dan's math@home - problem of the week - archives
Problem Archives page 14

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

131 - Back at the 7-11
132 - - The Top Five !
133 -- Walk, Eat, Bike
134 A Peachy Quintet
135 - Reciprocal Pairs
136 How Lo CanUGo
137 -- 2000 Shuffles !
138 The Twelve Coins
139 - A Funny Inverse
140 Swimr Loses Cap

Problem #131 - Posted Tuesday, October 16, 2001
When Jack went back for a late-night snack, he bought three items off the rack. Zack
rang up the snacks and said "5.70, Jack." "Wait, Zack, you multiplied the prices instead
of adding!" "Multiply, add; it still comes out the same. Pay up." What the hack is going
on here at the Snack'n'Shack?! (What were the prices of Jack's three items?)

Solution:  Let the three prices be p , q , and r. Then p q r = 5.70 and p + q + r = 5.70.
The first tells us that as fractions, p q r = (a/b)(c/d)(e/f) = 57/10 = (3*19) / (2*5).
We look for fractions containing 3 and 19 on top and 2 and 5 on the bottom, in different orders and with
extra cancelling factors, and see what they add up to. (I misplaced the solution to this one, so I found these!)
 prices (fractions) prices (decimals) product sum (1/2)(3/1)(19/5) (19/20)(5/4)(24/5) (3/4)(19/10)(4/1) (3/4)(2/1)(19/5) (19/20)(3/2)(4/1) (1/1)(3/2)(19/5) (1/1)(19/10)(3/1) (5/4)(8/5)(57/20) (6/5)(19/10)(5/2) (6/5)(19/10)(12/5) (3/2)(19/10)(2/1) (0.50)(3.00)(3.80) (0.95)(1.25)(4.80) (0.75)(1.90)(4.00) (0.75)(2.00)(3.80) (0.95)(1.50)(4.00) (1.00)(1.50)(3.80) (1.00)(1.90)(3.00) (1.25)(1.60)(2.85) (1.20)(1.90)(2.50) (1.25)(1.90)(2.40) (1.50)(1.90)(2.00) 5.70 5.70 5.70 5.70 5.70 5.70 5.70 5.70 5.70 5.70 5.70 7.00 7.00 6.65 6.55 6.45 6.30 5.90 5.70 5.60 5.55 5.40
The items Zack brought back from the Snack'n'Shack cost p = \$1.25 , q = \$1.60, and r = \$2.85.

Dan's notes: Quasi-C offered coupons and discounts; see negative price structure below.
Derek found close ones; all add to 5.70: a)1.06*2.25*2.39 = 5.70015 , b) 1.13*1.86*2.71 = 5.695878 ,
c) 1.20*1.69*2.81 = 5.69868 , d) 1.36*1.45*2.89 = 5.69908 ; all products round to 5.70.
Anirban got closest with 1.319*1.50*2.881 = 5.7000585 but 7-11 doesn't have prices in 'mils' (1/10 cents)

P.S. I found a version of this where the sum and product are both 7.11 ; I should have used that one!
Two of you sent me answers that involve four items, and claimed three can't be done for 7.11.
Bach soon with those bonus contest points!!

