dan's math@home - problem of the week - archives
 
 
Problem Archives page 22
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
 
211 Cogito Log-o-Sum
212 - C h e c k , P l e a s e
213 - Island Hopping...
214 - The Four Means!
215 - Put A Hex On It !
216 - Band in Boston ?
217- Baskets of Apples
218- Holy Sphere+circle
219 -- See Double Free
220 - - Cube in a Cone!
 
 
Problem #211 - Posted Wednesday, September 1, 2004
Next-to-last problem of 2003/04 = New season started with Problem 213
Cogito Log-o-Sum ! "I think about logs, so I exist !" (back to top)
Find the smallest value of the sum S; prove your answer is as small as possible.
S = loga(bc) + logb(ac) + logc(ab)
Note: loga(bc) means log base a of (b*c). Show reasoning clearly

Solution: Longtime entrant Hermen J. offers this: "Hi Dan, The smallest value is 6 (If a=b=c).
S=loga(BC) +logb(AC)+logc(AB) = log BC/logA+logAC/log B+log AB/log C
= log B/log A+logC/logA+log A/logB+logC/logB+logA/logC+log B/log C [A, B and C all pos]
This is three times X/Y + Y/X. I should like to prove the smallest value of X/Y+Y/X is 2
X/Y+Y/X-2=(X-Y)^2 / X*Y the smallest value is 0 (if X=Y) then smallest value of X/Y+Y/X=2.
The smallest value of S is 2 + 2+ 2 = 6." [This is what I thought the answer was - Dan]
Then Fanfan pointed out that as stated, there is no minimum "Taking a = b = e (= 2.718...),
[the above] becomes 1 + log c +1 + 1/log c + log c +1/log c = 2 + 2 log c + 2/log c. Taking c
tending to 0, we have log c tending to -infinity and the whole expression also tends to -infinity :
this means that the expression is unbounded from below! This might not be the answer you expected."
 
Ed Wern offers these choices: i) Assuming a, b, c are all [positive integers], the minimum sum
occurs when a=b=c, all >=2. In this case, the sum always = 6. [Dan's note: The ans if a,b,c > 1]
ii) Assuming instead that a,b,c can be any positive values, except 1, an extremely large negative
sum can be found. With d = a vanishingly small positive number, using a=d, b=d, and c = 1+d.
For example, for d = 1/10^6, S ~=~ -2.8e07; larger negative sums can be found for smaller d.
iii) The smallest positive value sum can be found as approx. zero, with d ~=~ 0.7628434129,
using a = d, b = 1-d, c = 1+d. I believe there are other sets that approach zero as well.
 
Prachai K. gave a nice alternative : "As the problem wants proof, surely use ArithM-GeomM
loga (bc) + logb (ca) + logc (ab) > or = 3 cuberoot ((loga (bc))(logb (ca))(logc (ab)))
= 3 cuberoot [((log b + log c)(log c + log a)(log a + log b))/log a log b log c]
To make things easier, let log a = x, log b = y and log c = z ; Consider (x+y)(y+z)(z+x)/xyz
= [(x^2)y + (x^2)z + (y^2)x + (y^2)z + (z^2)x + (z^2)y + xyz + xyz]/xyz
Again, by AM-GM, that is > or = [8 8throot ((x^8)(y^8)(z^8))]/xyz = 8xyz/xyz = 8
So (x+y)(y+z)(z+x)/xyz >= 8. [((log b + log c)(log c + log a)(log a + log b))/log a log b log c] >= 8
From first line, loga (bc) + logb (ca) + logc (ab) >= 3 cubert ((loga (bc))(logb (ca))(logc (ab)))
= 3 cuberoot 8 = 3(2) = 6 -- Equality occurs when x=y=z, that is log a = log b = log c
a = b = c ; The smallest value is 6."
 
WINNERS - Problem 211 . (back to top) . leader board
Hermen Jacobs . . . . . . .10 pts - Proving each part is at least 2 is sufficient!
Fanfan . . . . . . . . . . . . . . 8 pts - Surprising answer and only minutes after Hermen!
Radu Ionescu . . . . . . . . 6 pts - Good proof of "infinite descent" sort of! S --> -oo
Nikita Kuznetsov. . . . . 5 pts - Said to try a=e, b=e^d, c=1/e^10; then S ~ 10d -> -oo
Prachai K. . . . . . . . . . . . 5 pts - Nice proof using geom mean < arith mean; see above
Ed Wern. . . . . . . . . . . . . 4 pts - You got three choices, all correct! See above as well.
Mark Rickert. . . . . . . . . 4 pts - Right, finding the smallest abs value is another challenge
Allen Druze. . . . . . . . . . 4 pts - Nice completing the square proof min of x + (1/x) is 2.
Art Morris. . . . . . . . . . . 3 pts - Got S >=6, but too quick to assume 'symmetric => a=b=c'
Tim Poe . . . . . . . . . . . . . 3 pts - Right idea, putting ab = 1 and varying c; S -> -oo.
Vince LoCascio. . . . . . . 3 pts - You are correct sir; as stated has no min; unless a,b,c > 1
Nick McGrath . . . . . . . . 3 pts - Questioning the premise is a great place to start!
Alan O'Donnell . . . . . . 3 pts - Where do I get more problems is the question!
Jack Dostal . . . . . . . . . . 3 pts - Varying a, b, and c will put the sum equal to any real?
Phillipe Fondanaiche. . 3 pts - Oui, juste. En Anglais cette fois; quelle est l'occasion?
Zahi Teitelman. . . . . . . 3 pts - Nice to rewrite as x + y + (1+x)/y + (1+y)/x.
Marcello Cammarata. . 3 pts - Right to cover both bases (6 and -oo); good proofs
Denis Borris . . . . . . . . . 3 pts - Letting b = c = e and playing around with a did it!
Drew. . . . . . . . . . . . . . . . 3 pts - This time it's b = (1/2)^x, a = c = 2 ; let x -> oo.
T Miller . . . . . . . . . . . . . 2 pts - Hi Truls. I'm not sure of the two values of log1(1)
Phil Sayre . . . . . . . . . . . 2 pts - True ends up a=b=c ; bc = a^x interesting approach
Ajit Athle . . . . . . . . . . . 2 pts - Good calculus proof of minimum; 2nd deriv fixed in resub
Jeremy Galvagni. . . . . . 2 pts - Problem did have a smaller answer than I expected too!
Kirk Bresniker . . . . . . . 1 pt - Your later arrival of S = -2 wasn't a min or a max; local min?
 
