dan's math@home - problem of the week - archives
 
 
Problem Archives page 20
with Answers and Winners. For Problems Only, click here.
 
1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+ . prob index
 
191 -- The Disk Cutters
192 --- Slice Into Dice !
193 - Even That's Odd !
194 Uncommn Multiple
195 - Over-Committeed
196 - The Pentominator!
197 CommonRemainder
198 - Odd and Different
199 - Twice The Angle !
200- Run with the Dogs
 
 
Problem #191 - Posted Friday, January 2, 2004
The Disk Cutters (back to top)
From a smooth 30-inch-diameter circular plywood disk, two smaller circular disks
of diameters 20" and 10" are cut. What is the largest circular disk that can be cut
from one (either) piece of the remaining plywood?
Ignore width of sawblade when cutting. Explain your answer carefully.
 
Solution: Many of you recognized this as a Descartes / Soddy "Kissing Circle" problem, and some just used the
distance formula, the law of cosines, Heron's formula, or just the Pythag Thm. I like a resourceful group; keep it up!
 
Descartes' Kissing Circles Formula says for curvatures a, b, c, d (= reciprocals of the radii), there is a "close relatonship"
. . (a^2 + b^2 + c^2 + d^2) = (1/2)(a + b + c + d)^2. .The curvature is negative if you're inside the circle.
Descartes' formula comes from 1638; Fredrick Soddy's poem about it came 293 years later (sent in by Rick Montgomery):
Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.		 Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.
With this relation in mind, it's a simple enough matter to plug in the known and find the unknown:
Known radii : 5 , 10 , and -15 (neg sign for concave big circle), Unknown radius : r.
Curvatures : 1/5 , 1/10 , -1/15 , a = 1/r. Formula of Descartes:
[(1/5)^2 + (1/10)^2 + (-1/15)^2 + a^2] = (1/2)[(1/5 + 1/10 - 1/15 + a)^2]
(1/25 + 1/100 + 1/225 + a^2) = (1/2)(1/5 + 1/10 - 1/15 + a)^2 . . mult both sides by 900 = 30^2:
36 + 9 + 4 + 900 a^2 = (1/2)(6 + 3 - 2 + 30a)^2 ==> 2(49 + 900 a^2) = (7 + 30a)^2 ==>
98 + 1800 a^2 = 49 + 420 a + 900 a^2 ==> 900 a^2 - 420 a + 49 = 0 ==> (30 a - 7)^2 = 0
So a = 7/30 giving r = 30/7 ; the diameter of the circular disk is d = 60/7 = 8 4/7 inches.
The double root means the two circles are the same size, normally there's a 'big' and 'small' solution.
 
Another way to find r is to think of the center of the new circle as the solution to simultaneous eq'ns
in variables x, y, and r, where (x, y) is the center of the new circle. See Jeremy's answer for equations.
See below for hints about the other methods; challenge yourself to solve by all methods listed above.
 
WINNERS - Problem 191 . (back to top) . leader board . . .
Tim Poe . . . . . . . . . . . . . 10 pts - Descartes made day-head spin so used eqns of circles. Good nuff!
Nick McGrath . . . . . . . . 7 pts - Descates and Soddy teamed up nicely to give r = 30/7. Well done.
Marcello Cammarata . . 5 pts - Ingenious use of Heron's theorem (falls under law of cosines wing?)
Art Morris . . . . . . . . . . . 4 pts - Yes, 49r^2-420r+900 = 0 has double root at r = 30/7. 'Beecroft' too!
Quasi-C. . . . . . . . . . . . . . 4 pts - Law of Cosines is a good idea when there's 3 of anything. yep.
Rick Montgomery . . . . . 4 pts - You woulda gotten 3 pts for the next ans but bonus poem point!
Joe Alvord . . . . . . . . . . . 3 pts - Used Dan's Circles page to help with Dan's contest. How self-refereetial.
Ajit Athle . . . . . . . . . . . . 3 pts - Resub and decimal roundoff weren't ideal for pts, but good answer!
Ed Wern . . . . . . . . . . . . . 3 pts - Dropped a perp from unknown center to x-axis and used Pythm.
Jeremy Galvagni . . . . . . 3 pts - x^2+y^2=(15-r)^2, (x+5)^2 + y^2 = (10+r)^2, (x-10)^2 + y^2 = (5+r)^2
Ken Duisenberg . . . . . . . 3 pts - same idea as JG, gets 5+2r = 25-6r+4y = -25+4r-6y , r = 30/7.
Phil Sayre . . . . . . . . . . . . 3 pts - Center (0, 0), JG eqns, x = 45/7, y = 60/7, r = 30/7. Nailed it!
Hermen Jacobs . . . . . . . . 3 pts - I gave you 2 days predate because of e-mail delay, good answer!
Jim No-Spam . . . . . . . . . 2 pts - Welcome-back, No-Spam! Nice in-line pics but r = 30/7 not d.
S. Khoo . . . . . . . . . . . . . . 2 pts - Easier to deal with squares not square roots; good answer though
Yakov Macak . . . . . . . . . 2 pts - Nice series of steps, leading up to radius of 4 2/7 inches. Yes.
Allen Druze . . . . . . . . . . 1 pt - Five inch radius is the right hand circle, this one needs 2B smaller.
 
 
Problem #192 - Posted Sunday, January 18, 2004
Slice Into Dice! (back to top)
I painted a block of wood that's a x b x c cm (a <= b <= c whole numbers).
If I then cut it up into one-cm cubes, half of the cubes have paint
on them and half don't. What are all possible values of a, b, and c?
Ignore width of sawblade when cutting. Explain answer carefully.

