dan's problem archives 151-160

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



Problem Archives page 16
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

151 Friends, Wives, $
152 Nearly Isosceles?
153- Sharon & Karen
154 - Swim 2D Boat !
155 - Can You Digit ?
156 - This Old House
157- Chop,Skip,Chop
158- Digits in the Bag
159 Money in theHole
160 - Cubes Come 4th



Problem #151 - Posted Wednesday, August 14, 2002
Friends, Wives, and Money (back to top)
Lucky Jim won $1,000,000 (a million dollars), split it up and gave it all to three male friends
and their wives. The wives (together) received just $4,000 short of $400,000. Jane got
$10,000 more than Catherine, and Marcy got that same amount more than Jane.
John Green was given as much as his wife, Henry Brown got half again as much as his wife,
and Tom Cobalt received twice as much as his. What was the first name of each man's wife,
and how much money did each of the six receive?

Solution: It was pretty straightforward to figure the wives' shares; a couple of you avoided checking all six
spousal pairings by a clever odd/even observation. Here's Quasi-C's approach...
"We can determine the wives' amounts C+M+J=396 (thousand); J=C+10 :M=J+10
So C+(C+10)+(C+10+10)=396 or 3*C = 366 C=122, J=132, M=142.
At this point, the problem is very finite. The husbands need to add up to 604 (thou).
Rather than just doing cases, let's observe that theoretically pairing any guy with any wife gives an even amount
except Henry. The only way to take 1.5*(Wife amount) to get an even number is with Jane. So Henry must be
married to Jane. Henry gets 198. So either 2*122 + 142 = 406 or 122 + 2* 142 =406. It is the latter.
So Tom Cobalt ($284.000) is married to Marcy ($142,000), John Green($122,000) is married to Catherine
($122,000), and Henry Brown ($198,000) is married to Jane ($132,000)."

WINNERS - Problem 151 . (back to top) . leader board

Hermen Jacobs . . . . . 10 pts - First flawless answer; not a native English speaker; not a problem!
Arthur Morris . . . . . . 8 pts - First answer, correct paris and amounts, some mis-steps & a bit unclear
Tim Poe . . . . . . . . . . . 7 pts - Unique "excess" method, only one combo had $55000 over average.
Quasi-C . . . . . . . . . . . 6 pts - Parity in odd/even thousands if not in payoffs! Savings 67% of work
Steve Lawrie . . . . . . . 5 pts - Good answer and I like your Huxley quote. Aldous' little brother?
Ed Wern . . . . . . . . . . 4 pts - Yep, safe to assume there are no tax implications. Good odd/even.
Jack Dostal . . . . . . . . 4 pts - You win the prize for the most equations used (eleven); clear approach!
Ludwig Deruyck. . . . 3 pts - Sure, you might as well let EXCEL check all six possibilities!
Sue B . . . . . . . . . . . . . 3 pts - Returning from sabbatical, eh? Welcome back! 'Stupid' Jim, eh? ;-}
Joe Alvord. . . . . . . . . 3 pts - Yes, 6 matchups. "Some give 1/3 cent but defacing money is a fed offense!"
Les Billig . . . . . . . . . 3 pts - Welcome back, after a 49-problem rest! Show us your trials or errors!
AZ Runner . . . . . . . . 3 pts - Good strategy; 604 >> 396 so double largest wife's share (you know)
Drew . . . . . . . . . . . . . 2 pts - Your checking was thorough, but there really is a consistent sol'n.
Nick McGrath. . . . . . 2 pts - Good answer; had correct shares; but you mean 396 and not 496...
Ji Zheng . . . . . . . . . . 2 pts - Nice try; wives ok but there is a combination that works.
Paul Botham . . . . . . . 2 pts - Got right women's shares and spice (spouses), no men's amounts.
Mohamed Omar (new) 2 pts - Welcome to dansmath! Fine method but 132000 - 10000 not 1000
Michelle Lee . . . . . . . 2 pts - Another returning fan! How are things at 'real' Berkeley? ;=}
Phil Sayre . . . . . . . . . 2 pts - Got answer in under the wire. Catherine is still chums with others!
Nikita Kuznetsov . . . 1 pt - Got women's amounts right; you may have done twice and 3/2 backwards...



Problem #152 - Posted Sunday, September 1, 2002 Nearly Isosceles, Right? (back to top) Last problem of the 2001-02 contest ! New season starts with Prob 153! The most famous Pythagorean Triangle (a right triangle with all integer sides) is the < 3 , 4 , 5 > where 3^2 + 4^2 = 5^2. The 'legs', a=3 and b=4, are consecutive integers, making a nearly-isosceles right triangle. What are the next four smallest integral right triangles with consecutive legs?

Solution: (From Nick McGrath) "We have a^2 + b^2 = c^2; we are seeking solutions where b=a+1.
So, a^2 + (a+1)^2 = c^2 => 2a^2 +2a + 1 - c^2 = 0. Using the standard formula for solving a quadratic:
a = (1/4)*( -2 +or- sqrt(2^2 - 4*2*(1-c^2))) => a= (1/2)*(-1 + sqrt(2*c^2 - 1)) (eqn 1) [we only need pos sqrt]
For a solution in integers sqrt(2*c^2-1) is an integer, k say. Then k=sqrt(2*c^2-1) => k^2 - 2*c^2 = -1.
This is a well known Pell equation which can be solved using continued fractions for sqrt(2). (See my note - Dan)
Or at least, if we discover the first solution we can generate all others. First few integer solutions for (c,k) are: (1,1);(5,7);(29,41);(169,239);(985,1393);(5741,8119);(33461,47321);(195025,475807);(1136689,1607521);(6625109,9369319).
For the current problem we need only the first six solutions(the first being trivial and the second given in the question)
Substituting the next four results into (eqn 1) gives:(c,a) = (29,20);(169,119);(985,696) and (5741,4059).
So, the required Pythag. triangles are: (a,b,c) = (20,21,29); (119,120,169); (696,697,985) and (4059,4060,5741)."