 
Problem #212 - Posted Monday, September 13, 2004
= Last problem of 2003/04 = New season started with Problem 213 =
Check, Please ! (back to top)
At the bank Hank cashed a check but the teller gave him cents
instead of dollars, and dollars instead of cents. After buying a 5-
cent piece of gum, Hank had exactly twice as much as his original
check. How much was the check? Show reasoning clearly.
 
Solution: (From Fanfan) Assume the correct amount is d dollars + c cents, i.e. 100d+c cents.
Then Hank received c dollars + d cents, i.e. 100c+d cents, which gives the equation (in cents)
100c+d - 5 = 2 (100d+c) <=> 98c = 199d + 5 . This means 199d+5 = 0 (modulo 98)
which is equivalent to 3d+5 = 0 (mod 98) or 3d = -5 (mod 98) or 3d = 93 (mod 98).
Since 3 is not a divisor of 98, we can simplify by 3 and get d = 31 (mod 98)
which can also be written d = 31 + 98k with k integer. It is then simple to compute c as
(199d+5)/98, which is c = 63 + 199k. Assuming that c (and d) must be integer between
0 and 99, we see that the only solution is c=63 and d=31 : the check was $31.63
 
WINNERS - Problem 212 . . lots of you this time! (back to top) . leader board
Ed Wern . . . . . . . . . . . . 10 pts - Surprise: "I used simple algebra - and it worked!"
Nick McGrath . . . . . . . . 7 pts - Yes, 98b = 199a + 5, gives Diophantine solution
Hermen Jacobs . . . . . . . 5 pts - 'Basic' program included - thanks for your entry!
Drew. . . . . . . . . . . . . . . . 5 pts - Bonus point for extras: 31$ 63c or 58$109c etc.
Radu Ionescu . . . . . . . . 4 pts - Interesting; 98(c-2d) = 5 + 3d, smaller numbers!
Vince LoCascio. . . . . . . 4 pts - Good approach; 100x + y ; y = 2.0306x ~ 2x.
Tim Poe . . . . . . . . . . . . . 4 pts - Look for integer x = (98y - 5)/199, then $31.63
Fanfan . . . . . . . . . . . . . . 4 pts - Quotable solution again; I'm a fan of your style.
Mark Rickert . . . . . . . . 3 pts - Same successful approach; 100y+x-5 = 100x+y
Mario Roederer . . . . . . 3 pts - It's been a while, welcome back. SprShtSuccess
Marcello Cammarata. . 3 pts - I like your "usual continued fractions" method!
Quasi-C . . . . . . . . . . . . . 3 pts - Welcome back; yes you missed a few good ones!
Phil Sayre . . . . . . . . . . . 3 pts - Yes, u = (199d + 5)/98 gives d=31 after a few trials
Justin Rowsell . . . . . . . 3 pts - Euclid's Algorithm; sounds easier than cont frac...
David Huang (new) . . . . 3 pts - Welcome to my contest; yay graphing calc tables!
Art Morris . 2 pts- Slight misprint good ans Yakov Macak . 2 pts- No blunder down under
Alan O'Donnell 2 pts- Back fromWorldPuzzle Allen Druze . . 2 pts- $31.63 alright, good!
Jack Dostal . 2 pts- Attached sprsht proves it Jeremy Galvagni . 2 pts- Also used gra calc
Nats Kroy (new) 2 pts- Welcome, and thanks!
Phillipe Fondanaiche. 2 pts- Dioph. lives!
Zahi Teitelman . 2 pts- Noted x must be odd Ajit Athle . 2 pts - Yes, 100x + y originally
Prachai K. . . . . . 2 pts- Internet needed, yes! Jon Stearn . 2 pts - Welcome back, good ans!
Denis Borris . 2 pts - Ok, x < 49 is true Kirk Bresniker . 2 pts- Saw you in MathWorld!
Ken Duisenberg. 2 pts- 98y - 199x = 5 reduces

Kirk also scored 3 pts on Prob 209.
Joe Alvord . 2 pts- New hip, more hop!  Coyotek (new) . 2 pts- Hi; a new teacher too!
 
 
Problem #213 - Posted Thursday, September 23, 2004
= First problem of 2004/05 = Beginning of my eighth year! =
Island Hopping (back to top)
Three nearby islands: A, B, C, are isolated from the rest of the world.
Every year, 5% of the people of island A move to island B, 5% from
A to C, 15% from B to A, 10% from B to C, 10% from C to A, and
5% from C to B. In the long run, what fraction(s) of the population
will live on each island? Show reasoning clearly.
 
Solution (finally!): Thank you all for your patience; I have been madly revising my
Prealgebra textbook, and it's finally done; it will be out early 2005! Here's Prob 213:
(psych! link you later.)
Many of you used three equations from movement percentages: Alan O'Donnell's answer:
When the population of each island is stable, we have Pop = Pop + immigrants - emigrants
A = 0.90 A + 0.15 B + 0.10 C . . . . . . . 10A = 15B + 10C
B = 0.75 B + 0.05 A + 0.05 C .===>. . 25B = 5A + 5C ===>. A=13/24, B=1/6, C=7/24
C = 0.85 C + 0.05 A + 0.10 B . . . . . . . 15C = 5A + 10B
 