@ All of the following got the exact answer of 1.25, 1.60, and 2.85:
Lisa Schechner . . . . . 10 pts - Good method: x+y=5.7-z; xy=5.7/z, search for sol's w/2 dec.
Tim Poe . . . . . . . . . . . 8 pts - Virtual Basic to Lisa's Excel; bonus pt for lagging by only 1 minute!
Steve Lawrie . . . . . . . 5 pts - Tried 1.00,1.50,2.00 just to see, decided one above 1.5, one below.
Michelle Lee (new) . . . 4 pts - Hello! Got approx ans 1.25, 1.50, 2.95 then zoomed in on correct one
Punam Bhullar (new) . 3 pts - Welcome to the contest ; guessing, checking and adjusting: tried and true!
Arthur Morris . . . . . . 3 pts - Got it on the resubmission; good original try with factor triplets.
Phil Sayre. . . . . . . . . . 3 pts - That Python language seems to get a 'grasp' on a variety of problems!
Zahi Yeitelman . . . . . 3 pts - Nice: xy(5.7-x-y)=5.7; I'll have to graph the curve and look at it!
Hermen Jacobs . . . . . 3 pts - Welcome back; programming can involve (or substitute for) thinking!
@ Here's the group that got some amazingly close (or interesting) answers:
Quasi-C . . . . . . . . . . . 3 pts - 1.20,1.90,2.50 add to 5.60 but I liked (10.15)*(-4.32)*(-0.13).
Derek Chin . . . . . . . . 3 pts - Incl bonus pt for first and largest list of close calls (see above)
Anirban Bhattacharyya . 2 pts - That is incredibly close to 5.70 but still not 'exact' mathematically.
Navi Sidhu . . . . . . . . 2 pts - Those were good numbers; (1.10)(1.97)(2.63) = 5.69921, still no cigar
Tim Bock (new) . . . . . 2 pts - I enjoyed the 'saga' of a, b, and c ; you settled on Navi's Numbers.
Alex Chan . . . . . . . . 1 pt - Setting up the equations is usually half the battle!
Sabina Paradi . . . . . 1 pt - Good idea, trying case where 2 prices are = ; might lead to the answer

Problem #132 - Posted Friday, October 26, 2001
Five bicyclists tried to predict their order of finish in a race. Alice, Bert, Charlie, Daria, &
Ernie, the only entrants, spoke . . . Alice: "Bert will finish two places higher than Charlie."
Bert: "I'm gonna finish in third place, you just watch me." Charlie: "Daria will be first."
Daria: "Bert'll finish second." Ernie: "Charlie will be three places lower than Alice."
It turned out only one was correct: the eventual winner. There were no ties.
What were their places in the race?

Solution:  Most of you noticed that Bert has to be lying, so did not finish first or third. Or second, unless Daria wins...
Here's Lisa's short but sweet explanation: "Clearly B or C can not be the winners. This implies that D can not win.
So, it is between A and E. If you assume A wins and try to place the rest you fail (can't be done) So, try E as a winner.
Bingo- it works. They must have finished in this order: E, A, D, B, C. (only E has a correct prediction)."

Lisa Schechner . . . . . 10 pts - I can see what your thinking is by the sufficient signposts.
Nikita Kuznetsov . . . 7 pts - Yes, congratulations to Ernie, and to you too for such a high finish!
Arthur Morris . . . . . . 5 pts - Right: Bert can't be 1st or 3rd, C not 1st, D not 1st , etc!
Michelle Lee . . . . . . . 4 pts - I like your 5 x 5 chart idea , rows for people, cols for places
Sudipta Das . . . . . . . 4 pts - Good solid logic; so many responses you get an extra pt for earliness
Katherine Dostal (new) 3 pts - Systematically elim all the wrong chioces, good strategy, welcome!
Steve Lawrie . . . . . . . 3 pts - Good answer, nice logic, thanks for the link on your website!
Hermen Jacobs . . . . . 3 pts - D can't B true and C or B neither ; A elim so that leaves E.
Tim Poe . . . . . . . . . . . 3 pts - An Excel spreadsht with all 120 orders, followed up by pure logic!
Erin Keithley (new) . . 3 pts - Well said, correct answer, and welcome to the contest!
Willette (new) . . . . . . . 3 pts - Good reasoning, nice to have you; are you THE Willette from DVC?
The rest are either later & correct, or gave partial answers, or had partial problems:
 Quasi-C . . . . . . 2 pts - E=1 but C=5 and 3 both? Tai Dang (new) . 2 pts - It's not who you believe... Derek Chin . . . 2 pts - Second pt for resubm's'n Phil Sayre . . . . 2 pts - Correct answer, good job Punam Bhullar 2 pts - E does win; good expl. Omar Gymboski (new) 2 pts - Hi! nice reasoning Alex Chan . . . . 2 pts - Right, B & C not both rt. Anirban Bhattacharyya 2 pts - Elim A,B,C then D. Tim Bock . . . . . 2 pts - Quick look solved lots! Sabina Paradi . 2 pts - Is Ernest his real name? Payman Parang 2 pts - Worked thru the possib's. Tony Sun (new) . 2 pts - Good, 'none refer to E' Drew (new) . . . . . 1 pt - Only 1 rider can be right Sukhmine Bains 1 pt - Daria can't be first... Cyril Virtusio (new) 1 pt Can't believe more'n one Navi Sidhu . . . 1 pt - Right order, no reasons. Dugan McGloughlin 0 pts , 0 hints , try again! Zahi Yeitelman . 0 pts - You sent #129; sorry!