Solution:
Alan O'D. noted I didn't say I painted ALL the cubes on the outside, so there could be many more solutions. Finitely many?
Nick McG and others wrote clean programs in EXCEL, C, C#, BASIC, Visual Basic, and some were attached. Thanks!
The main equation is that the total volume is twice the number of inside cubes ; Thus abc = 2(a-2)(b-2)(c-2).
Expanding gives (a headache and): abc- 4ab - 4ac + 8a = 4bc - 8b - 8c + 16 ; a[bc- 4(b+c) + 8] = 4[bc - 2(b+c) + 4] ;
we get Phil's Equation that could be a 3D surface in (a, b, c) space with lattice points crawling all over it; dozens of them.
The twenty solutions in the triangular sector a <= b <= c are given by lots of you! (Volumes given in parentheses)
 
5,13,132 (8580) - 5,14,72 (5040) - 5,15,52 (3900) - 5,16,42 (3360) - 5,17,36 (3060)
5,18,32 (2880) - 5,20,27 (2700) - 5,22,24 (2640) - 6,9,56 (3024) - 6,10,32 (1920)
6,11,24 (1584) - 6,12,20 (1440) - 6,14,16 (1344) - 7,7,100 (4900) - 7,8,30 (1680)
7,9,20 (1260) - 7,10,16 (1120) - 8,8,18 (1152) - 8,9,14 (1008) - 8,10,12 (960) ;
The last one can be made from one big box of 1000 sugar cubes ; 480 inside and 480 outside.
 
WINNERS - Problem 192 . (back to top) . leader board . . .
Nick McGrath . . . . . . . . 10 pts - Writing programs to avoid formulas is like using a sugar substitute!
Joe Alvord . . . . . . . . . . . 7 pts - The fancy counting method of faces, edges and corners works well!
Hermen Jacobs . . . . . . . . 6 pts - Bonus pt for being just minutes behind Joe. Expanding search found all.
Ed Wern . . . . . . . . . . . . . 5 pts - Correct list of 20 as well as dramatic unfolding story of their discovery.
Tim Poe . . . . . . . . . . . . . 4 pts - Thanks for the nice list and VBM (visual basic macro, not 'very bad man')
Jeremy Galvagni . . . . . . 4 pts - Nice idea: 9x9x9 cube too small for 1/2, 10x010x010 too big; limits search.
Albert Heinrich (new) . . 4 pts - Welcome! Great answer and great taste in Prealgebra textbook too %;=}
Alan O'Donell . . . . . . . . 4 pts - Yes, 7,7,100 and 8,8,18 ok under a<=b. Bonus pt for finding flaw!
S. Khoo . . . . . . . . . . . . . . 3 pts - Safe statement: "There are at least 20". Nice program to generate 'em.
Quasi-C. . . . . . . . . . . . . . 3 pts - You claimed to get 18 sol'ns but I counted 19 ; "in all" =/= "all of them".
Marcello Cammarata . . 3 pts - Impressive to do this on a hand calculator- what , no spreadsheet this time?
Phil Sayre . . . . . . . . . . . 3 pts - Yes, a = 4[bc - 2(b+c) + 4] / [bc- 4(b+c) + 8] I'd like to see that surface!
Art Morris . . . . . . . . . . . 3 pts - Our awesome all-time leader blinked and went on vacation! Good ans.
Allen Druze . . . . . . . . . . 3 pts - Included interesting steps and observations about prime factors of 140/(28-a).
Yakov Macak . . . . . . . . . 3 pts - Glad you're enjoying the problems series; apparently you're not alone!
Vince LoCascio . . . . . . . 2 pts - Early entry thanks for alertness - coulda found smore soltutions
Ajit Athle . . . . . . . . . . . . 2 pts - Got some of the many a,b,c that work in abc = 2(a-2)(b-2)(c-2).
Ken Duisenberg . . . . . . . 2 pts - You did find all the solutions that had c = b+2 for all possible a.
Mario Roederer . . . . . . . 2 pts - Good to get 90% of the solutions, thanks for "ordering in volume"
Zahi Teitelman . . . . . . . 1 pt - The equations were right; glad you got some of the solutions.
 
 
 
Problem #193 - Posted Friday, February 6, 2004
Even That's Odd ! (back to top)
In a valid base ten multiplication problem, all the
even digits were replaced by E's, and the odd digits
were written as O's, giving the result at the right.
What must the original problem have been ?
Explain answer carefully. (Time's up on this one, winners soon.)
Solution: Dan's note: I originally had the graphic as preformatted text, which looked fine in Netscape but apparently
misaligned in Internet Explorer or Windows Mozilla or whatever most of you were looking at, so I re-did it as a GIF.
Here's Vince's excellent solution:
 
First of all, I'm assuming that the numbers are inadvertently shifted and the problem should read: 
 OEE
  EE
EOEE
EOE
OOEE
Since the three digit number times the two digit number is a 4 digit
number, the hundreds digit of the 3-digit number must be 1 or 3, but since a1 will not allow a computation in line 3 greater than 2000, the first number must be 3:		 
 3EE
  EE
EOEE
EOE
OOEE
Since the units digit must produce a 4 digit result on line 3 and the
tens digit must produce a 3 digit result on line 4, the multiplier must be 28:		 
 3EE
  28
EOEE
EOE
OOEE
For 2 times 3EE to produce a number of the form EOE, the units digit of 3EE must be either 6 or 8 and for 8 times the tens digit to have and odd carry, the middle digit must be 4. A little experimentation demonstrates that a 6 in the units digit of the multiplicand does not produce a high enough value in the units digit of line 4. Therefore, 8 is needed and the full specification becomes: 
 348
x 28
2784
696 .
9744
WINNERS - Problem 193 . (back to top) . leader board . . .
Art Morris . . . . . . . . . . 10 pts - Art learned long mult in the 30's, got me beat by (at least?) 3 decades!
Vince LoCascio . . . . . . . 7 pts - Nice answer, may I quote you on that? Keep on entering!
Nick McGrath . . . . . . . . 5 pts - You jumped on the free resubmission offer and cashed in!
Rick Montgomery . . . . . 4 pts - Good answer; bonus 128 x 26 = 0768+2560 = 3328 if 0 allowed.
Ken Duisenberg . . . . . . . 4 pts - Yes, you were right about my temporary 'scattered' misalignment
S. Khoo . . . . . . . . . . . . . . 4 pts - Some of the problems that are easier to discover are hard to explain!
Zahi Teitelman . . . . . . . 4 pts - Right, 1st o is 1 or 3, so 1st e on 2nd is 2; that limits your search!
Marcello Cammarata . . 3 pts - Right, now that you can see the actual problem it's not a problem.
Jeremy Galvagni . . . . . . 3 pts - Thanks for doing a search for the EOE at right; no solution to that.
Tim Poe . . . . . . . . . . . . . 3 pts - Yes, you were missing something, namely a space at the end of row 4.
Ed Wern . . . . . . . . . . . . . 3 pts - You were right to be mystified, and not alone btw, good recovery.
Quasi-C. . . . . . . . . . . . . . 3 pts - I wasn't misaligned but apprently my problem was! Good ans resub.
Omar Gymboski . . . . . . 3 pts - Good to hear from you and nice solution to this odd one.
Jack Dostal . . . . . . . . . . 3 pts - right, 1st e=/=0 and 1st o < 5 limits search, u found the right one.
Joe Alvord . . . . . . . . . . . 3 pts - Good creative use of spreadsheet to handle the digits of the mult.
Ravi Raja . . . . . . . . . . . . 3 pts - Nice to see you enter again; good "process of elimin." answer.
Ajit Athle . . . . . . . . . . . . 3 pts - Good approach : let M1 = oee = abc etc, to simplify expl'n.
FanFan (new) . . . . . . . . . . 3 pts - Welcome, thanks for your entry. Good clear answer too; keep it up!
Hermen Jacobs . . . . . . . . 2 pts - Woulda been 3 pts but the 9744 was replaced by 9688 = 346*28
Allen Druze . . . . . . . . . . 2 pts - Good solution, I liked expl'n but no middle rows shown in solution
 