I also liked Vince's solution: To reach a steady state,
a 10% reduction in A must equal a 15% increase in B and a 10% increase in C;
a 25% reduction in B must equal a 5% increase in A and a 5% increase in C;
a 15% reduction in C must equal a 5% increase in A and a 10% increase in B. Therefore,
(1) 10A=15B+10C; 10A-15B-10C=0
(2) 25B=5A+5C; -5A+25B-5C=0
(3) 15C=5A+10B; 5A+10B-15C=0
(4) multiplying equation 2 by two and adding to equation 1 produces the equation 4C=7B;
(5) subtracting the two equations in step 4 produces the equation 4A=13B;
If A=13, B=4, and C=7. Therefore, the steady state fractions are:
A = 13/24 , B = 4/24 = 1/6 , C = 7/24
 
WINNERS - Problem 213 . . lots again this time! (back to top) . leader board
First problem of the 8th season.......
Nick McGrath . . . . . . . . 9 pts - Sorry 2 dock a point 4 skipping solving & rounding
Fanfan . . . . . . . . . . . . . . 7 pts - Good answer and I liked the proof of convergence!
Alan O'Donnell. . . . . . . 6 pts - Same equations, as quoted above. Good job.
Ed Wern. . . . . . . . . . . . . 5 pts - I like your calm but utterly correct style there.
Art Morris. . . . . . . . . . . 4 pts - Used Excel with arb. start nos; good (.9 not .09)
Quasi-C . . . . . . . . . . . . . 4 pts - I like the word equilibrium; it's got balance.
Denis Borris . . . . . . . . . 4 pts - Sets up the "moves to" = "moves from" eqns.
Tim Poe . . . . . . . . . . . . . 3 pts - First answer was good, resub was better. Thanks!
David Huang. . . . . . . . . 3 pts - Good to point out the moves happen simultaneously.
Jack Dostal. . . . . . . . . . . 3 pts - Good: An+1 = An - .05 An - .05 An + .15 Bn + .1 Cn
Vince LoCascio. . . . . . . 3 pts - This was an original approach, no matrices etc.
Radu Ionescu . . . . . . . . 3 pts - Got 2a = 3b+2c, 5b = a+c, 3c = a+2b. Very clean.
Jeremy Galvagni. . . . . . 3 pts - My 8th yr of this contest, your 8th yr of teaching!
OsmTechie (new) . . . . . . 3 pts - Welcome! I applaud your use of Excel for good, not evil.
Drew. . . . . . . . . . . . . . . . 3 pts - True, you proved 13,4,7 works, not that it's unique.
Coyotek . . . . . . . . . . . . . 3 pts - Nice language, transition matrix, M^n --> answer
Phillipe Fondanaiche. . 3 pts - Good to diagonalize to raise matrix to a high power!
Phil Sayre . . . . . . . . . . . 3 pts - "The natives are restless" Nice to notice rows go constant.
Mark Rickert. . . . . . . . . 2 pts - Your B equation was off and caused island panic.
Ken Duisenberg . . . . . . 2 pts - Good steps shown but orig eqn was off, .1 not .05.
Marcello Cammarata. . 2 pts - Right, the determ. is 0. Proportions ok but no fractions
Gilad Skyte (new). . . . . . 2 pts - Welcome to my contest, thanks 4 good solution.
Mario Roederer. . . . . . . 2 pts - Correct answer; works with any starting values!
Allen Druze. . . . . . . . . . 2 pts - Right, An - An-1 does approach 0, making it stable.
Ajit Athle . . . . . . . . . . . 2 pts - Glad you enjoyed the problem. (1300, 400, 700)
Kirk Bresniker . . . . . . . 2 pts - Good use of this popular trio of equations.
Zahi Teitelman. . . . . . . 1 pt - Answer was approx. right; missing eqns and steps.
Albert Alvarez . . . . . . . 1 pt - Good answer, (came after prob 214 was up)
Hermen Jacobs . . . . . . . 1 pt - Welcome back from vacation; I love traveling too!
Yakov Macak . . . . . . . . 1 pt - Working backwards from # 215; I'll see you there.
 
Problem #214 - Posted Monday, October 4, 2004
The Four Means ! (back to top)
Given two positive real numbers a and b ,here are four ways of
computing the "mean":
am = arithmetic mean = (a + b) / 2
gm = geometric mean = [a b]
hm = harmonic mean = 2 / (1/a + 1/b)
rms = root mean square =[(a^2 + b^2) / 2]
a) Rank these from smallest to largest, proving your result. If one type
is not always less than another, say when it's equal and when it's more.
b) Use the diagram to compare at least three of the means using the
lengths of segments. Assume E is at the center of the semicircle.
Show reasoning clearly; one point penalty for resubmissions.
Which to use, and how?
 
Solution : (From Prachai K.) hm <= gm <= am <= rms
a) Prove gm<=am : (a-b)^2 >= 0 ; a^2 + 2ab + b^2 >= 4ab
(a+b) >= 2sqrt(ab) ; (a+b)/2 >= sqrt(ab)
Prove hm <= gm : (a-b)^2 >=0 ; a^2 + 2ab + b^2 >= 4ab
(a^3)b + 2(a^2)(b^2) + a(b^3) >= 4(a^2)(b^2) ; ab(a^2 + 2ab + b^2) >= (2ab)^2
(a+b)sqrt(ab) >= 2ab ; sqrt(ab) >= 2ab/(a+b) ; sqrt(ab) >= 2/(1/a + 1/b)
Prove rms >= am ; (a-b)^2 >= 0 ; a^2 - 2ab + b^2 >= 0
2a^2 + 2b^2 >= a^2 + 2ab + b^2 ; (a^2 + b^2)/2 >= (a^2 + 2ab + b^2)/4
sqrt((a^2 + b^2)/2) >= (a+b)/2 ; They can all be equal if a=b (quite easy to see)
b) From your diagram, let CD = a and DF = b ; HE = am because a+b is diameter
and (a+b)/2 is radius ; GD = gm because triangles CGD and GFD are similar and
a/GD = GD/b ; HD = rms because HE = (a+b)/2 and DE = DF - FE = (b-a)/2.
From Pythagorean, HD is the rms. Is there hm?
(Dan's note: Yes! Fanfan and others report that dropping a perpendicular from D to GE
at X, the distance GX is then the harmonic mean hm! I dare you to Pythm-prove it!
See a nice PDF at this link supplied to the 'dansmath community' by Ed W. Thanks Ed!
And Vince L points out the geom mean of a and b is also the geom mean of hm and am!)
 