Problem #133 - Posted Saturday, November 3, 2001
A woman walks to her friend's house at R mph. It takes her T hours, and she spends S
hours having lunch with her friend. She borrows a bicycle for the trip home, but must
follow a path that's U times as long as the walking path. If she bikes N times as fast as
she walks, what is her total elapsed time for the round trip? (Including lunch!)
Clues for integers R , T , S , U , and N : RTSUN = 72 ; R+T+S+U+N = 13 ; U > N > S ;
a majority of R , T , S , U , N are prime numbers ; RT = S + UN ; {R,T,S,U,N} has four elements.

Solution: by semi-new contestant "Drew":

Her total elapsed time for the round trip was 10 hours. Here is why:
I know that R+T+S+U+N = 13 I also know that {R,T,S,U,N} has 4 elements so their must be one and only
one number that is repeated once and only once. so I make a list of all possible combinations.
1+2+3+3+4 ; 1+2+2+3+5 ; 1+1+2+3+6 ; 1+1+2+4+5 . I know that RTSUN=72 so I multiply the number
in my 4 combinations.1*2*3*3*4=72 ; 1*2*2*3*5=60 ; 1*1*2*3*6=36 ; 1+1+2+4+5=40 (You mean *)
Only 1*2*3*3*4 works so there are my numbers.

I know that RT=S+UN, 3*3=1+2*4 is the only combination here that works. So I know that R=3,R=3 and S=1
so the time getting there was 3 hours and the time at lunch was 1 hour. The last thing me know is that U>N>S so
U must = 4 and N=2. With this in mind. The walking path in 9 miles long (3 hours at 3 MPH = 9 miles) so the
bike path is 36 miles (9 miles * 4 is 36 miles) she biked 6 MPH (3 MPH * 2 = 6 MPH) so she biked for 36 miles
at 6 MPH which means she biked back for 6 hours. So 3 hrs getting there, 1 hr at lunch and 6 hrs getting back
= 10 total hours for the trip.

Dan's Note: Thanks, kudos, congrats, and an extra point each go to Steve L. and Lisa S. for their answer
of 7.11 = 3.16 + 1.50 + 1.25 + 1.20 = 3.16 * 1.50 * 1.25 * 1.20. In fact 7.11 can't be done with 3 items!

Steve Lawrie . . . . . . . 10 pts - Good approach ; thanks for the 7.11 answer as well! (see above)
Sudipta Das . . . . . . . . 7 pts - I like it when people use prime factors to limit the possibilities!
Ludwig Deruyck . . . . 5 pts - Might have filled in last steps, but thanks for quick response!
Hermen Jacobs . . . . . . 5 pts - Right, orig. 4 poss. for U, N, S: 321,432,431, 421. Only 431 works.
Anirban Bhattacharyya 4 pts - That's right; at least 3 primes from 2,2,2,3,3. Time = T+S+TU/N.
Nikita Kuznetsov . . . . 4 pts - Good: why ride bike if it takes twice as long? (Different muscle groups!)
Lisa Schechner . . . . . . 3 pts - You hit 2, 3, 3 on first try, then total time = T(1+U/N)+S = 10 hrs.
Tim Poe . . . . . . . . . . . . 3 pts - The majority of 1, 2, 3, 3, 4 are primes; 2, 2, 2, 3, 3, is a big majority!
Drew . . . . . . . . . . . . . . 3 pts - Good approach; 4 elements means one repeat; write out the poss.
Quasi-C . . . . . . . . . . . . 3 pts - S<N<U ? What SNU with U? Glad to see UR on top of it!
Omar Gymboski . . . . . 3 pts - Welcome back, good logical answer with the right result!
Phil Sayre . . . . . . . . . . 3 pts - There are five sums with a total of 13; you knew to chose the best!
Gleb Podkolzin (new) . 3 pts - Yes, 3.3.1.2.4 = 72 and add to 13. All other cond's ok. Welcome!
Cindy Tang . . . . . . . . 3 pts - Right, good result, 10 hrs, (but don't put 1 on the list of primes)