Dan's note: Many of you knew that {a = m^2 - n^2, b = 2mn, c = m^2 + n^2} forms a Pythag triple (a,b,c).
You can generate the (a,b,c) by using m=2, n=1 ==> (3,4,5) ; then m=prev c = 5, n= prev m = 2 ==> (21,20,29), etc.
These give (as Art, Nikita, Sue noticed) a ratio of successive perimeters approaching 5.828 = 3 + 2 sqrt[2].
This helps explain Nick's reference to 'continued fraction solution of Pell's equation x^2 - d y^2 = +/- 1.
Special thanks to Ludwig who gave a formula for generating all (m,n,a,b,c) from a single parameter!
. . . He included reference to Robert Simms' page www.math.clemson.edu/~rsimms (see formulas below)
And as Phil remarked, "... I thought I knew all there was to know about Pythag. triangles!"

WINNERS - Problem 152 . (back to top) . leader board

Steve Lawrie . . . . . . . 10 pts - Brute force method 'did the job'; what if you limited it to sqrt = integer...
Hermen Jacobs . . . . . 7 pts - BASIC lives on! You got all four; no penalty for near-ans (29978,29979, 42396).
Lisa Schechner . . . . . 5 pts - Early (3rd) entry might have gotten bonus point with fuller explanation
Nick McGrath. . . . . . 5 pts - Very quotable; I edited your solution just a bit; thanks for the Pell reference!
Ji Zheng . . . . . . . . . . 4 pts - Used Python to squeeze out the solutions where sqrt[a^2+b^2] = Int[it]
Joe Alvord. . . . . . . . . 4 pts - Worked his 'spreadsheets to the bone' for this one! Sure worked!
Phil Sayre . . . . . . . . . 4 pts - Never too late to learn more trix! Another Python user too.
Ludwig Deruyck. . . . 4 pts - Bonus pt: k=1,2,3,...; m=floor[(3+sqrt[8k-7])/2], n=k-(m^2-3m+2)/2
Tim Poe . . . . . . . . . . . 3 pts - Using a 'minor macro' from Visual Basic... is that an oxymoron?
Quasi-C . . . . . . . . . . . 3 pts - Nice approach: a sqrt[2] < c < a sqrt[2] + 1 and have Excel crank it out!
Sue B . . . . . . . . . . . . . 3 pts - Pulling out the Java; that's where you've 'bean'! Ratio appr. 3 + 2 sqrt[2]
Jack Dostal . . . . . . . . 3 pts - Iterative C++ program that picks out the integer values of sqrt[a^2 + (a+1)^2]
Paul Botham . . . . . . . 3 pts - Noticed that n^2+n even --> c odd --> n or n+1 is mult of 4. Ok!
Arthur Morris . . . . . . 2 pts - Legs were to be consecutive, not long leg and hypotenuse; related family!
Nikita Kuznetsov . . . 2 pts - Glad you 'relaxed' over this programming exer; good use of TI-82!
Les Billig . . . . . . . . . . 2 pts - First 3 tri's ok; near-sol'ns (3074,3075,4348) and (15611,15612,22078)
Zahi Yeitelman . . . . . 1 pt - Same thing as Art; your long leg was 1 less than your hypotenuse.
Jon Stearn . . . . . . . . .1 pt - A returning fan! Method seems ok but your OLE didn't Link or Embed.
AZ Runner . . . . . . . . .1 pt - Another hyp = leg + 1 entry; (3,4,5), (7,24,25), (9,40,41) etc. do work.
Th. Sarojkumar (new) 1 pt - Welcome to my contest, 'right' triangles! Now you'll be on two years' lists.


Problem #153 - Posted Wednesday, September 11, 2002
Sharon & Karen . . . for each other (back to top)
First problem of the 2002-03 contest! : : This begins my SIXTH season!!
Two algebra students decide to save time on their homework by sharin' the work equally.
But after a while Karen has only done three-fifths of the problems that Sharon has left,
which in turn is four-sevenths of the amount that Sharon has done. How much faster
must Karen work than Sharon,if they're carin' to finish simultaneously?

Solution: Dan's note: I tried to make the wording clear, but as Phil said, "This illustrates the difficulty
translating language into math." There were two interpretations of what the "which" (in turn) referred to...
I gave a bonus point (or rounded up) to those of you that pointed out the ambiguity and gave both ways,
and didn't take off or count as wrong the interpr'n I list below as #2; There were more points awarded this week.

1. (From Jon Stearn) Assume sharing work equally means each does the same number of problems (w)
k is the number of problems Karen has completed ; s is the number of problems Sharon has completed
To finish at the same time, Sharon must work (w-s) problems in the same time that Karen works (w-k).
The ratio of (w-k)/(w-s) will tell us how much faster Karen must work. What we know: k = (3/5)(w-s) ;
(w-s) = (4/7)s . . . substituting for (w-s) in the first equation -> k = (3/5)*(4/7)s = (12/35)s ;
solving for w in the second equation -> w = (11/7)s . . . use these in our solution ratio (w-k)/(w-s) =
((11/7)s - (12/35)s) / ((4/7)s) = ((55/35) - (12/35))/(20/35) = (55-12) / 20 = 43/20 = 2.15 times faster!