WINNERS - Problem 214 . . lots again! (back to top) . leader board
 
Marcello Cammarata. . 10 pts - First place for Marcello (specifically, and Italy generally.)
Nick McGrath . . . . . . . . 7 pts - Nice answer beating out earlier responses with correctness
Fanfan . . . . . . . . . . . . . . 6 pts - Bonus point for showing the GX = harmonic mean
Prachai K. . . . . . . . . . . . 5 pts - Good algebra proofs and quotable solution!
Jeremy Galvagni. . . . . . 5 pts - Nice approach proving b) first then invoking geom
Phillipe Fondanaiche. . 4 pts - Thanks for your continued global support; bien fait!
Coyotek . . . . . . . . . . . . . 4 pts - Square both sides, subtract, get each comparison!
Zahi Teitelman. . . . . . . 4 pts - Interesting use of Trig to get distances; good one!
Ed Wern. . . . . . . . . . . . . 4 pts - Later answer; bonus point for awesome link (above)
Art Morris. . . . . . . . . . . 3 pts - Skipped the geometry, extra pt for being so early & right
Drew. . . . . . . . . . . . . . . . 3 pts - Good proofs on first part, looser logic on second...
Vince LoCascio. . . . . . . 3 pts - Points out that gm of a and b = gm of hm and am!
Quasi-C . . . . . . . . . . . . . 3 pts - Amazing what follows from the simple (a - b)^2 > 0
Radu Ionescu . . . . . . . . 3 pts - Algebraically sound proofs; geometry a bit 'sketchy'.
Ajit Athle . . . . . . . . . . . 3 pts - Nice use of formula ab > 4 a^2 b^2 / (a + b)^2
Allen Druze. . . . . . . . . . 3 pts - Using similar triangles gets HE > GE > GD etc.
Mario Roederer. . . . . . . 3 pts - Unique use of plots of am vs gm etc for 0 < a/b < 1
Albert Alvarez . . . . . . . 3 pts - Nice proofs and logical geometric statements.
Hermen Jacobs . . . . . . . 3 pts - Good to hear from you again, long trip, no math?
Akifumi . . . . . . . . . . . . 3 pts - Welcome back, A.I. Good solid answers & reasoning
Mark Rickert. . . . . . . . . 2 pts - Answer and ranking correct; needed to see more proof
Tim Poe . . . . . . . . . . . . . 2 pts - Good to run a numeric check, sometimes misleading
Jack Dostal . . . . . . . . . . 2 pts - Lots of contestants laid off part b) this time...
Denis Borris . . . . . . . . . 2 pts - Nice idea to square all means then compare em.
David Huang. . . . . . . . . 2 pts - Good to have you as a repeat enterer, nice alg prfs.
Phil Sayre . . . . . . . . . . . 2 pts - Another geo-non-metric response, but nice first part
Alan O'Donnell. . . . . . . 2 pts - Back from the World Puzzle Champ's in Croatia!
Ralph Nguyen (new). . . . 1 pt - Welcome to my contest, good try! True all = if a = b.
Yakov Macak . . . . . . . . 1 pt - Lagged others by a month but so did I! Good on ya.
 
 
Problem #215 - Posted Saturday, October 23, 2004
Put A Hex On It ! (back to top)
Find the side length and area of the largest regular hexagon that
can be inscribed in a square of side 1 meter. Prove your answer.
Simple gif attachments ok, or just describe it! Show reasoning; 1 pt penalty for resubs.
 
Solution : Sorry, mates, I forgot to post the solution last time! Here's Alan O'D's:
"Rotate the hexagon so that a 45/45/90 triangle is formed in one corner, and the
hex touches the square with 4 of its 6 vertices. ie, 2 diameters of the hexagon
are touching the square. This must be the optimal positioning, since rotating the
(fixed sized) hexagon either way will increase the size of square required for it.
If we now look at 1/4 of the square, we have an equilateral triangle inclined at
15 degrees, with one of its vertices touching a corner of the square.
Looking at one edge, we have r sin(15) + r sin(45) = 1/2 or
r = 1/(2 (sin(15) + sin(45)) ; sin(15) = (sqrt(6) - sqrt(2))/4 ; sin(45) = sqrt(2)/2
Side = r = 2/(sqrt(6) + sqrt(2)) = 0.517638... Area of eq. triangle = r^2 sqrt(3)/4
Area of Hexagon = 6 r^2 sqrt(3)/4 = 6 sqrt(3)/(8+4 sqrt(3)) = 0.6961524...
 