The rest are either later & correct, or partially correct :
 Payman Parang . . 2 pts - Knowing U=4 's the key. Zahi Yeitelman . . 2 pts - Good, all < 6, then solve. Tim Bock . . . . . . . 2 pts - Yep, long day, 1 hr lunch Sabina Paradi . . . 2 pts - You got it all-right, 10 hr. Anne Sufka (new) . 2 pts - Good, add 1 to 2,2,3,4. Michelle Lee . . . . 2 pts - First & best of the close. Katherine Dostal .1 pt - Got the RTSUN part. Alex Chan . . . . . . 1 pt - So far, good formulas Punam Bhullar . . 1 pt - Right, d=rt , use N*T. Adel Fahramand . 1 pt - Starts ok but 4hrs 2 short

Problem #134 - Posted Monday, November 12, 2001
Here are five sweet number puzzles, each is a 'peach'.
a) If 3^n = 4 and 4^m = 8, then how much is 9^(n - m) ?
b) What number is three times the sum of its two digits ?
c) This is the smallest factorial that's a multiple of 2^22 .
d) A fraction between 34/45 and 43/54 with denom < 100 ?
Dan's note : This part was too easy. I meant 34/45 and 40/53.
e) The number of primes whose squares have 4 or 5 digits ?

Solution: Thanks to Robin P. and Jeff W. for some of these vintage problems!
a) 4^m = 8 means (2^2)^m = 2^(2m) = 2^3 so m = 3/2.
Then 9^(n-m) = 9^n / 9^m = [(3^2)^n] / [9^(3/2)] = [(3^n)^2] / [3^3] = 16/27 as 3^n = 4.
b) If it's three times the sum of its digits, it's a multiple of 3 so its digits add up to a
mult of 3 so the number is three times this sum so it's a mult of 9 so its digits add to
a mult of 9 so the number is a mult of 27. The number 27 works, so this is it!
c) Go along in the factorials 1 * 2 * 3 * 4 ... ; counting 2's contributed by each even factor:
2 (1; 1 so far) , 4 (2; 3) , 6 (1; 4) , 8 (3; 7) , 10 (1; 8) , 12 (2; 10) , 14 (1; 11) , 16 (4; 15) ,
18 (1; 16) , 20 (2; 18) , 22 (1; 19) , 24 (3; 22). So 24! is the first mult of 2^22.
d) You could do 34/45 = 0.75555... and 43/54 = 0.7962962... so that two obvious answers
are 0.76 = 19/25 and 0.78 = 39/50 , also there are many dozens more , for instance
7/9 , 10/13 , 11/14 , 13/17 , 15/19 , . . . , 76/99 . I count a total of (about) 134 answers!
What I meant was the "cheap average" where you add numerators and add denoms:
(34 + 43) / (45 + 54) = 77/99 = 7/9 . This way you always get a fraction between the two!
So for 34/45 and 40/53 we get 74/98 = 37/49 ; the only other is 71/94. (see bonus winners)
e) The square roots of 1000 and 99999 are 31.6228 and 316.226 ; there are 54 primes in
this range, starting at 37 and ending at 313. Here 37^2 = 1369 and 313^2 = 97969.

