=632
Phil Sayre . . . . . . . . . . . 1 pt - Also got A = 730 and 632, (without all the fanfare of explanation ;-} )
Allen Druze. . . . . . . . . . 1 pt - The 600 was min but unreachable; resub 10x10x10; 8x13x10 close (A=628)
K Sengupta . . . . . . . . . . 1 pt - Nice color proof that 10x10x10 won't work; no b) but thanks for entry!


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