WINNERS - Problem 215 . . thanks for entering! (back to top) . leader board
 
Nick McGrath . . . . . . . 10 pts - Nice proof that the optimal rotation is 15 degrees
Fanfan . . . . . . . . . . . . . . 7 pts - Beautiful idea that hex is in intersection of 3 squares
Akifumi. . . . . . . . . . . . . 5 pts - A few answers were earlier but lacked as much proof
Radu Ionescu . . . . . . . . 4 pts - Right idea with rotation a bit too big with side & area
Marcello Cammarata. . 4 pts - Equiv. prob: smallest square around hexagon - good!
Mark Rickert . . . . . . . . 4 pts - Movin up the charts with good proof about best rot'n
Art Morris. . . . . . . . . . . 3 pts - All-time leader comes close with sqrt[1+(arctan15)^2]/2
Denis Borris . . . . . . . . . 3 pts - Thanks for providing relevant link to an old problem
Tim Poe . . . . . . . . . . . . . 3 pts - 'Logically' true isn't always a logical path to proof
Alan O'Donnell. . . . . . . 3 pts - Rotate hex so that 45/45/90 in one corner - it works!
Quasi-C . . . . . . . . . . . . . 3 pts - Good job to realize c = hyp = x sqrt[3] then solve
Vince LoCascio. . . . . . . 3 pts - Enlarge, then rotate. A good strategy in this case!
Ayane Buccini (new). . . . 3 pts - Welcome to the contest, good answer, nice word doc!
Jeremy Galvagni. . . . . . 3 pts - side = 0.5/cos(15) slightly more than 0.5 true
Phil Sayre . . . . . . . . . . . 3 pts - Nice 'coordinated' approach, eqn has 15 deg as soln.
Kirk Bresniker . . . . . . . 3 pts - Becoming a regular now, with exact radical answers
Hermen Jacobs . . . . . . . 2 pts - There is a better solution by rotating around a bit
Coyotek . . . . . . . . . . . . . 2 pts - Again, by rotation there is more area & longer side
David Huang. . . . . . . . . 2 pts - You can use trig to find the side in terms of the angle
Ajit Athle. . . . . . . . . . . . 2 pts - Symmetric hex of side 0.5486 won't quite fit in sq
Mario Roederer. . . . . . . 2 pts - Solution is correct ans but no prf of = width or maximal
Phillipe Fondanaiche. . 2 pts - Good answer for length, logically true; needed area
Allen Druze. . . . . . . . . . 1 pt - Rotation a little bit increases area, then decreases it
Ravi Raja . . . . . . . . . . . 1 pt - L = 2 - sqrt[2] is a bit too big, not same as Ajit's ans
Yakov Macak . . . . . . . . 1 pt - Good answer, just after #216 had started coming in.
 
 
Problem #216 - Posted Saturday, November 6, 2004
Band in Boston ? (back to top)
Red Sox, Patriots, John Kerry, lots to celebrate! A marching band goes up
Main Street in single file. Then one band member sits down, and the rest of
them march on in twos. One more sits and the rest march in threes, and so on,
until one sits and the rest march in rows of ten. What is the smallest positive
number of members that could have started this big band march on Boston?
Show reasoning clearly; one point penalty for resubmissions.
 
Solution : Prachai K. sends this in from Thailand; let's all hope things are ok with Prachai! (He says he's ok!)
"Say there are x members. So 2 divides x-1 , 3 divides x-2 , . . . , 10 divides x-9
That is, x = 1(mod2) , x = 2(mod3) , . . . , x = 9(mod10). Cut out the unnecessary ones.
These are left: x = 4(mod5) ; x = 6(mod7) ; x = 7(mod8) ; x = 8(mod9)
At last! The GCD of the mods is 1. We can use the Chinese Remainder Thm.
The CRT tells me that there's only one solution mod(5*7*8*9), which is mod 2520.
I don't need anyone to tell me the answer is 2519. [Details sketchy there...]
[Dan's note; Search Chinese Remainder Theorem at mathworld.com for relatively prime moduli.]
Others pointed out x = -1 (mod m), so LCM of 2, 3, . . . , 10 is 2520; ans 2519.
 
WINNERS - Problem 216 . . HAPPY NEW YEAR! (back to top) . leader board
Arthur and others point out that there's no Main Street in Boston; making this a "fictitious arithmeticus."
Marcello Cammarata . 10 pts - Nice "filtering" sieve spreadsheet settled on 2519
Art Morris. . . . . . . . . . . 7 pts - True; "No Main St. in Boston" => marching difficult!
Nick McGrath. . . . . . . . 5 pts - Correct to look at it as N = -1 (mod x) , x any 2 to 10
Fanfan . . . . . . . . . . . . . . 5 pts - Last group has 10k, orig 10k+9, 10k+2 mult of 8, etc.
Hermen Jacobs . . . . . . . 4 pts - N-1 divisible by 2, N-2 by 3, etc. Basic prog says 2519
Prachai K. . . . . . . . . . . . 4 pts - Glad you feel the need to use Chinese Remainder Thm
Mark Rickert . . . . . . . . 4 pts - True, the ans is 1 less than LCM of 1 thru 10; why?
There were so many responses this week that I'll "table" the rest of you; arranged mostly but not
only chronologically ; welcome to several new contestants; sorry about the lag time these days.
Many correct answers got just 2 pts this week so incorrect was 1 pt in most cases. - Dan ;-}
 Akifumi. . . . . . . . . . . . . 3 pts  Phillipe Fondanaiche. . 3 pts
 Denis Borris . . . . . . . . . 3 pts  Alan O'Donnell. . . . . . . 3 pts
 Vince LoCascio. . . . . . . 3 pts  Ed Wern. . . . . . . . . . . . . 3 pts
 Jeremy Galvagni. . . . . . 3 pts  Quasi-C . . . . . . . . . . . . . 3 pts
 Tim Poe . . . . . . . . . . . 2 pts  Kirk Bresniker . . . . . 2 pts  Santo (new) . . . . . . . . . 2 pts
 Allen Druze. . . . . . . . 2 pts  Coyotek . . . . . . . . . . . 2 pts  Yakov Macak . . . . . . 2 pts
 Andy Oats (new). . . . . 2 pts  David Nadler (new) . . 2 pts  Ajit Athle . . . . . . . . . 2 pts
 Phil Sayre . . . . . . . . . 2 pts  Mario Roederer. . . . . 2 pts  Drew. . . . . . . . . . . . . . 2 pts
 Zahi Teitelman. . . . . 2 pts  Anandi (Rao) (new) . . 2 pts  Nick Sherrard. . . . . . 2 pts
 Ken Duisenberg . . . . 2 pts  Mahul Bhattacharya (new) 2 pts  Pam Deutsch (new) . . 2 pts
 The Bozzball . . . . . . 2 pts  Justin Rowsell. . . . . 1 pt  Radu Ionescu . . . . . . 1 pt
 Ravi Raja . . . . . . . . . 1 pt  Henry Yang (new) . . . 1 pt  Zeke&Ruth Moore (new)1 pt
 
 
Problem #217 - Posted Thursday, December 2, 2004
Baskets of Apples ! (back to top)
I walked by a holiday display, and noticed there were nine straw baskets,
each one having a whole number of apples, at most 9, and possibly none.
Then I also noticed the mean number of apples was 4, the median was 4,
and the mode was 2. Is this possible, and if so, how many solutions are there?
(not counting the positions of the baskets) Show reasoning clearly; 1 pt penalty for resubmissions.
 