 
Problem #194 - Posted Thursday, February 26, 2004
An Uncommon Multiple ! (back to top)
Prove that for all integer values of x,
is an exact multiple of the number 8640 . . Explain answer carefully. (Time's up on this one.)
 
Solution: Notice that our polynomial f(x) is very factorable, in fact f(x) = (x-2)(x-1)^2(x)^3(x+1)^2(x+2).
This reminds me of the problem that says that the product of any n consecutive integers is divisible by n!. - Dan
 
As Nick puts it, 8640 = 2^6 * 3^3 *5 So we need to show that the given poly is divisible by 2^6(=64) 3^3(=27) and 5.
Abbreviating x^k to xk we have: x9 - 6x7 + 9x5 - 4x3 = x3(x6 - 6x4 + 9x2 -4) = x3 (x2 - 4) (x4 - 2x2 +1)
= x3 (x2 - 4) (x2 - 1)^2 = x3 (x+2) (x-2) (x+1)^2 (x-1)^2 = (x-2) * (x-1)^2 * x^3 * (x+1)^2 * (x+2)
= (x-2)(x-1)(x-1).x.x.x.(x+1)(x+1)(x+2)
5 is a factor of (x-2)(x-1)x(x+1)(x+2) since those 5 factors contain all residues mod 5
3 is a factor of (x-2)(x-1)x and of (x-1)x(x+1) and of x(x+1)(x+2) and therefore 3^3=27 is a factor of our polynomial
8 is a factor of (x-2)(x-1)x(x+1) and (x-1)x(x+1)(x+2) (because in both sets of factors we have a set of all residues mod 4
and therefore one factor is divisible by 2 and one of the others by 4) and therefore 8^2= 2^6 is a factor of the polynomial.
We have shown the polynomial is divisible by 5, 27 and 64 and therefore by 8640 for any integer x.
 
And newcomer Prachai K. from Thailand had a nice solution as well:
Since x^9 - 6x^7 + 9x^5 - 4x^3 = (x-2)((x-1)^2)(x^3)((x+1)^2)(x+2) and 8640 = (2^6)(3^3)(5), so first I assume that x is even.
x-2,x and x+2 is divisible by 2. There are five of them altogether, so five 2s divide into that big stuff. But either x-2 or x must be
divisible by 4, so that means x^9 - 6x^7 + 9x^5 - 4x^3 is divisible by at least six 2s.
The other case is that x is odd, so x-1 and x+1 must be divisible by 2 and there are four of them altogether. But again, either x-1 or
x+1 must be divisible by 4, then two more 2s appear and x^9 - 6x^7 + 9x^5 - 4x^3 is divisible by six 2s.
Now either x-2, x-1 or x must be divisible by 3. If x-2 is, then x+1 is also. There are 3 of them. If x-1 is, then x+2 is also, and there
are 3 of them. And there are 3 "x" as well. So three 3s divide into x^9 - 6x^7 + 9x^5 - 4x^3.
x-2, x-1, x, x+1 and x+2 are five straight numbers so one of them is divisible by five. x^9 - 6x^7 + 9x^5 - 4x^3 is divisible by five.
From all of that, it can be concluded that x^9 - 6x^7 + 9x^5 - 4x^3 is divisible by (2^6)(3^3)(5) = 8640.
 
WINNERS - Problem 194 . (back to top) . leader board . . .
Nick McGrath . . . . . . . . 10 pts - There's the "book" proof, as Erdos would say (see above).
Vince LoCascio . . . . . . . 7 pts - Got a nice proof in on the first try, just after Hermen's 2nd try
Hermen Jacobs . . . . . . . . 6 pts - The second submission was indeed better; second correct ans
Art Morris . . . . . . . . . . . 5 pts - I couldn't fill in all your reasoning with primes but good framework
Quasi-C. . . . . . . . . . . . . . 5 pts - This is kind of tailored to anumber theory jock, isn't it? %;-}
Joe Alvord . . . . . . . . . . . 4 pts - I like your notation x/3 has rem 0, 1, or 2, etc. Good expl, thanx.
Phillipe Fondanaiche (new) 4 pts - Bienvenue, comment va Paris? Good answer, tres bien explique'.
Zahi Teitelman . . . . . . . 3 pts - I agree that for x = 3 and 4 it's a mult of 8640 but I don't see further
Jack Dostal . . . . . . . . . . 3 pts - That's the right idea, maybe another point with a tighter proof...
Allen Druze . . . . . . . . . . 3 pts - Well explained answer and factored polys and integers, thanks.
S. Khoo . . . . . . . . . . . . . . 3 pts - Right, it might be 2^7 or 2^8 dep on x mod 4, 8640 is min.
Marcello Cammarata . . 3 pts - Early entry got an extra point but 2^6 needed not 2^5.
Carl Chenard (new) . . . . 3 pts - Welcome to my little contest; good explanation! Keep on entering.
Jeremy Galvagni . . . . . . 3 pts - You're welcome; I'll try to keep tossing these fun probs out there!
Alan O'Donnell . . . . . . . 3 pts - The induction proof idea oughta work, your diff polys were good.
Ed Wern . . . . . . . . . . . . . 3 pts - Worked hard for that fact'z'n, and it paid off 8640 times.
FanFan . . . . . . . . . . . . . . 3 pts - That's another nicely factored answer. Keep em coming!
Prachai K. . . . . . . . . (new) 3 pts - Welcome, thanks for your entry. Good clear answer too; keep it up!
Phil Sayre . . . . . . . . . . . . 3 pts - I liked the table of factors approach; 123234345, 234345456, etc.
Radu Ionescu (new) . . . . . 3 pts - Nice to have you with us; nice clear explanation too, thanks!
Ajit Athle . . . . . . . . . . . . 2 pts - Good answer and explanation; the two resubmissions cost you a point
 