2. (From Anirban B.) Let there be a total 2n problems and let Sharon and Karen indiv. solve n problems.
Let Sharon solve x fraction of the problems. Then by the problem, 3/5*n*(1-x) = 4/7*n*x (Aha! - Dan) Solving for x we get x=21/41. Hence Sharon has solved 21/41 fraction of the problems and is left with
20/41 fraction of her problems. Now Karen has completed 3/5*(1-21/41)=12/41 fraction of the problems
and is left with 1-12/41 = 29/41 part of the problems. Let now Sharon solve at the rate of a problems per
unit time and Karen at the rate of b problems per unit time. As Sharon and Karen are to finish together,
we get 20*n/(41*a)=29*n/(41*b) so b/a=29/20. Karen must work 29/20 times as fast as Sharon.

WINNERS - Problem 153 . (back to top) . leader board

Arthur Morris . . . . . 10 pts - Asked if 'which' was 'Karen has done' or 'Sharon has left'; did both.
Anirban Bhattacharyya . 7 pts - Did 2nd interpr. (29/20), good explanation & formulas listed above
Tim Poe . . . . . . . . . . . 6 pts - That's the way I thought of the problem, 43/20 as fast for the rest.
Joe Alvord. . . . . . . . . 5 pts - Good, divide the work into 41 equal parts for each worker; 43/20 club.
Jon Stearn . . . . . . . . . 5 pts - Thanks for carin' to let me be sharin your answer, movin up the charts!
Jack Dostal . . . . . . . . 4 pts - I liked your descriptive variables SR, SW, KR, KW ; KR/SR = 43/20.
Nikita Kuznetsov . . . 4 pts - Our Cornell freshman pointed out the uncertainty & did both right.
Uwe Buenting . . . . . . 3 pts - Early entry, got 29/20 but then found how much Sharon should speed up.
Nick McGrath. . . . . . 3 pts - The 'after a while' was the 'before' part; you got the 29/20 ans.
Charles Lo (new) . . . . 3 pts - Welcome! As you said, K = (3/5)(P - S) = (4/7)S gives 20/41 vs. 29/41.
Phil Sayre . . . . . . . . . 3 pts - My ambiguity, not language in general! 43/20 OK, not sure abt 35/20.
Zahi Yeitelman . . . . 3 pts - S did x of 2z, K did 4x/7, yes. This brings you to 29/20 like Charles.
Les Billig . . . . . . . . . 3 pts - A fellow Maths teacher, must have read the 'which' as I did; 43/20.
Ed Wern . . . . . . . . . . 3 pts - Thanks for Darin' to try this (hehe) good eqns; N/2 = 11/7 S etc.
Th. Sarojkumar . . . . 3 pts - Good setup, equations led you to an answer of 29/20 like others.
Ji Zheng . . . . . . . . . . 2 pts - Your amount 29/20 compared their speeds; 253.75% was Sh's speedup.
Sue B . . . . . . . . . . . . . 2 pts - Good start; you lost me at Sharon's negative rate; does she erase probs?
Hermen Jacobs . . . . . 2 pts - Nice, your 6/55 and 7/22 were equal to mine but subtr from 1/2 not 1.
Quasi-C . . . . . . . . . . . 2 pts - Agree with KT = (3/5)(W/2 - ST) but 187/100 might come from alg miscue.
Allen Druze . . . . . . . .2 pts - Welcome back! Just under the wire; get 20/29 reverses to 1.45 ok.
Renee Prasad (new) . . .1 pt - I understand your first eqn, although 9/41 is a bit off by subtracting.



Problem #154 - Posted Saturday, September 21, 2002 Swim 2D Boat ! (back to top) You (A) see a boat (B) tied up across the river and want to reach it in a straight line. The total distance straight across to C & then downstream to the boat is 1/5 mile longer than the half mile directly from A to B, and AC > CB. The current is 2 mph and in still water you swim at 3 mph. a) At what angle (to the nearest degree, from your edge of the river) should you point so you drift and exactly reach the boat? b) How long (in min. and sec.) will it take you to swim there? Please explain your reasoning. current : 2 miles per hour

Solution: I decided to put up my own solution this time; there were lots of excellent ones submitted too!
My original statement AC > AB was clearly impossible; Tim Poe tried to explain it by saying the river might
not be a constant width or the flow might not be constant, I fixed the problem after noticing the misstatement.
Let b = AC, a = BC, and c = AB (standard triangle notation). We are given c = 0.5 and a + b = 0.5 + 0.2 = 0.7.
Then a^2 + b^2 = c^2 = (0.5)^2 = 0.25; substitute b = 0.7 - a and get b = 0.4 or 0.3 (quadratic equation).
But AC > CB so b > a and so a = 0.3, b = 0.4. Now we aim at a point D between C and B . We set up a vector
sum AD + DB = AB as vectors; the lengths are proportional to the swim speed of 3 mph and the current flow
of 2 mph |AD| = 3x and |DB| = 2x. Then |CD| = 0.3 - 2x ; by the Pythag thm (0.4)^2 + (0.3 - 2x)^2 = (3x)^2.
This leads to 4 x^2 - 1.2 x - 0.25 = 9 x^2 ; 100 x^2 + 24 x - 5 = 0 ; x = (-6 +/- sqrt[161])/50 ; so x can
be -0.3737 or 0.133772 (the latter makes sense). Then angle CDA has cosine = to (0.3 - 2x)/(3x) = 0.080874
so the angle we seek is 85.3611 degrees or approx. 85 degrees from the downstream direction.
Time to swim is dist / rate = (3x mi)/(3 mph) = x hours = 0.133772 hrs = 8 min 1.5776 sec ~ 8 min 2 sec.
WINNERS - Problem 154 . (back to top) . leader board (I was more generous with pts for partially correct ans.)