*** Thank you all for your patience the last couple of weeks; brutal publishers deadlines for my book! - Dan ***

 

Solution : Let's assume the 9 baskets are arranged from smallest to largest no. of apples.
Key points: Mean = 4, and 9 baskets ==> 36 apples; Median = 4; 5th basket has 4 apples.
(Single) Mode = 2; at least 2 of first 4 baskets have 2 apples. Possible first 4 baskets:
0122, 0222, 1222, 1223, 2222, 2223, 2224. (Turns out 1223 won't give any final solutions.)
Then fill out the 36 apple total with fewer of any other number than 2's, and end up with: 
012245679
022244589
022244679
022244688
022244778
022245579
022245588
022245669
022245678
022246677
 
122244579
122244588
122244669
122244678
122245569
122245578
122245668
122245677
 
222244479
222244488
222244569
222244578
222244668
222244677
222245559
222245568
222245577
222245667
 

222344559
222344568
222344577
222344667
222345567
222445566
Some of you got 33, some found all 34 solutions , some found fewer, thanks to all for entering!
I meant that 2 was the only mode (Type 1); Nick and Quasi also counted Type 2's (multimodal)!
 
WINNERS - Problem 217 . . (back to top) . leader board
 
Nick McGrath. . . . . . . . 10 pts - First with 34 baskets; also says 89 of type 2 (multimodal)!
Tim Poe . . . . . . . . . . . . . 7 pts - Thanx for Visual Basic code; I'll crash my Mac with it.
Mark Rickert . . . . . . . . 5 pts - I like your strategy of filling the first 4 or 5 all ways, etc.
Akifumi. . . . . . . . . . . . . 4 pts - You got it; middle basket has 4, at least two 2's, yep!
Marcello Cammarata . 4 pts - Where you are it might be baskets of Roma tomatoes!
Quasi-C . . . . . . . . . . . . . 4 pts - Early entry; got 33 baskets type 1 (and 69 type 2 too)
Alan O'Donnell. . . . . . . 3 pts - Yes single modal; sorry I don't know when next prob posts.
Denis Borris . . . . . . . . . 3 pts - Quite right to note it wasn't required to list all 34.
Kirk Bresniker . . . . . . . 3 pts - Made a 'quick count' of the progressions & got all 34.
Joe Alvord . . . . . . . . . . . 3 pts - Your thoughts were correct; fill the first 4 baskets...
Phillipe Fondanaiche. . 3 pts - a<b<...< i ; a+b+...+i = 36 ; count up; trente-quatre.
Vince LoCascio. . . . . . . 3 pts - Nice use of 'underage' (not under-age) to find 33
Prachai K. . . . . . . . . . . . 3 pts - Glad things are ok up in Bangkok; good answer too
Ed Wern. . . . . . . . . . . . . 2 pts - Your resub appended 222445566; still a couple short
 
Some of you found 33 combos, some got nearly as many...
Ken Duisenberg . . . . 2 pts  Jeremy Galvagni . . . . 2 pts  Allen Druze. . . . . . . . 2 pts
 Zahi Teitelman. . . . . 2 pts  Art Morris . . . . . . . . . 2 pts
Hermen Jacobs . . . . . 2 pts
 Ravi Raja . . . . . . . . . 2 pts  Jack Dostal . . . . . . . . 2 pts Mario Roederer. . . . . 2 pts
   Coyotek. . . . . . . . . . . . 2 pts
 Here are some folks with a few  fewer combinations but thanks  for entering, keep it up!
 Barb Owca (new) . . . . . 1 pt  Phil Sayre. . . . . . . . . . 1 pt  Radu Ionescu . . . . . . 1 pt
  Zeke&Ruth Moore . . . .1 pt  
 

Problem #218 - Posted Friday, December 17, 2004
Holy Sphere (and circle) (answer both for max pts) (back to top)
 
a) A hole is drilled through a wooden sphere (top). If the
hole is 6 cm long, find the volume of wood that remains.
 
b) A chord of a circle is tangent to an inner, concentric
circle (bot). If the chord is 6 cm long, find the area of the ring.
 
Show reasoning clearly; one point penalty for resubmissions.

Solution : Dan's note: The first part was already in my archives, problem #94. I thought it looked familiar!!
a) The main idea is that I don't say what the radii of the sphere or the hole are,
so if the problem is to have a solution the answer should be independent of those radii.
If we then assume that r = rad of cyl hole = 0, then R = rad of sphere = 3. Thus the volume
is V = (4/3) pi R^3 = (4/3) pi 3^3 = 36 pi , or about 113.04 cubic cm of wood. This is the
volume, whether it's a small ball with a tiny hole, or a large "napkin ring/bracelet" shape.
This can also be done using calculus, integrating the area of an annulus times thickness.
Here's a poem sent in by Denis Borris:
"Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six cm long.
Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!"

 

 
b) This part can be done directly using the Pythagorean Theorem; the area of the ring
(annulus) is pi R^2 - pi r^2, from the picture R^2 = r^2 + 3^2 so we get
A = pi R^2 - pi (R^2 - 9) = 9 pi or about 28.27 sq cm
Also, by an argument similar to the first, the R and r aren't given, so we can assume that
r = 0, making R = 3 since L = 6. Thus A = pi (3^2) = 9 pi as before.
 