 
Problem #195 - Posted Saturday, March 6, 2004
Over - Committeed ? (back to top)
Fifteen college teachers (A,B,...,O) must serve on a total of twenty committees (1-20), such that:
i) Each teacher is on exactly four committees,
ii) Each committee has three teachers on it,
iii) No two committees have more than one teacher in common.
List (or accurately describe) such a committee structure, or else prove it can't exist.
Explain answer carefully. One pt penalty for resubmissions.
 
Solution: This is a combinatorics problem, from an interesting area called "block design."
Here's new contestant Carl Chenard's solution:
"To form the first 15 committees, Let A be on the first committee, B on the second and proceed
alphabetically through ) which would be on the 15th committee. For the second member of the
committee, begin with B on the first committee, C on the second and proceed alphabetically
placing O in the14th and A in the 15th. To avoid repeat members, start with D for committee 1,
E in committee 2 and proceed alphabetically placing A in the 13th, B in the 14th, and c in the 15th.
To form the final 5 committees, Place every fifth letter in committee 15 (A,F,and K).
Same with the 16th committee (B,G,L) The final list would be:
ABD - BCE - CDF - DEG - EFH - FGI - GHJ - HIK - IJL - JKM
KLN - LMO - MNA - NOB - OAC - AFK - BGL - CHM - DIN - EJO
Further combinations of lists can be generated by starting with a different letter for the third member."

Tim Poe's table (at right) was slick; teachers 1-5 across the top and 6-10 at left, put x's if 2 teachers are on a comm; then groups of 4 x's creating the last 5 teachers and which orig 2 they share a comm with.

Dan's Note: The solution I saw used a 3D geometric pattern: Put a teacher at each corner of a cube, a teacher at the center of each face of the cube, and a teacher at the middle of the cube. Then form the committees by: the two diagonals on each face (12), the main space diags (4) and then the 6 center spots IJKLMN are only on 2 committees; make the last 4 comms by alternate triangles on this octahedron; IJK, ILM, NKL, NJM.

 
WINNERS - Problem 195 . (back to top) . leader board . . .
Carl Chenard . . . . . . . . 10 pts - Overachiever wins a round in his second attempt; film at 11. ;=}
Tim Poe . . . . . . . . . . . . . 7 pts - I like that table way of showing committees; it's 2 1/2 dimensional.
Nick McGrath . . . . . . . . 5 pts - Submitted a list; "not much math but it works"; 60 pairs occur.
Vince LoCascio . . . . . . . 5 pts - Easier to form the first 15 committees, then use 8 unused pairs.
Prachai K. . . . . . . . . . . . 4 pts - Good way to organize; first 15 into 3 grps, this simplifies it.
Phillipe Fondanaiche . . 4 pts - I like the comment 'here's one of many ways.' I wonder how many?
Omar Gymboski . . . . . . 4 pts - Welcome back; "ABC, ADE, . . . , KMN, QED" (Last not a comm.)
Ed Wern . . . . . . . . . . . . . 4 pts - Later entry earned extra point by thorough explanation of thinking
Ajit Athle . . . . . . . . . . . . 4 pts - Nice duality; list of all comms and list of what comms each tchr is on.
Art Morris . . . . . . . . . . . 3 pts - Your top down approach missed 4 of the 60 pairings, 93%.
Hermen Jacobs . . . . . . . 3 pts - Early enough entry but a few pairs shared committees; typos?
Quasi-C. . . . . . . . . . . . . . 3 pts - Table testing tells the tale; the teachers take trivial travails.
Marcello Cammarata . . 3 pts - Trial and error is a great system, especially if you catch the errors!
Rick Montgomery . . . . . 3 pts - One set of (how many I wonder) possible committees, yes.
Yakov Macak . . . . . . . . . 3 pts - Your Aussie timetable seems fine; you see the problem tomorrow!
Alan O'Donnell . . . . . . . 3 pts - Yes; I checked a random sample of 7 teachers; all on 4 comm's.
Monika Reynolds (new) . 2 pts - Thanks for e-mailing it in; all committees worked out right.
Allen Druze . . . . . . . . . . 1 pt - Good to express an opinion ; but it turned out to be possible.
 
 
Problem #196 - Posted Wednesday, March 17, 2004.
The Pentominator ! (back to top)
"Pentominoes" are objects made from five congruent squares
attached along full edges. The 12 possible shapes are as shown:
Use a simple grid of text with rows like "TTTUUFF..." (no need for attached pictures) to show how
to use all 12 to: a) Make any 6 x 10 rectangle (there are over 1000 solutions), b) Make two 6 x 5
rectangles (using all twelve pieces once!), c) Make into an 8 x 8 square, missing its four corners.
Each shape can be rotated or flipped over at will. Show your answer carefully.
 