Steve Lawrie . . . . . . . 10 pts - Nice proof of a) and thanks for the list of 54 squared primes!
Arthur Morris . . . . . . 6 pts - All were fine & early, but 0.59259 isn't definitively 16/27
Sudipta Das . . . . . . . . 6 pts - Good formula [gr.int.]: [n!/2] + [n!/2^2] + ... = 22 --> n = 24.
Lisa Schechner . . . . . . 5 pts - Yes, {n/(n+11)} is increasing; 2nd frac was actually 35/46.
Anirban Bhattacharyya 4 pts - One of the earliest entries but 54 primes not 74, others very good!
Quasi-C . . . . . . . . . . . . 4 pts - Fine terminology; "24! is exactly divisible by 2^22"; thanks.
Hermen Jacobs . . . . . . 4 pts - Thanks for carrying & thinking about my problem throughout Holland!
Michelle Lee . . . . . . . . 3 pts - First answer received! a) 9^(ln4/ln3 - ln8/ln4) right but unsimp; no c).
Tim Poe . . . . . . . . . . . . 3 pts - First 4 were fine then somehow 396^2 instead of 316^2; good try!
Derek Chin . . . . . . . . . 3 pts - a) was approx right; prove it's 16/27. Others were ok!
Phil Sayre . . . . . . . . . . 3 pts - Nice b): 10a + b = 3(a+b) ; 7a = 2b ; 27. Thanx 4 complete PF of 24!.

These are either later & correct, or partially correct :
 Jack Dostal (new) . . . 2 pts - 23! only has 2^19 . . . Nikita Kuznetsov. . 2 pts - a-d ok ; not 143 primes Erin Keithley . . . . . 2 pts - b,e ok; a,d approx. ok. Ludwig Deruyck . . 2 pts - Thanks 4 Matica table Zahi Yeitelman . . . 2 pts - All were ok; thanks! Ernesto von Ruckert (new) 2 pts - Welcome! Ok abde. Payman Parang . . . 1 pt - Missed 3 primes & others Navi Sidhu . . 1 pt - a) approx; 4 extra primes Cindy Tang . . 1 pt - Some extra small primes Alex Shen (new) 1 pt - bde ok; 18! needs defense Sabina Paradi 1 pt - Th parts u did wr ok. Thx! Adel Fahramand 1 pt - Multiple of 2^22 not factor Punam Bhullar 1 pt - 41/54 ok; 4 others see above Alex Chan . . . . 1 pt - Do .592bar = 16/27 . . .

Bonus point for 134d) : Art Morris (First with 37/49 = 74/98) , Phil Sayre (first with 71/94) , and
Sudipta Das (first with both). Also answering but no pt: Navi Sidhu , Lisa Schechner , Tim Poe , Payman Parang.

Problem #135 - Posted Thursday, November 22, 2001
Find all solutions (that you can) to : 1/x + 1/y = 1/14 :
a) where x and y are positive integers, x <= y ,
b) where x and y are any integers and x <= y .

Solution: Good old Egyptian fractions! (over 4000 years old, to be approximate!)
Write 1/x + 1/y = 1/a as ax + ay = xy ; xy - ax - ay + a^2 = a^2 ; (x - a)(y - a) = a^2 ;
(x - 14)(y - 14) = 196 so for the denoms add 14 to different divisors of 196. Cool, huh!
a) I count five solutions to the first one: (Egyptian rules would not have allowed the last one)
1/15 + 1/210 = 1/16 + 1/112 = 1/18 + 1/63 = 1/21 + 1/42 = 1/28 + 1/28 = 1/14
b) Use subtraction : four more appear (Egyptians would have freaked at negative numbers!)
1/13 -- 1/182 = 1/12 -- 1/84 = 1/10 -- 1/35 = 1/7 -- 1/14 = 1/14