WINNERS - Problem 218 . . (back to top) . leader board
Some of the 2 pt were early errors, others were later correct responses.
Radu Ionescu . . . . . . . . 10 pts - Good, a bit light on steps, but passed the dansmuster test
Ed Wern. . . . . . . . . . . . . 7 pts - Noticed I used the first part in Problem #94 long ago!
Mark Rickert . . . . . . . . 5 pts - "Radius of sphere not stated so say 3." That's the way!
Nick McGrath. . . . . . . . 4 pts - Does the whole task with an integral of thin rings!
Ravi Raja . . . . . . . . . . . 4 pts - That's the key, that the result is indpt of R and r!
Art Morris. . . . . . . . . . . 3 pts - Since diameter of hole is not given, I choose 0. Yes.
 Zahi Teitelman . . . . . . 3 pts - Vol of wood is Vsphere - Vcyl - 2 Vendcap. then simplify
Tim Poe . . . . . . . . . . . . . 3 pts - That's the idea; "not much info here"; make some up
Marcello Cammarata. . 3 pts - Direct approach like Zahi, Vcap = pi h^2(R - (h/3)).
Mario Roederer. . . . . . . 3 pts - Sphere link: www.wiskit.com/marilyn/sphere.html
Phil Sayre . . . . . . . . . . . 3 pts - Area of annulus * thickness => integral R drops out
Zeke Moore . . . . . . . . . . 3 pts - Good sol'n from a teacher who uses my problems!
Jeremy Galvagni. . . . . . 3 pts - True; calculus is harder to type than algebra. Good ans.
Phillipe Fondanaiche. . 3 pts - We'll all look for your site en francais (pas de cedille)
Alan O'Donnell. . . . . . . 2 pts - Good work; although wrote 12 pi instead of 36 pi
Hermen Jacobs . . . . . . . 2 pts - The 70 pi was close; did 54 + 16 s.b. 54 - 18.
Denis Borris . . . . . . . . . 2 pts - Great poem; see comments above; (no numer. ans for b)
Juan Carlos Carrara (new) 2 pts - Bienvenidos - good ans. all the way from Argentina
Frank Mullin (new). . . . 2 pts - Welcome to the contest, good answer, nice bicycle
Kirk Bresniker . . . . . . . 2 pts - Right; subtract those two Vcaps and get 36 pi.
Joe Alvord . . . . . . . . . . . 2 pts - John Campbell, early sci-fi writer, reasoned it out.
Ajit Athle. . . . . . . . . . . . 2 pts - Attributes logic to Martin Gardner (where I first saw it)
Nats Kroy. . . . . . . . . . . . 2 pts - Good integral job and Pythagorean annulus...
Coyotek . . . . . . . . . . . . . 1 pts - Part b was correct; "conceded defeat" on part a.
Allen Druze. . . . . . . . . . 1 pt - The 9 pi(e) was right; the hole can be any fraction
Yakov Macak . . . . . . . . 1 pt - Good answer; came in after some 219's so just 1 pt.
 

 
Problem #219 - Posted Monday, December 27, 2004
See Double Free (back to top)
Call a set of numbers "double-free" if there is no pair {m, 2m} in the set.
For example, {1, 3, 4, 5} is double-free. Find the size of the largest double-free
subset of {1, 2, 3, . . . , 2^n} for each n = 3, 4, 5, 6, 7, and 8; give actual
maximal sets for n = 3, 4, 5. Show reasoning clearly; one point penalty for resubmissions.
 
Solution :
Ok, this is not an original problem but you guys still analyzed it brilliantly!
Math teacher Zeke figured you use each number that qualifies in order from 1 to n;
Use 1, not 2, 3, 4, 5, not 6, 7, not 8: DF(2^3) = {1, 3, 4, 5, 7}. size = 5
The next is DF(2^4) = {1, 3, 4, 5, 7, 9, 11, 12, 13, 15, 16}. size = 11
The next is DF(2^5) = DF(2^4) U {17, 19, 20, 21, 23, 25, 27, 28, 29, 31}. size = 21.
The recursions for F(n) = DF(2^n) are F(n+1) = 2 F(n) + (-1)^(n+1) = F(n) + 2 F(n-1)
The sizes are, starting at DF(2^1): 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, . . .
Another method is this: Use F(1) = {1}, F(2) = {1, 3, 4}; then form new sets:
F(3) = F(1) U {5, 6, 7, 8} ; F(4) = F(2) U {9, 10, 11, 12, 13, 14, 15, 16}, etc.
 
WINNERS - Problem 219 . . (back to top) . leader board
Some of the 2 pt were early errors, others were later correct responses.
Art Morris. . . . . . . . . . 10 pts - All-time great with another weekly win. Good expl.
Marcello Cammarata. . 7 pts - Good strategy to elim the second quarter of the nos.
Zeke Moore . . . . . . . . . . 5 pts - "Go in order, use what you can, no use skipping..."
Hermen Jacobs . . . . . . . 5 pts - Nice recurs.: F(n) = 2^n - F(n-1) = 2^n - 2^(n-1) + ...
Nick McGrath. . . . . . . . 4 pts - Include the odds, and 4*the odds, and 16* the odds,...
Coyotek . . . . . . . . . . . . . 4 pts - Exclude one element from pairs 1-2, 2-4, 3-6, etc.
Denis Borris . . . . . . . . . 4 pts - Size ~ (2/3)*(2^n); glad you got way into this one!
Tim Poe . . . . . . . . . . . . . 3 pts - Recovered the right answer with good resub, thanks!
Jeremy Galvagni. . . . . . 3 pts - Nice you threw in sol'ns with the largest nos. too
Mario Roederer. . . . . . . 3 pts - "No reason not to have all the odds > (2^n)/2" true!
Joe Alvord . . . . . . . . . . . 3 pts - 1,3,4,5,7: 1 elim 2, 3 elim 6, etc. Can go in order!
Phillipe Fondanaiche. . 3 pts - Ouais, this is On-Line Encyc. Seq #A001045.
Alan O'Donnell. . . . . . . 3 pts - Good answer; thanks for sequence reference too
Quasi-C. . . . . . . . . . . . . 3 pts - Maximal set can include all odds; no I didn't get 218
Radu Ionescu . . . . . . . . 2 pts - Earlier entry did solve the prob for 2n but not 2^n
Vince LoCascio. . . . . . . 2 pts - Missed a few integers there but good entry, thanks
Prachai K. . . . . . . . . . . . 2 pts - Include the odds above half, then got all up to 85.
Allen Druze. . . . . . . . . . 2 pts - I like Fibonnesque recursion a(n+2) = a(n+1) + 2 a(n)
Kirk Bresniker . . . . . . . 2 pts - Resub fixed orig "awk" program, not awk-ward!
Ed Wern. . . . . . . . . . . . . 2 pts - Remove 1st half, keep 2nd half, use 1st quarter, etc
Phil Sayre . . . . . . . . . . . 2 pts - Uses odds + 4 * (subset from 2 previous). Yep!
Ravi Raja . . . . . . . . . . . 2 pts - Missed the 43 and the 171; flaw in evens, others ok
Mark Rickert . . . . . . . . 2 pts - Correct later entry; nice but not maximal points
Zahi Teitelman . . . . . . 1 pt - There's some dropoff from doubling each time
Ajit Athle . . . . . . . . . . . 1 pt - Your intuition was right about both parts of yr answer!
 