Solution: One of my first math books as a kid was 'Polyominoes' by Solomon Golomb (not Gollum).
Many of you sent in solutions for a) that worked in b); I personally preferred ones for a) that
were 'simple' in terms of no subrectangles or fault lines all the way across. But that's just me. - Dan
Here are the solutions copied from Philippe's reponse, I mean answer: (Notice 'a' has no fault lines.) 
a) 6 x 10 rectangle 
L L N N Y Y Y Y U U 
I L T N N N Y F F U 
I L T T T X F F U U 
I L T W X X X F Z V 
I P P W W X Z Z Z V 
I P P P W W Z V V V
b) two 6 x 5 rectangles 
U U U P P Y Y Y Y I 
U X U P P L L Y Z I 
X X X N P L Z Z Z I 
V X F N N L Z W T I 
V F F F N L W W T I 
V V V F N W W T T T
c) an 8x8 square missing its 4 corners 
  N N N Y P P 
N N Z L Y P P I 
Z Z Z L Y Y P I 
Z W W L Y F F I 
W W L L F F T I 
W X U U V F T I 
X X X U V T T T 
  X U U V V V
WINNERS - Problem 196 . (back to top) . leader board . . .
Jeremy Galvagni . . . . . 10 pts - Back in the fray and moving up the charts! (Or just in it for fun.)
Fanfan . . . . . . . . . . . . . . 7 pts - Keep on watching for new problems; I'm on a semi-regular schedule!
Ed Wern . . . . . . . . . . . . 6 pts - Thanks for the link for us be able to play with these! (See above)
Phillipe Fondanaiche. . 5 pts - La 'solution proposee' etait bonne; merci pour votre participation
Tim Poe . . . . . . . . . . . . . 4 pts - Managed to (have to) open your excel doc in word! Funny but ok.
Ajit Athle . . . . . . . . . . . . 4 pts - A fan of these tessellation type problems, eh? I'll remember that.
Nick McGrath . . . . . . . . 3 pts - A 5 x 12 from 2 5 x 6 is kind of a 6 x 10 but not quite; fixed in resub.
Jack Dostal . . . . . . . . . . 3 pts - Yep, those'll work all right. Thanks for coming back for more!
Marcello Cammarata . . 3 pts - I'm sure Martin Gardner had a lot to say about this subject too!
Alan O'Donnell . . . . . . . 3 pts - You cut out a set of pentominoes; I can't find the ones from my book
Carl Chenard . . . . . . . . 3 pts - "Trial and error is much better than the error of not trying" (c)me
Monika Reynolds . . . . . 2 pts - Another good answer, all three parts ; a bit after the previous ones
S. Khoo . . . . . . . . . . . . . . 2 pts - Good to see you back after a brief absence - thanks
Allen Druze . . . . . . . . . . 1 pt - Did come up with list that make a 5x6: INLXUV.
Albert Alvarez (new) . . . . 1 pt - Welcome to my contest; good answer. (I see you already sent 197.)
 
 
 
Problem #197 - Posted Saturday, March 27, 2004
A Common Remainder (back to top)
480608 ,- 508811 , 723217. These three numbers, when divided by a certain
natural number > 1 , all yield the same remainder. What is that divisor and that remainder?
Show your reasoning carefully.
 
Solution: Brute force approach: Divide all three numbers by everything under the sun
and see which one first gives the same remainder. Not illegal, but somewhat inelegant.
Ed's mathematical approach:
"Having found the solution by brute force, I wanted to find it a bit more mathematically.
So I wrote each number as follows: 480608 = xD + C ; 508811 = yD + C ; 723217 = zD + C
Now, the difference between any two nmumbers will be a multple of D (D is our desired divisor):
723217 - 508811 = 214406 = (z-y)D ; 508811 - 480608 = 28203 = (y-x)D
Next, I just found the Greatest Common Factor by an old method I learned long ago (Newton? Euclid?):
continually subtract until zero is reached; the preceding number is the GCF:
214406 , 28203 , 16985 , 11218 , 5767 , 5451 , 316 , 79 , 0 .
(Philippe & others noted prime fac of diffs: 28203 = 3*7*17*79 ; 214406 = 2*23*59*79)
So the GCF of (z-y)D and (y-x)D is 79. Since 79 is prime, D must be 79.
A little division revealed that this works, and the remainder in each case is 51.
 
WINNERS - Problem 197 . (back to top) . leader board . . .
Marcello Cammarata . 10 pts - There were non-Excel ways, but not against the rules!
Nick McGrath . . . . . . . . 7 pts - That's using the all-powerful prime common factor.
Ed Wern. . . . . . . . . . . . . 6 pts - The spreadsheet guilt got to you and you produced a proof.
S. Khoo. . . . . . . . . . . . . . 5 pts - 'That's our divisor' all right, the 79 that appears in differences.
Tim Poe . . . . . . . . . . . . . 4 pts - Nice and factual, the 79 working; where'd it appear from?
Art Morris . . . . . . . . . . . 4 pts - MATLAB strikes again; I haven't used it much but it works.
Phillipe Fondanaiche. . 4 pts - Yes, 480608 = AN + R ; etc (B-A)N a factor of diff'ces.
Fanfan . . . . . . . . . . . . . . 4 pts - Good ans, thanks for mention'g Euclid's algorithm by name.
Carl Chenard . . . . . . . . 3 pts - Nice idea; what kind of simple loop program was it?
Hermen Jacobs . . . . . . .. 3 pts - Good to have you back; I'll answer your other q soon!
Joe Alvord . . . . . . . . . . . 3 pts - Sprsht w/cols B,C,D = rem when nos div by A, look for 0,0,0.
Radu Ionescu . . . . . . . .. 3 pts - Good equations, like Philippe's. They do the job.
Vince LoCascio. . . . . . . 3 pts - Another essential difference engine to solve the prob
Ajit Athle. . . . . . . . . . . . 3 pts - Thanks for reminding me that QED = Quite Easily Done
Mark Moyer . . . . . 2 pts - Vive les differences!
Jeremy Galvagni . .2 pts - Brute finesse good
Quasi-C. . . . . . . . . .2 pts - Hey hey NCAA!
Jack Dostal . . . . . . 2 pts - Eqns of diffces!
Ken Duisenberg . . .2 pts - Rem is the same.
Prachai K. . . . . . . . 2 pts - A good solution
Allen Druze . . . . . 2 pts - I got you twice ok
Ravi Raja . . . . . .2 pts - Welcome back RR
Phil Sayre . . . . 2 pts - Congruence vs search
Albert Alvarez . . . .2 pts - Subtract & conquer
Monika Reynolds . .2 pts - Well-explained
Drew . . . . . . . . . . . . 2 pts - Blast from 2002!
Mike Kosciuk (new) . 2 pts - Wlcme2dnsmth
Andy Oats (new) . . . . 2 pts - Fellow teacher!
Zahi Teitelman . . . . 2 pts - Good use GCD
Dennis Molter . . . . . 2 pts - Hi again right
Robert Dostal . . . . . 1 pt - Bit later but ok...
Lisa Schechner . . . . 1 pt - Also 2 pts #189
Yakov Macak . . . . . 1 pt - Better late than never
 