Arthur Morris . . . . . . 10 pts - Fourth entry but first fully correct one; Art for Art's sake!
Hermen Jacobs . . . . . . 7 pts - Good logic; nice to see that BASIC programs live on!
Michelle Lee . . . . . . . . 5 pts - Simple but effective: let x = 1, 2, 3, ..., solve for y.
Lisa Schechner . . . . . . 4 pts - Wisely limited search to 15 <= x <= 28 and so y = 28, ... , 210.
Anirban Bhattacharyya 4 pts - Yes, 1/x >= 1/y so 2/x >= 1/14 so x <= 28. That helps!
Jack Dostal . . . . . . . . . 3 pts - I liked your flow of logic; got all 9... any relation to another contestant?
Sudipta Das . . . . . . . . 3 pts - You got all nine! No penalty this time but please show reasoning.
Phil Sayre . . . . . . . . . . 3 pts - Good : y = 14x / (x - 14) and go from there. Thanks also for #131.
Allen Druze . . . . . . . . 3 pts - Welcome back! Good expl'n about only testing x = 15 , . . . , 28.
Navi Sidhu . . . . . . . . 3 pts - You got em; no eqns or steps this time; thanks for 134d.
Ludwig Deruyck . . . . 3 pts - Another Matica Table[Solve[14(x+y)==x*y],{x,15,100}]. Cool!
Quasi-C . . . . . . . . . . . . 3 pts - All 9 right, and nice idea to graph a solution curve on TI-whatever
Zahi Yeitelman . . . . . . 3 pts - I liked your expl'ns and eq'ns -- please no more attchm'ts!
These are either later and/or partially correct :
Nikita Kuznetsov . . . . 3 pts - First entry, but missed two of the subtractions. Show more steps!
Steve Lawrie . . . . . . . . 2 pts - Also 1/15 + 1/210, etc. Good use of prime factors.
Drew . . . . . . . . . . . . . . 2 pts - Good to mult thru by 14 ; 1 pt deduc for resubmission , 7 of 9.
Tim Poe . . . . . . . . . . . . 2 pts - Thorough discussion; left out 1/18 + 1/63 but resubmitted it.
Derek Chin . . . . . . . . . 2 pts - Needed more on the subtractions, but liked your chart.
Tim Boomer . . . . . . . . 2 pts - You got 8 of 9, 28, 28 was legal; could I see your steps next time?
Matt Situ (new) . . . . . . . 2 pts - Welcome to the contest! See above for extra answers
Marcin Mika . . . . . . . 2 pts - Welcome back! Good method: (x-14)(y-14)=196, x=15 is ok.
 Sukhmine Bains . . 1 pt - Yes, <= is less or =. Katherine Dostal . . 1 pt - Got nearly half of 'em. Tony Sun . . . . . . . . 1 pt - Good job, got 7 of 9. Uri Andrews (new) . 1 pt - Welcome! See above soln Sabina Paradi . 1 pt - Yep, one denom is mult of 7 Punam Bhullar 1 pt - Right, can't do that alg step Payman Parang 1 pt - Some good ans, x=0 not. Alex Chan . . . . 1 pt - Three or four ain't bad!

Bonus point for 134d) : Art Morris (First with 37/49 = 74/98) , Phil Sayre (first with 71/94) , and
Sudipta Das (first with both). Also answering but no pt: Navi Sidhu , Lisa Schechner , Tim Poe , Payman Parang.

Problem #136 - Posted Tuesday, December 4, 2001
If each of the letters A, B, and C represents a different digit, what is the MINIMUM value of
(ABC) / (A+B+C) ? NOTE: In ABC, A is the hundreds digit, B is the tens digit, and
C is the ones digit - - they are not multiplied.

Solution: The first time I goofed in the key formula on my spreadsheet and got something like 809, while
we now know the key number is 189 (if you insist the first digit isn't zero, otherwise it's 019; both accepted).
Quasi-C pointed out that A<B<C means there are only 10C3 = 120 triples to check!
And as Art Morris said, "If you can't think, compute." (But some of us like to do both! ;-)
"Ok. (ABC)/(A+B+C), where A,B, and C are different digits ranging from 0-9. So, to minimize the
numerator, A < B < C, which only makes sense. However, to maximize the denominator, you want
the three digits to be very big. But, if that is true, creating a big denominator will compromise
minimizing the numerator. So, the solution that I choose is that:
A = 0, B = 1, C = 9 --> (019)/(0+1+9) = 19/10 = 1.9
But, if you don't count 0 as a digit, then I'll make a quick adjustment:
A = 1, B = 8, C = 9 --> (189)/(1+8+9) = 189/18 = 10.5
But, I still stick with 1.9 as my answer." -- Derek Chin