Nick McGrath. . . . . 10 pts - Good use of exact quadratic formula and solving clearly for AC, etc.
Les Billig . . . . . . . . . 6 pts - Nice outline of a solution; I shaved off a point for the shaved answer.
Joe Alvord . . . . . . . . 6 pts - Bonus point for explaining both solutions including AB > AC.
Phil Sayre. . . . . . . . . 4 pts - Right, the 'problem had a problem'; thanks for both answers to cover it.
Hermen Jacobs . . . . 3 pts - Yes I had a mistake; early entry got an extra point but ans a bit off.
Arthur Morris . . . . . 3 pts - Not sure how tan of angle is 2.5; time was too long for swim
Tim Poe. . . . . . . . . . . 3 pts - Extra point for great scramble to explain AC > AB but resub still off
Jack Dostal. . . . . . . . 3 pts - The 8 min to swim 0.4 mi was assuming headed straight for C...
Nikita Kuznetsov. . . 3 pts - Yes, 2 + 3 cos x = (CB/AB) ; x = 85.36, but time was too long.
Quasi-C. . . . . . . . . . . 3 pts - Aiming upstream won't reach the boat but your neg angle worked.
Anirban Bhattacharyya . 2 pts - The angle of 48 degrees didn't work, but using components was a good idea
Troy Butler (new) . . . 2 pts - Welcome to the contest; glad you re-entered; interesting use of imag nos.
Yonlawan Chirawatcharadej (new) 2 pts - Good try; you don't need to go to C first. Welcome to ProbOfWk!
Zahi Yeitelman . . . . 2 pts - Nice partial answer, another 48 degree vote gave too slow a time...
Allen Druze . . . . . . . 1 pt - Later entry ; good try ; proportions help but don't tell the whole story
Uwe Buenting. . . . . . 1 pt - Thanks for the C++ code; you weren't late if the next prob isn't up.


Problem #155 - Posted Monday, October 7, 2002
Can You Digit? (A few puzzles about numbers with special digit properties) (back to top)
a) Find two three-digit numbers, not containing zero, whose squares end in the same 3 digits
(as the number, in the same order). (One-digit example: 5^2 = 25; both end in 5.)
b) Find two pairs of consecutive three-digit numbers whose squares have the same digits
(for each pair). (Two-digit example: 13^2 = 169, 14^2 = 196, same digits.)
c) Find a set of six non-zero 1-digit numbers such that the sum of three of them equals the sum
of the others, and the sum of the squares (of the same three) is the sum of the squares of the others.


Solution: (Dan's Note: What could be simpler than this first program? Besides omitting line 10, 'clear screen!')

a) (From Hermen Jacobs): "Hi Dan, Here you have my answers of #155. I found these with basic-programs:

10 cls ::: 20 for i=111 to 999: a=i*i: b=a-i: c=b/1000 ::: 30 if c=int(c) then print i, a ::: 40 next i "
. . . 376^2=141376 625^2 = 390625
b) (From Phil Sayre): "I wrote a slightly more complicated program, which (in English) did this:
. . . . << for i from 100 to 999 ::: n=i*i, m=(i+1)*(i+1) ::: make strings of n and m ::: convert the strings to

lists of characters ::: sort the lists ::: if the lists are the same length ::: check that the lists are identical :::
if so, got a solution >> . . I found these pairs: 157 and 158 ; 913 and 914."

c) (From Ed Wern) "With the assumption that all digits were distinct, I tried to use Excel for this case...

I figured there would be 4200 possibilities to check (10!/3!7! * 7!/4!3! = 4200) ... and that there
would need to be a couple near the middle of the range (5, 6), with one high number (8 or 9), on
one side, and a couple relatively high numbers (7,8) and a small number (2, 3, or 4) on the other.
I started with 6, 5, 9 vs 8 7 x, then 'saw' that I should change to 4, 5, 9 vs 3, 7, 8.
Upon checking: 4 + 5 + 9 = 18 = 3 + 7 + 8. Also, 16 + 25 + 81 = 122 = 9 + 49 + 64.
Only saving grace - I did NOT do the math for this problem in Excel, just paper and brain....
Dan's Note: Other triplet pairs were 1, 6, 8 vs 2, 4, 9 ; 1, 5, 6 vs 2, 3, 7 ; and 2, 6, 7 vs 3, 4, 8.


WINNERS - Problem 155 . (back to top) . leader board (I was more generous with pts for partially correct ans.)