 
Problem #198 - Posted Friday, April 9, 2004
Odd and Different! (back to top)
 5+1  3+3  3+1+1+1  1+1+1+1+1+1
 6  5+1  4+2  3+2+1
There are four ways to make 6 as a sum of odd numbers,
and four ways to make 6 as a sum of different numbers.
a) Write each number from 1 to 10 in all possible ways (arranged as shown from
largest to smallest) as the sum of odds, sum of distincts, and sum of any.
b) Prove (for general n) that the number of sums of odd numbers giving n is the
number of sums of distinct numbers giving n, or find the first counterexample.
Show your reasoning carefully.
 
Solution: Once again I bow to the awesome Philippe F. (qui habite Paris, France):
Let A(n),B(n),C(n) the number of all possible ways of writing each number from 1 to n, respectively
as the sum of odds,the sum of distincts and the sum of any. We have the following table:
n A(n) B(n) C(n)
1 1 1 1
2 1 1 2
3 2 2 3
4 2 2 5
5 3 3 7
6 4 4 11
7 5 5 15
8 6 6 22
9 8 8 30
10 10 10 42
Hereafter the list of the possible ways: in blue with odds, red with distincts and green with any..
n=2 : 1+1 ; 2 ; 2, 1+1 . . . . n=3 : 3, 1+1+1 ; 3, 2+1 ; 3, 2+1, 1+1+1
n=4 : 3+1, 1+1+1+1 ; 4, 3+1 ; 4, 3+1, 2+2, 2+1+1, 1+1+1+1
n=5 : 5, 3+1+1, 1+1+1+1+1 ; 5, 4+1, 3+2, 5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1
n=6 : 5+1, 3+3, 3+1+1+1, 1+1+1+1+1+1 ; 6, 5+1, 4+2, 3+2+1 ;
6, 5+1, 4+2, 4+1+1, 3+3, 3+2+1, 3+1+1+1, 2+2+2, 2+2+1+1, 2+1+1+1+1, 1+1+1+1
n=7 : 7, 5+1+1, 3+3+1, 3+1+1+1+1, 1+1+1+1+1+1+1 ; 7, 6+1, 5+2, 4+3, 4+2+1
7, 6+1, 5+2, 5+1+1, 4+3, 4+2+1, 4+1+1+1, 3+3+1, 3+2+2, 3+2+1+1, 3+1+1+1+1,
2+2+2+1, 2+2+1+1+1, 2+1+1+1+1+1, 1+1+1+1+1+1+1
n=8 : 7+1, 5+3, 5+1+1+1, 3+3+1+1, 3+1+1+1+1+1, 1+1+1+1+1+1+1+1 ;
8, 7+1, 6+2, 5+3, 5+2+1, 4+3+1 ;
8, 7+1, 6+2, 6+1+1, 5+3, 5+2+1, 5+1+1+1, 4+4, 4+3+1, 4+2+2, 4+2+1+1, 4+1+1+1+1, 3+3+2,
3+3+1+1, 3+2+2+1, 3+2+1+1+1, 3+1+1+1+1+1, 2+2+2+2, 2+2+2+1+1, 2+2+1+1+1+1,
2+1+1+1+1+1+1, 1+1+1+1+1+1+1+1
n=9 : 9, 7+1+1, 5+3+1, 5+1+1+1+1, 3+3+3, 3+3+1+1+1, 3+1+1+1+1+1+1, 1+1+1+1+1+1+1+1+1
9, 8+1, 7+2, 6+3, 6+2+1, 5+4, 5+3+1, 4+3+2
9, 8+1, 7+2, 7+1+1, 6+3, 6+2+1, 6+1+1+1, 5+4, 5+3+1, 5+2+2, 5+2+1+1, 5+1+1+1+1, 4+4+1,
4+3+2, 4+3+1+1, 4+2+2+1, 4+2+1+1+1, 4+1+1+1+1+1, 3+3+3, 3+3+2+1, 3+3+1+1+1,
3+2+2+2, 3+2+2+1+1, 3+2+1+1+1+1, 3+1+1+1+1+1+1, 2+2+2+2+1, 2+2+2+1+1+1,
2+2+1+1+1+1+1, 2+1+1+1+1+1+1+1, 1+1+1+1+1+1+1+1+1
n=10 : 9+1, 7+3, 7+1+1+1, 5+5, 5+3+1+1, 5+1+1+1+1+1, 3+3+3+1, 3+3+1+1+1+1,
3+1+1+1+1+1+1+1, 1+1+1+1+1+1+1+1+1+1
10, 9+1, 8+2, 7+3, 7+2+1, 6+4, 6+3+1, 5+4+1, 5+3+2, 4+3+2+1
10, 9+1, 8+2, 8+1+1, 7+3, 7+2+1, 7+1+1+1, 6+4, 6+3+1, 6+2+2, 6+2+1+1, 6+1+1+1+1, 5+5, 5+4+1, 5+3+2,
5+3+1+1, 5+2+2+1, 5+2+1+1+1, 5+1+1+1+1+1, 4+4+2, 4+4+1+1, 4+3+3, 4+3+2+1, 4+3+1+1+1, 4+2+2+2,
4+2+2+1+1, 4+2+1+1+1+1, 4+1+1+1+1+1+1, 3+3+3+1, 3+3+2+2, 3+3+2+1+1, 3+3+1+1+1+1, 3+2+2+2+1,
3+2+2+1+1+1, 3+2+1+1+1+1+1, 3+1+1+1+1+1+1+1, 2+2+2+2+2, 2+2+2+2+1+1, 2+2+2+1+1+1+1,
2+2+1+1+1+1+1+1, 2+1+1+1+1+1+1+1+1, 1+1+1+1+1+1+1+1+1+1
 