Derek Chin. . . . . . . . . 10 pts - Thanks for covering all your basses, if you get my spelling.
Tim Poe . . . . . . . . . . . . 7 pts - I did a 'sort' on your prob136.xls and I agree, 019 is legal.
Drew . . . . . . . . . . . . . . 5 pts - Yes, 199 is smallest ratio at 10.43... and 189 rules w diff digs.
Quasi-C . . . . . . . . . . . . 5 pts - Woulda been 2nd place but your ratio of 1.05 needed fixin'.
Hermen Jacobs . . . . . . 4 pts - Nice point: the digit A has the largest influence, so make it small.
Anirban Bhattacharyya 3 pts - First answer of 199 was corrected to 189 - thanks!
Arthur Morris . . . . . . 3 pts - I always give style points! (Usually mentally, or karmically)
Ludwig Deruyck . . . . 3 pts - Last week Mathematica, this time an HP48G+! What next?
Steve Lawrie . . . . . . . . 3 pts - One of the first answers but 189 gives lower than 198.
These are either later and/or partially correct , approx in order rec'd :
Lisa Schechner . . . . . . 2 pts - Spreadsheeting is going pretty low; join the crowd ;-) (No ratio given.)
Michelle Lee . . . . . . . . 2 pts - You can do better than 012 -> ratio of 4, even with a leading 0.
Nikita Kuznetsov . . . . 2 pts - Right, answer is 1.9 for 019; you can't go lower.
Sudipta Das . . . . . . . . 2 pts - Right; 189 does it (with no 0's). What was your secret method?
Marcin Mika . . . . . . . 2 pts - Thanks for the C code, you limited search to 648 numbers!
Phil Sayre . . . . . . . . . . 2 pts - You almost got a bonus pt for thoroughness but no 189.
Zahi Yeitelman. . . . . . 2 pts - Good explanation and equations; 189 is best, ratio 10.5 yes.
Tim Boomer . . . . . . . . 2 pts - Good approach, a multivariable-calculus-constrained-min!
Allen Druze . . . . . . . . 2 pts - Nice proof using 100a+10b+c = x, a+b+c=y, min x/y.
Matt Situ . . . . . . . . . . 2 pts - A regular customer! How's my coffee? Yes, 189/(1+8+9)=10.5
Jorge Ramirez (new). . 2 pts - Thanks for taking the challenge! 199 is best but 3 diff. digits p.fav.
Payman Parang . . . . . 2 pts - You joined the 189 crowd. The 'no-leading-zeroes' group.
Anne Sufka . . . . . . . . 2 pts - 'After trial and error' you got the treasured 189; good trials!
 Uri Andrews. . 1 pt - 00c -> 1 ok but no repeated digits Fabio Garcia (new) 1 pt - 1 could be a ratio as in Uri's Padmaja (new). . 1 pt - Welcome! Needed more guesses Katherine Dostal . 1 pt - 109->10.9 pretty close. Alex Chan . . . . . . 1 pt - Three or four ain't bad!

Problem #137 - Posted Saturday, December 15, 2001
A shuffle of 2n cards puts the first n cards in the odd positions and the last n cards
in the even positions.[For example shuffle (1,2,3,4,5,6) and you get (1,4,2,5,3,6).]
Heather has 10 cards, 1-10, Briana has 12 cards, 1-12. Each shuffles her deck 2000
times. "Hey, my deck is back to its original state!" Who said that, and which card
does the other deck have in position #5?