 
Problem #220 - Posted Sunday, January 9, 2005
Cube in a Cone (back to top)
A cone has a circular base of radius 1 ft, and vertex at height 3 ft directly
above the center of the circle. A cube has four vertices on the base and
four on the cone's lateral side. a) What length is a side of the cube?
b) What cone height (r = 1) makes the cube side 1 ft? Show reasoning clearly.
 
Solution : Newly updated; answer to 219 was here for a while. From Kirk B.:.
For a cone of height h and radius r, a cube of side length a sitting centered on the base
will touch the lateral surface at four points. If we cut through two diagonal corners of
the cube, we'll find two triangles similar to h:r : h-a:a/sqrt(2) or a:r-a/sqrt(2).
Solving either gives the same answer: h:r = h-a:a/sqrt(2) , giving:
==> a * h / sqrt(2) = rh-ra ==> a * (h/sqrt(2) + r) = rh ==> a = rh / (r + h/sqrt(2)
The first question is to find a for r=1, h=3, so a = 3/(1 + 3/sqrt(2)) ~= 0.9611
The second question is to find h for r=1, a=1, so 1 = h/(1 + h/sqrt(2))
==> 1+h/sqrt(2)=h ==> 1=h(1-1/sqrt(2) ==> h = 1/(1-1/sqrt(2)) = ~ 3.414
 
WINNERS - Problem 220 . . (back to top) . leader board
I enjoyed proving the equivalence of many of your radical answers! - Dan
Nick McGrath. . . . . . . . 10 pts - Recent perennial contest leader gives (3/7)(3\/2 - 2)
Zeke Moore . . . . . . . . . . 7 pts - Strong relative newcomer gives us (-6 + 9\/2)/7
Marcello Cammarata. . 5 pts - Our friend from Italy chooses the form 6/(3\/2 + 2)
Art Morris. . . . . . . . . . . 4 pts - Art lets us do the subst of c=3 into s=2/(\/2+2/h)
Mark Rickert . . . . . . . . 4 pts - This answer was 'marked down' as 2/(\/2 + 2/3)
Kirk Bresniker . . . . . . . 4 pts - Cube touches cone at 4 pts, and so's your score!
Hermen Jacobs . . . . . . . 3 pts - Octogenarian gives my favorite form 3\/2/(3 + \/2)
Zahi Teitelman . . . . . . 3 pts - Yes; x = 6/(2 + 3\/2); I admire your refusal to approx.
Alan O'Donnell. . . . . . . 3 pts - This Scotsman gives it correctly as y = 3/(1 + 3/\/2)
Yakov Macak. . . . . . . . . 3 pts - I liked your answer well enough to steal it; see above
Quasi-C. . . . . . . . . . . . . 3 pts - h = r \/2 = 3\/2/(3 + \/2), as usual nice explanation
Tim Poe . . . . . . . . . . . . . 3 pts - Nice algebra giving s = 3(1 - N) = 6/(2 + 3\/2)
Jeremy Galvagni. . . . . . 3 pts - Gave both 6/(2 + 3\/2) and (9\/2 - 6)/7 to be safe!
Denis Borris . . . . . . . . . 3 pts - Cube diag = c \/2 leads to 6/(2 + 3\/2) then 2 + \/2
Phillipe Fondanaiche. . 3 pts - Chooses two : 1/(\/2/2 + 1/3) and 3(3\/2 - 2)/7
Ajit Athle . . . . . . . . . . . 3 pts - Good eqns give us 3\/2/(3 + \/2) and b) 2 + \/2
Jerome Mahadeo (new) . 3 pts - All the way from South Africa, welcome to my contest!
Joe Alvord . . . . . . . . . . 2 pts - Later resub changing \/3's into \/2's fixed the problem.
Phil Sayre . . . . . . . . . . . 2 pts - Resub was correct, thanks for fixing first attempt
Prachai K. . . . . . . . . . . . 2 pts - Good expl but miswrote as (9\/2 - 6) / 5 not /7.
Ken Duisenberg . . . . . . 2 pts - Correct final measures but steps/reasons too thin
Coyotek . . . . . . . . . . . . . 2 pts - Nice x/y = 1/\/2 and y = 3 - 3x ; x = (3/7)(3\/2 - 2)
Irina Kolesnik . . . . . . . 2 pts - Welcome back after years! Beautiful diagram and ans!
Ed Wern (update) . . . . . . 2 pts - I ended up finding your excellent ans; thanx 4 remindg
James Wan (new) . . . . . . 2 pts - Thanks for finding my contest; good ans 3/(1+3\/2/2)
Radu Ionescu . . . . . . . . 1 pt - Didn't understand step to get 2\/2 / 3 ; better on part b.
Luis deBrito (new) . . . . . 1 pt - Didn't receive your reasoning but ans s.b. irrational
Vince LoCascio. . . . . . . 1 pt - Yes, diam = \/2 x ; first part ok; 2nd part off a bit
 
 THANKS to all of you who have entered, or even just clicked and looked.
My site is now in its 8th season - OVER 60,000 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 - 2005 A.D.
 
Problem Archives Index
 
Probs & answers . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
Problems only . . . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
Probs & answers . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
Problems only . . . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
Probs & answers . 181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231-240 . 241-250 . 251+
Problems only . . . . 181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231-240 . 241-250 . 251+
 
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or go back and work on this week's problem!
 
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