For part b) I show Phil Sayre's (re)submitted proof:
These examples confirm Euler's Partition Theorem: The number of partitions of an integer n in which all parts are
odd equals the number of partitions of n in which all parts are distinct. The proof consists of showing a one-to-one
correspondence between the two types of partitions. The algorithm is: express the parts of a partition with distinct
(unequal) parts as sums of odd numbers times powers of 2; collect coefficients of each odd number, and add them
together; use the sum as the repeat factor for the odd number. The right hand sides are the partitions into odd parts.
7 =7x2^0 =7^1 = 7 . . . . . . . . . . . . . . . . . . (This is for n=7...)
6+1=3x2^1+1x2^0 =3^2+1 = 6 + 1 . . . . . . 5+2=5x2^0+1x2^1 =5+1^2 = 5 + 1 + 1
4+3=1x2^2+3x2^0 =3+1^4 = 3 + 1 + 1 + 1 + 1 . . . . . 4+2+1=2^2x1+2^1x1+2^0x1 = 1 + 1 + . . . + 1
WINNERS - Problem 198 . (back to top) . leader board . . .
Phillipe Fondanaiche . . 10 pts - Bien fait ; thanks for Euler's title : 'De Partitione Numerorum'
Jeremy Galvagni . . . . . . 6 pts - Lists good ; 1-1 corresp should be clear but I wasn't sure
Nick McGrath . . . . . . . . 5 pts - I'm glad to 'expose' you to the world of partitions; good job!
Hermen Jacobs . . . . . . . . 3 pts - Not everyone got part b, and you were the first. See above proof
Tim Poe . . . . . . . . . . . . . 3 pts - First ans received, I didn't get attchmt but ans seemed to imply ok
Marcello Cammarata . . 3 pts - Your lists of sums were perfec & idea of 1-1 corresp was right.
Phil Sayre . . . . . . . . . . . 3 pts - Good 2nd answer , thanks; (woulda been 4 pts on 1st try)
Ed Wern. . . . . . . . . . . . . 3 pts - The spreadsheet guilt got to you and you produced a proof.
Ravi Raja. . . . . . . . . . . . 2 pts - It's possible to get any number as the sum of odds, 3+1+1=5
Vince LoCascio. . . . . . . 2 pts - Didn't get attchmt; your lists ok up to 13; theorem says equal
Drew . . . . . . . . . . . . . . . 2 pts - Right, no neg or 0 allowed ; infinite partitions. Thanks for joke!
Allen Druze. . . . . . . . . . 2 pts - Was after the next prob wzup but extra point for great answer
Ajit Athle. . . . . . . . . . . . 1 pt - Your 'donkey labor' almost paid off; forgot 10 = 5+5 but theorem true
 
 
Problem #199 - Posted Sunday, April 18, 2004
Twice The Angle (back to top)
Many triangles have one angle double, or twice, another (e.g. angles 30,60,90).
a) Find the smallest integral triangle with one angle twice another.
b) Find general expressions that can generate infinitely many such triangles (a, b, c).
In part a), 'smallest' means smallest perimeter a+b+c. 'Integral triangles' have all three sides integers.
Show reasoning carefully.
 
Solution: You guys got a lot of mileage from the Laws of Sines and Cosines ;
Angles A,B,C are A, 2A, 180-3A ; sin(180-x) = sin x ; so a/sinA = b/sin2A = c/sin3A
gives b/a = sin(2A) / sinA = 2 cosA and c/a = sin(3A)/sinA = 4 cos^2(A) - 1
leading to b^2 = a(a + c), which generates (all the) solutions. Note b > a so search
b=2, 3, etc. get b=6, a=4, c=5 so (4, 5, 6) is the smallest triangle, perim = 15.
Fanfan managed to relate the sides (a, b, c) to a right tri : (2b, c, 2a+c) = (v2,uv,u2-v2)
Arthur found the oft-overlooked "inverse half angle cosine rule" (see below), while Radu
decided to bisect the 2A and get simila triangles (my NY accent hanging on); c/a = (a+d)/c etc.
Most of you found the smallest solution (4, 5, 6), then (7, 9, 12) and (9, 15, 16), . . .
 
WINNERS - Problem 199 . (back to top) . leader board . . .
Marcello Cammarata . . 10 pts - Right; Law-O-Sines and the restriction cos x > 1/2 gives it.
Art Morris . . . . . . . . . . . 7 pts - Interesting: arcos x = 2 arcos(sqrt[(1+x)/x]) does the trick!
Tim Poe . . . . . . . . . . . . . . 5 pts - Good job locating the resource for the solution; 2-param family.
Nick McGrath . . . . . . . . 4 pts - Nice use of the triple-angle formula courtesy of deMoivre.
Phillipe Fondanaiche . . 4 pts - Good 1-param fam. Oui, il se peut repondre en francais, s'il te plait!
Fanfan . . . . . . . . . . . . . . . 4 pts - Nice conversion to Pythag triple and familiar formulas
Drew . . . . . . . . . . . . . . . . 3 pts - Thanks for stream-of-consciousness ans. and using my lessons
Ed Wern. . . . . . . . . . . . . 3 pts - First ans got to me; nice algebra, good list of all perim < 200
Quasi-C. . . . . . . . . . . . . . 3 pts - Trig leads to b^2 = a(a+c); incr b from 1 until sol'n at b=6.
Hermen Jacobs . . . . . . . . 3 pts - Smallest is 4, 6, 5; yes, & b2 - a2 = ac is a good way to get 'em.
Jeremy Galvagni . . . . . . 3 pts - Did a TI-83 program to test law o cosines; got 4,5,6 & 7,9,12
Prachai K. . . . . . . . . . . . 3 pts - Good answer; you didn't need any hints! Yes, a+b>c, etc.
Vince LoCascio . . . . . . . 2 pts - Did a search, finding angles & doubles with rational cosines
Ajit Athle . . . . . . . . . . . . 2 pts - 16, 28, 33 perim 77 works, but you had the 4, 5, not the 6.
Zahi Teitelman . . . . . . . 2 pts - Yes, A, 2A, 180-3A; c = (b2-a2)/a ; 4,6,5 or 4x, 6x, 5x.
Albert Alvarez . . . . . . . 2 pts - Good generator b^2 = c(a+c) gives 16, 28, 33 (maybe 4,6,5?)
Radu Ionescu . . . . . . . . 2 pts - Interesting to bisect one angle and get a similar triangle!
Phil Sayre . . . . . . . . . . . 2 pts - Managed to use both Law of Sines and Heron's area formula
Ravi Raja. . . . . . . . . . . . 1 pt - a=2RsinA, b=2RsinB, c=2RsinC where R is the circumradius!
Allen Druze. . . . . . . . . . 1 pt - Yep, (4,5,6) is the smallest 'all-around' triangle (perimeter joke)
 