Nick McGrath. . . . . 10 pts - Had "no time for logic, brain power or math ability" so used Excel!
Uwe Buenting. . . . . . 7 pts - Good response; included C++ code; 'trivial sol'ns' 1,2,3 & 1,2,3 don't count.
Jon Stearn . . . . . . . . 6 pts - The first person to get all four answers to c); I awarded you a bonus point!
Tim Poe. . . . . . . . . . . 5 pts - Also got all 4 c)'s, with visual basic macro (.vbm). Sets must have 6 diff elems.
Ed Wern . . . . . . . . . . 4 pts - 'Excel junkie' shows some good algebra: x^2 = 1000y + x ; x = ... from q.f.
Jack Dostal. . . . . . . . 4 pts - It's good 2B 'iterative,' especially in programming! Got all 4 to c); sums, ss?
Phil Sayre. . . . . . . . . 4 pts - Got three of the c)'s; I liked your semi-translation into 'program-glish.'
Hermen Jacobs . . . . 3 pts - Got a,b,c); didn't charge resub. pt. for iffy answers '1+7+7 = 3+3+9 ; sqs add too.'
Sue B. . . . . . . . . . . . . 3 pts - Yes, all of em! B might b for bean, w/your reliance on Java (shouldn't we all b!)
Zahi Yeitelman . . . . 3 pts - a), b) good. I like your reasoning of odd/even in part c). There were others too
Joe Alvord . . . . . . . . 3 pts - Nice use of spreadsheet: col 1 = n, col2 = n^2, col 3 = n^2 - n, scan col 3 for 000's.
Troy Butler . . . . . . . 3 pts - Bonus pt. for a): 2 dig: 25, 76; 3-dig: 376, 625; 4-dig: 9376; 5-and 6-? **
Drew . . . . . . . . . . . . . 3 pts - Good a), c); no part b); very early entry moved you up (or less down)
Arthur Morris . . . . . 2 pts - You found cubes with the same last 3 digits: 125 and 249; see my slogan *
Quasi-C. . . . . . . . . . . 2 pts - Nice answer on a and c; u mizundaztood b) but it's cool that 486 and 513 share 6.
Anirban Bhattacharyya . 2 pts - Getting one on c) was ok, but wanted two pairs for part b). Nice logic on a).
Les Billig . . . . . . . . . 2 pts - Again, one on b and none on c) but early enough to get 2 pts. Go(,) teach!
Allen Druze . . . . . . . 2 pts - Good step: (ABC)^2 = 1000k + ABC ; ABC(ABC - 1) = 1000k ; etc.
Nikita Kuznetsov . . . 1 pt - A bit later & without b) but busy as ur I'm happy to hear from you at all!
Th. Sarojkumar . . . . 1 pt - You got the right answers, good! Submitted after 156 was up but before this ans!

* Dan's Slogan: There are very few 'wrong answers'; mostly they're correct answers to other questions.
** Bonus: 1 pt each to the first two non-Troys that send me the answers to the 5-and 6-digit versions!




Problem #156 - Posted Saturday, October 19, 2002
This Old House! (back to top)
"Wow, cool house!" my friend said one day. "How old is it?" "Well, my dad was born in it,
and the house was fifteen years old then. And the funny thing is, if you square the house's age,
the first half is my dad's age and the second half is my age!" How old is the house, how old is
my dad, and how old am I? Warning: Style points for Solving, not just Searching!


Solution: (From returning solver Dennis Borris) Analysis :
H = House's age, F = Father's age, M = My age ; H^2 is a 4-digit number (has 1st half and 2nd half)
H = F + 15, so F = H - 15 [1] And of course F - M > 0 ! H^2 = 100F + M
[1] H^2 = 100(H - 15) ; H^2 - 100H - M + 1500 = 0 ; Quadraticize: H = 50 +- sqrt(1000 + M)
H = 82 or 18, 83 or 17 So H = 82, F = 82 - 15 = 67, M = 24 (82^2 = 6724) 17, 18 and 83 all out-of-range.
(Dan's note: You other contestants can fill this in slightly; this was pretty concise but effective!)


WINNERS - Problem 156 . (back to top) . leader board

Anirban Bhattacharyya 10 pts - Yes; first entry and nice job, 10x + y + 15 > 32 and note x = 6, y = 7.
Joe Alvord . . . . . . . . 7 pts - Good work; (x+15)^2 = x^2 + 30x + 225 is the winning ingredient!
Hermen Jacobs . . . . 6 pts - Slick algebra showing 35^2 - 225 + m is square; 68 dad, 89 son not possible
Denis Borris . . . . . 5 pts - Canadian algebra works just like down here! Welcome back to my contest!
Arthur Morris . . . . . 4 pts - Inspector Morris, you rounded up the suspects, and I don't mean decimally.
Nick McGrath. . . . . 4 pts - Assume Dad is 17 (proved from house > 32 bec 4 dig sq); deduce the rest!
Chris Pentacoff . . . . 4 pts - Strong showing from a long absence... welcome back, howya doin'!
Tim Poe. . . . . . . . . . . 4 pts - Also good approx ; (H - 50)^2 = 1000 ; H ~ 82 seems good, zero in!
Uwe Buenting. . . . . . 4 pts - Bonus pt for noticing H = D+16 is possible, depending on birthdays!
Quasi-C. . . . . . . . . . . 3 pts - Did the trick with pure reasoning mixed with a nice pinch of algebra
Jack Dostal. . . . . . . . 3 pts - D^2 - 70D + 225 = 0 has D = 66.666... but m isn't 0; goood approx!
Jon Stearn . . . . . . . . . 3 pts - Nice; (d-a)(d-b)=0 ; a+b=70 & a b = 225 - m ; a = 67 , b = 3 is right
Ed Wern. . . . . . . . . . . . 3 pts - You remembered the quadratic formula two weeks in a row & it pays!
Yonlawan Chirawatcharadej 3 pts - I agree D < M would mean eliminate the case H = 83 so 82^2 = 6724
Troy Butler . . . . . . . . . . 3 pts - Nice thought path: 69+15 = 84; 84^2 = 7056 ; close! Try 67, check!
Ritwik Chaudhuri . . 3 pts - Good to see you enter again after a long break; hope you had fun!
Phil Sayre. . . . . . . . . . 3 pts - Right you are; r = sqrt[1000+m] has to be an integer; that;'s the key.
Charles Lo . . . . . . . . 3 pts - I agree the house could be 15 to 99 but has to be over 32 to have 4 dig sq
Zahi Yeitelman . . . . 2 pts - The only solution is 82^2 = 6724 because you want the son to be younger!
Ahearn Bizmah (new) 2 pts -Hi Ahearn, Welcome to my contest; good use of algebra; meant 24 not 27
Allen Druze . . . . . . . 2 pts - Good step: (ABC)^2 = 1000k + ABC ; ABC(ABC - 1) = 1000k ; etc.
Nikita Kuznetsov . . 2 pts - I can see what you meant & you got the ages right, nice use of graphing!
Wait (new) . . . . . . . . 1 pt - The dad's got to be exactly 15 years younger than the house; house rules!


