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Problem Archives page 16

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 index

151 Friends, Wives, \$
152 Nearly Isosceles?
153- Sharon & Karen
154 - Swim 2D Boat !
155 - Can You Digit ?

Problem #151 - Posted Wednesday, August 14, 2002
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)."

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...
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 . . . . .
. . . . 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!"

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 . . . . .
. . . . 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
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.

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.
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 . . . . . . . . . . . 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.
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 aren't late if the next prob isn't up.

Problem #155 - Posted Monday, October 7, 2002
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.

 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 dansmath.com - Dan the Man Bach - 2*7*11*13 A.D.

Problem Archives Index

Probs & answers . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90
Problems only . . . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90

Probs & answers . . . 91-100 . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+
Problems only . . . . . 91-100 . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+

or go back and work on this week's problem!