 
Problem #200 - Posted Friday, April 30, 2004
Run With The Dogs (back to top)
The writer Jack London had to go from Skagway to a remote camp in a sled
pulled by 5 husky dogs. For 24 hours the dogs pulled the sled at full speed.
Then 2 dogs ran off with a pack of wolves, and the sled was slowed down
proportionally. Jack reached camp 48 hours later than he planned. "If those
2 huskies had held on for 50 more miles, I would have been only half as late."
How far was the camp from Skagway and how long did it take Jack to get there?
Show reasoning carefully.
 
Solution: Speaking of mileage (cheap pun; see solution to 199); a good way to approach
this one is to figure out how many miles one dog could pull the sled in an hour or a day...
Here's this year's leader Nick McG's solution:
"Congratulations on reaching 200! . . . Let speed of sled (with 5 dogs) = v;   let distance = d
We are told that if the dogs had stayed for another 50 miles, he would have arrived 24 hrs earlier.
So 24 hours were lost in those 50 miles or: 50/v = 50/(3v/5) ­ 24 => 24v = 100/3 => v = 25/18.
In the first 24 hours he travels 24v = 24*25/18 = 100/3 miles
In the remaining d ­ 100/3 miles he loses 48 hours by traveling at 3v/5 instead of v or:
(d ­ 100/3)/ v = (d ­ 100/3) / (3v/5) - 48; subst for v and rearranging: 
d ­ 100/3 = 5D/3 ­ 500/9 ­ 48*25/18  =>   d = 400/3 . . . Time taken (with 5 dogs) would have
been d / v = 400/3 * 18/25 = 96 hours. Actual time = 96 + 48 = 144.
So, the total distance is 400/3 = 133 1/3 miles  and time taken = 144 hours." (6 days)
(Drew doubts the speed would really be proportional, and I agree - Dan.)
WINNERS - Problem 200 . (back to top) . leader board . . . (some later correct answers just 2 pts)
Ed Wern. . . . . . . . . . . . . 10 pts - Hey is this the first first place for you? Good eqns and ans.
Nick McGrath . . . . . . . . 7 pts - Nice concise solution; I think I'll copy & paste & not type!
Tim Poe . . . . . . . . . . . . . . 6 pts - Asks what's the conversion 1 dogpower <--> 1 horsepower?
Ken Duisenberg . . . . . . . 5 pts - Welcome back Ken. 133.333 is pretty close but shorter!
Marcello Cammarata . . 4 pts - Breaks time up into T1 + T2 after loss of dogs.
Quasi-C. . . . . . . . . . . . . . 4 pts - Good series of equations, no number theory this time!
Drew . . . . . . . . . . . . . . . . 4 pts - Yes, there's sled and human resistance so prob. not prop.
Art Morris . . . . . . . . . . . 3 pts - Our all-time leader blinked for a while this time but recovered!
Vince LoCascio . . . . . . . 3 pts - True, I didn't say Jack planned to use all the dogs all the time.
Hermen Jacobs. . . . . . . . 3 pts - Not a lot of snow in Holland but good answer anyway!
Jack Dostal . . . . . . . . . . 3 pts - Right; 5 dogs woulda been 96 hours, deserters slowed it 48.
Phil Sayre . . . . . . . . . . . 3 pts - Good system of equations with nice series of steps, thanks!
Ajit Athle. . . . . . . . . . . . 3 pts - Says Canadian Huskies can go 70 mi/day; Alaskans are wimps?
Albert Alvarez . . . . . . . 3 pts - Being 'problematic' is a good thing with this website!
Jeremy Galvagni . . . . . . 2 pts - Mileage a bit off, but 'woulda' time of 96 hrs was right
Robert Dostal . . . . . . . . 2 pts - That's three from one family (counting Katherine); cool!
Phillipe Fondanaiche . . 2 pts - T'as raison: V = 25/18 mph and v = 5/6 mph works.
Denis Borris. . . . . . . . . . 2 pts - Good to see you again, or at least read your typing!
Prachai K. . . . . . . . . . . . 2 pts - Breaks down as above; 1 dog = k mi/day; then 3k and 5k...
Yakov Macak . . . . . . . . 2 pts - Your time functions seemed to do the trick I believe.
Radu Ionescu . . . . . . . . 1 pt - Yes 96 hrs but incorrect mileage and resub; keep entering!
Ravi Raja. . . . . . . . . . . . 1 pt - Only one answer to this one; 'half as late' = as he woulda been
Allen Druze. . . . . . . . . . 1 pt - The time was 6 days but the mileage was 4/10 of yours
(S.G.) Roy (new). . . . . . . . .1 pt - Welcome to my contest! Later entry, dist ok, time not.
 
 THANKS to all of you who have entered, or even just clicked and looked.
My site is in its seventh season - OVER 52,000 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 - 2004 A.D.
 
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