Problem #157 - Posted Wednesday, October 30, 2002
Chop, Skip, Chop ! (back to top)
"Only the smartest one shall survive," said the executioner. "All of you prisoners will be seated
around this round table. I will chop off the head of the prisoner in seat #1, skip seat #2, chop #3,
etc. Beheaded bodies (and chairs) will be cleared away post haste. When I get to the end I will
continue to chop, skip, chop, until there is but one prisoner remaining, who will then be freed."
a) If there are 13 prisoners, which is the lucky seat number? b) In the morning, you will be told
how many prisoners, n. You must figure out the best seat quickly, just knowing n. Which seat,
k, should you 'head' for, in order to be freed?


Solution: I admit, this is a classic problem; some of you found the solution online.
Hermen J. sent in the link: http://mathforum.org/library/drmath/view/55862.html
Nick McG. rephrased the problem in terms of a deck of cards: Discard 1, put 2 on bottom, discard 3, etc.
Nikita K.+ new contestant Robert Dostal implied+supplied the sweet formula k = 2*(n - 2^(int(log2(n))).
Ji Zheng tells me this is called the "Josephus Problem" which he read in "Concrete Mathematics".
Quasi-C sent in a water-tight induction proof that the formula works.

a) The first answer for n = 13 is to run and sit at seat #k = 10, or else your head will roll...

b) The general solution? Some of you did it the way I did over a double latte: draw a circle numbered 1 to n,
then cross out the odds, etc. The idea of a power of 2 quickly creeps in and the lucky seat number is always
even (for n > 1), and goes k = 2, 2, 4, 2, 4, 6, 8, 2, 4, 6, 8, 10, 12, 14, 16, 2, 4, . . . where the formula becomes:
Let 2^m <= n < 2^(m+1); sit at seat #k = 2*(n - 2^m). (The - is a minus.) For powers of 2 sit in the last seat.
For example if n = 13 then 2^m = 2^3 = 8, and so k = 2*(13 - 8) = 2*5 = 10 as done in part a).


WINNERS - Problem 157 . (back to top) . leader board

Tim Poe. . . . . . . . . . . 10 pts - As long as you don't need to run your macro while axeman is waiting!
Joe Alvord . . . . . . . . . 7 pts - I like the "run to seat 10" image. Nice: n in binary, drop leading 1, add a 0.
Uwe Buenting. . . . . . 6 pts - Early entry; recursive nature of your solution wasn't 'instant' enough
Hermen Jacobs . . . . . 6 pts - Nice use of good old BASIC; check your formula at powers of 2...
Arthur Morris . . . . . 5 pts - I believe your result but you didn't divulge your secret method.
Ed Wern. . . . . . . . . . . 5 pts - Example with 165 prisoners was accurate; is there a story for that n?
Troy Butler . . . . . . . . 4 pts - First entry, your solution didn't clear away the bodies & skip the live ones
Quasi-C. . . . . . . . . . . 4 pts - First 'sort of induction' then the real thing. Nice job.
Jack Dostal. . . . . . . . 4 pts - That's how I did it, on paper with numbers, circles and x's!
Anirban Bhattacharyya . 3 pts - Same problem with clearing bodies gave you k = 2^m for all n
Phil Sayre. . . . . . . . . 3 pts - Now #2 all time, you simulation runner! Yes, less 'grisly' as a card trick.
Nick McGrath. . . . . . 3 pts - Nice that you noticed the procedure is 2n - 3 steps. Check powers of 2
Jon Stearn . . . . . . . . . 3 pts - I'm not sure every round starts w chop and ends w skip but yr formula is ok.
Charles Lo . . . . . . . . 3 pts - Nice to hear from you and the Paradigmic Escape, is that your band?
Nikita Kuznetsov . . . 2 pts - Did you remember to clear away the bodies? It's only fair to the next seat.
Ritwik Chaudhuri . . 2 pts - That's right, work with the smaller power of 2 and subtract.
Ji Zheng . . . . . . . . . . . 2 pts - Pretty good expln but I think you need a 2^m in there rather than m(m+1)/2
Robert Dostal (new) . . . 2 pts - Good ans. Welcome to the contest; three from one extended family!
Yonlawan Chirawatcharadej 1 pt - See the above for explanation for n = 13; also need subtraction
Allen Druze . . . . . . . . 1 pt - Partly ok, is your k the lucky seat? Also what abt the case n=13...



Problem #158 - Posted Saturday, November 9, 2002
Digits In The Bag ! (back to top) Problem contributed by Sudipta Das
A math wizard has a bag containing the digits 0 through 9, and has used six of them to stick
two different three-digit perfect squares on the foreheads of Ann and Ben; A and B.
Both Ann and Ben know this fact, but each person can see the other person's number only.
The wizard asks Ann: "How many of the digits remaining in my bag can you exactly tell me?"
Ann replies:"Three." If the wizard now asks the same question to Ben, what should he reply?


