**dan's math@home - problem of the week - archives****Please visit my new page location at****www.dansmath.com/probofwk/probar16.html****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+ .**prob index151 Friends, Wives, $ 152 Nearly Isosceles? 153- Sharon & Karen 154 - Swim 2D Boat ! 155 - Can You Digit ? - 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**.**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...

**Zi Jheng**. . . . . . . . . . 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**. . . . .**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* - 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!

**Zi Jheng**. . . . . . . . . . 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**. . . . .**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**.**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**. . . . .**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.**Zi Jheng**. . . . . . . . . . 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**.**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**. . . . .**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**.**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**.- 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 = 2__5__; 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.
**THANKS to all of you who have entered, or even just clicked and looked.****My site is now in its sixth season - OVER 30,000 HITS so far!**(Not factorial.)**Help it grow by telling your friends, teachers, and family about it.**YOU CAN ALWAYS FIND ME AT **- Dan the Man Bach**- 2*7*11*13 A.D.

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