Solution: This fine problem was sent to me by Sudipta Das a while ago! Here's my ex-student Charles's solution:
The squares are 169, 196, 256, 289, 324, 361, 529, 576, 625, 729, 784, 841, 961; 0 isn't used in any of the squares, it
must be one of the four digits left in the bag. Ann has to conclude knowledge of two digits from Ben's forehead. Ben's forehead shows 169 (or 196, 961) 256 (or 625) 289 324 361 Ann's number can be 324 , 784 784 , 841 361 , 576 169 (196, 961), 576 289 , 529 , 729 , 784 Ben's forehead shows 529 576 729 784 841 Ann's number can be 361 , 784 , 841 289 , 324 , 841 361 , 841 169(etc.), 256(625), 361, 529 256(625), 529, 576, 729 From that list, only if ben's forehead shows 256 (or 625) can Ann know three digits. 8, 4 must be on her head and 0, 1 , 3, 7, 9 may be in the bag. she doesn't know if 784 or 841 is on her head, therefore she doesn't know if 1 or 7 is in the bag. The digits she knows are in the bag are 0, 3, and 9. Ben of course being equally intelligent figures
this out; figures he must have 256 on his head for Ann to answer "three." and can piece together the digit puzzle
from what he can see on Ann's forehead. He'll know all four digits in the bag : 0, 3, 9 and either 1 or 7 depending
on what Ann's forehead shows. Ben's answer is "four."


WINNERS - Problem 158 . (back to top) . leader board . . . . . . . . .. I'm back from vacation! Sorry if you guys got antsy - Dan

(Judge's decision: No 10-point winner - Hermen had a resubmission) . .. I got many excellent responses for this problem!
Hermen Jacobs . . . . . 9 pts - Right ; 13 perfect non-rep squares; 625 as well as 256, thanks for resub.
Nick McGrath. . . . . . 7 pts - Good, there are ten digit combn's, only 256/625 leave correct choices.
Charles Lo . . . . . . . . 5 pts - Ben says 'four' all right, thanks for the explanation, keep entering!
Anirban Bhattacharyya . 4 pts - Good use of trial and error to find which leaves 3 digits known
Phil Sayre. . . . . . . . . 4 pts - I like yr approach that the 'complementary group' has 2 digits in common.
Joe Alvord . . . . . . . . 4 pts - Good logic; Ben knows he has 2,5,6, and can see Ann's, so knows 4.
Tim Poe . . . . . . . . . . . 4 pts - Good proof that 064, etc. don't work as a solution, per Ann's statement.
Denis Borris . . . . . . . 3 pts - That's it; 40 Ann-Ben combos. Ben says 3..., no 4! Right, on second try!
Jon Stearn . . . . . . . . 3 pts - Good point: Assume Ben knows Ann said 3. You need that.
Quasi-C. . . . . . . . . . . 3 pts - You've introduced the concept of DSITB; definitely still in the bag!
Ed Wern. . . . . . . . . . 3 pts - Good explanation and table, and yes, we all agree on the digit zero!
Jack Dostal. . . . . . . . 3 pts - Join the .xls club! I like the comment that Ben coulda been blind.
Lindsay Dann (new). . 3 pts - Welcome to the contest; good answer. A friend of Les is a friend of more!
Uwe Buenting. . . . . . 2 pts - Thanks for answer; Ben can still say 4 because he does know his.
Nikita Kuznetsov . . 2 pts - Ben's sure about the zero, yes, but more after hearing Ann.
Arthur Morris . . . . . 2 pts - Ben can't have 529 beacuse Ann's answer wouldn't be confident.
Sue B . . . . . . . . . . . . 2 pts - Ben should say '4' in each possib. not depending on what he sees.
Zahi Yeitelman . . . . 2 pts - The number Ben knows doesn't depend on what he sees, based on Ann.
Les Billig . . . . . . . . . 2 pts - Also coulda been 625 not just 256, not clear if you meant both.
Sudipta Das. . . . . . . 2 pts - Good problem , thanks for helping me out with the answer too.
Vince LoCascio (new) .2 pts - Welcome! Just under the wire, with reasoning mostly given!
vze4bwtf (new) . . . . . .1 pt - (Is that your real name? Have you entered before?) Ann had to have 2,5,6.
Allen Druze. . . . . . . 1 pt - Not sure how you interpreted the question, more than the 0 is DSTIB.



Problem #159 - Posted Monday, November 25, 2002
Money In The Hole? (back to top)
Fearless Frank decided to play a fair coin-flip game with probability 1/2 of winning each
bet, and risked 1/m of his fortune (originally A dollars, m>1) at every flip. After 2n games,
Frank has won n games and lost n. Choose and explain the correct answer from this list: a) Frank has broken even; he still has his A dollars b) Frank is predictably ahead by a certain amount c) Frank is predictably behind by a certain amount (in case b or c give the exact formula in terms of m and n) d) Frank is now ahead, behind, or even, depending on the order in which the wins and losses occurred e) He is ahead, behind, or even, depending on m and n

Solution: The correct answer is c). The reason is as follows (from Tim Poe):

"For a wager of 1/m, the result of any particular coin toss beginning with fortune X will be
X + X(1/m) = X + X/m = X(1 + 1/m) = X*((m+1)/m) if you win, or X*((m-1)/m) if you lose.
For two consecutive tosses, with one win and one loss, the outcome will be X*((m+1)/m)*((m-1)/m).
Similarly, with 2n tosses with n wins and n losses (using associative and commutative properties of
multiplication), the result will be X *((m+1)/m)^n * ((m-1)/m)^n. ( = X*((m^2 - 1)/m^2)^n )

Dan's note : This simplifies to a final balance for Frank of A*[1 -- 1/m^2]^n dollars.
Dan's note to Tim: Your 'A' evolved into an 'X'. Don;t AX me why! $;-}


WINNERS - Problem 159 . (back to top) . leader board . . . .

Joe Alvord . . . . . . . . 10 pts - Good explanation there, Joe. That's the correct formula all right.
Nikita Kuznetsov . . . 7 pts - Sure the problem seems easy to a really smart guy like you . . .
Nick McGrath . . . . . . 5 pts - Yes, and you included a formula for x wins and y losses.
Anirban Bhattacharyya . 4 pts - Nice job and thanks for the examples of n = 2 and WWLL vs LWWL
Tim Poe . . . . . . . . . . . . 4 pts - Mighta been 3 pts but I liked your expl of why the others were false.
Jayavel Sounderpandian (new) 3 pts - Welcome to my contest. Good steps, reasons, and answer!
Quasi-C. . . . . . . . . . . . 3 pts - Right, the amout he has isn't the same as the amount he's behind.
Jack Dostal. . . . . . . . 3 pts - Thiorough solution, important reference to commut. law of multip.
Arthur Morris . . . . . 2 pts - Correct for n = 1 ; wanted steps and a formula for general case.
Uwe Buenting. . . . . . 2 pts - Good try, almost ok but you needed a 1/m^2 in last term.
Jon Stearn . . . . . . . . . 2 pts - You used the term 'more' but meant 'as much', some steps missing
Les Billig . . . . . . . . . . 2 pts - Close call, had to dock you a pt for putting an n in place of an m
Vince LoCascio (new) . 2 pts - That's the right call, now let's see how you got your answer.
Zahi Yeitelman . . . . 2 pts - Also stuck an n where an m shoulda been, good attempt though.
Phil Sayre. . . . . . . . . 2 pts - A later entry, but correct and before I posted an answer, so ok!
Drew (Besse) . . . . . . .1 pt - Are you the same as just 'Drew'? Not sure, send me an e-mail
Denis Borris . . . . . . . 1 pt - I think you might have had percentages in mind but s.b. no 100's
Hermen Jacobs . . . . . 1 pt - Welcome back from Thailand, Land of Sun, Sand, and Great Food!


Problem #160 - Posted Saturday, December 7, 2002
Cubes Come Fourth ! (back to top)
A long time ago, I noticed that 65 was the smallest integer that could be written as
the sum of two (different) squares in two different ways: 65 = 8^2 + 1^2 = 7^2 + 4^2.
a) A very famous number is the smallest integer that can be written as the sum of
two cubes in two different ways. What is this number, and why is it famous?
b) What is the smallest integer that's the sum of two different fourth powers in two
different ways? Be sure to say what all those cubes & fourth powers are.
It's OK to enter even if you can't do all parts of the problem!


Solution: (From 2000/01 winner and all-time leader Art Morris) a) The smallest number [expressible as
the sum of two cubes in two ways] is 1729 = 1^3 + 12^3 = 9^3 + 10^3 . It is famous because [English
mathematician] G. H. Hardy took Taxi #1729 to hospital to visit Srinivasa Ramanujan who was dying of
tuberculosis. Hardy said to his friend (Ramanujan) that the taxi number #1729 was a rather dull number.
(Ramanujan) then (immediately) told him of the above properties. b) I didn't derive the number, but found
in Science News Online of July 27, 2002 that the number was discovered by Leonard Euler [in 1772
according to Hermen]: 59^4 + 158^4 = 133^4 + 134^4 = 635,318,657.

[ Dan's Notes: (1) Art M. defines these as 'limousine' numbers; he says the 21st is 2216304050432.
(2) Jayavel S. tells us you can find another property of the number 1729 in 'Godel Escher Bach' by
Douglas Hofstader, and at www.mathpages.com/home/kmath028.htm (3) I recommend these books:
Godel, Escher, Bach (Hofstader) , A Mathematician's Apology (G.H.Hardy), with taxicab story, and
The Man Who Knew Infinity, A Life of the Genius Ramanujan (Robert Kanigel)
And Ed Wern's mail got bounced but finally reached me; he sent in a fine link to MathWorld ]


WINNERS - Problem 160 . (back to top) . leader board . . . .

Arthur Morris . . . . . . 10 pts - Thanks for the Euler reference for fourth powers, and the 'limos'
Denis Borris . . . . . . . . 6 pts - Started by making a bunch of equations, but rethought it!
Jayavel Sounderpandian . . 6 pts - Thanks for the references, and good answer besides.
Les Billig . . . . . . . . . . . 5 pts - Good: 4th powers end only in 0, 1, 5, 6 so sums in 0, 1, 2, 5, 6, 7.
Hermen Jacobs . . . . . . 5 pts - Right, the hospital was in Putney, England. Nice BASIC program!
Nick McGrath . . . . . . 4 pts - Thanks for using one of my favorites: www.mathworld.wolfram.com
Anirban Bhattacharyya . . 4 pts - Yes; if x^4+y^4=a^4+b^4 then (x^2+a^2)/(y^2+b^2) = (b^2-y^2)/(x^2-a^2)
Tim Poe . . . . . . . . . . . . 3 pts - Wow, a devious manipulation of a 160 x 160 Excel spreadsheet!
Quasi-C. . . . . . . . . . . . 3 pts - Maybe Gauss did do something notable in 1729; he was pretty busy!
Phil Sayre. . . . . . . . . . 3 pts - Right, 'A Mathem's Apology' by Hardy is a classic. Go Python!
Jack Dostal. . . . . . . . . 3 pts - Thanks for details on the taxicab story; 4th powers not known to Ram.
Gaurav Agrawal (new) . 3 pts - Welcome to my contest, good answer (Hardy, taxi, etc); keep it up!
Nikita Kuznetsov . . . . 2 pts - Not sure if 1729 = 19 * 13 * 7 will help; 62 quadrillion too big.
Nic Hargreaves (new) . . 2 pts - Hi there Nic; a couple of weeks after the others, but welcome!
Ed Wern . . . . . . . . . . . 2 pts - Good solution, thanks for the link, I added your points to your total!



THANKS to all of you who have entered, or even just clicked and looked.
My site is now in its sixth season - OVER 32,000 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 - 2003 A.D.


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