Solution: Each shuffle is a permutation with a certain 'order', the smallest number of shuffles to get back
to the original order. Figure out the orders with 10 cards, and with 12 cards; only one goes into 2000.
I used Mathematica to write out the shuffles:
```f[{a1_, a2_, a3_, a4_, a5_, a6_, a7_, a8_, a9_, a10_}] :=
{a1, a6, a2, a7, a3, a8, a4, a9, a5, a10}
s = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ;
Do[Print[s]; s = f[s], {7}]```
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{1, 6, 2, 7, 3, 8, 4, 9, 5, 10}
{1, 8, 6, 4, 2, 9, 7, 5, 3, 10}
{1, 9, 8, 7, 6, 5, 4, 3, 2, 10}
{1, 5, 9, 4, 8, 3, 7, 2, 6, 10}
{1, 3, 5, 7, 9, 2, 4, 6, 8, 10}
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
After 6 shuffles Heather's 10-card deck is back to its starting position, so after 2000 shuffles,
since 2000 = 333*6 + 2 , it's in the state {1, 8, 6, 4, 2, 9, 7, 5, 3, 10} with a '2' in position 5.
But for a 12-card deck, here's the code:
```g[{a1_,a2_,a3_,a4_,a5_,a6_,a7_,a8_,a9_,a10_,a11_,a12_}] :=
{a1, a7, a2, a8, a3, a9, a4, a10, a5, a11, a6, a12}
s = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} ;
Do[Print[{i, s}]; s = g[s], {i, 1, 11}]```
{1, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}}
{2, {1, 7, 2, 8, 3, 9, 4, 10, 5, 11, 6, 12}}
{3, {1, 4, 7, 10, 2, 5, 8, 11, 3, 6, 9, 12}}
{4, {1, 8, 4, 11, 7, 3, 10, 6, 2, 9, 5, 12}}
{5, {1, 10, 8, 6, 4, 2, 11, 9, 7, 5, 3, 12}}
{6, {1, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 12}}
{7, {1, 6, 11, 5, 10, 4, 9, 3, 8, 2, 7, 12}}
{8, {1, 9, 6, 3, 11, 8, 5, 2, 10, 7, 4, 12}}
{9, {1, 5, 9, 2, 6, 10, 3, 7, 11, 4, 8, 12}}
{10, {1, 3, 5, 7, 9, 11, 2, 4, 6, 8, 10, 12}}
{11, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}}
This deck returns to start every 10 shuffles so it's Briana who said her 12-card deck
had returned after 2000 shuffles to its original state.

Moral: Everyone plays with a full deck; it's just that decks come in different sizes! ;-}

Nikita Kuznetsov . . . . 9 pts - You're the winner but I'd like to see the idea of 6 & 10 in there
Lisa Schechner . . . . . . 7 pts - Yes, a cycle of length 6 won't go into 2000; prove other's a 10?
Arthur Morris . . . . . . 5 pts - That's it, Briana said it after sets of 10 shuffles.
Ludwig Deruyck . . . . 4 pts - Adapted the HP48G+to a permutation problem-- wow!
Sudipta Das . . . . . . . . 4 pts - Woulda been 3 but bonus for first thorough perfect proof
Tim Poe . . . . . . . . . . . . 3 pts - Fixed up Heather's response in second answer- good save
Hermen Jacobs . . . . . . 3 pts - All the right reasons plus the right answer; good job.
Phil Sayre . . . . . . . . . . 3 pts - I'll have to look up the number theory reference - thanks!
Marcin Mika . . . . . . . 3 pts - Nice setup using recursive / periodic function notation
Quasi-C . . . . . . . . . . . . 3 pts - Throw the group theory jargon around; make me homesick! ;-}
Jack Dostal . . . . . . . . . 3 pts - nice answer among those showing all shuffles explicitly
Sue B . . . . . . . . . . . . . . 3 pts - Welcome back from your educational hiatus! Bzz!
Omar Gymboski . . . . 3 pts - Later answer but I like your observation abt 1 9 8 .. 2 10 after 3.
Katherine Dostal. . . . . 2 pts - Heather takes 6 yes but Briana is a 10 cycle not 5 I believe
Zahi Yeitelman. . . . . . 2 pts - Good job, (the answers from here on were a bit later than others)
Steve Lawrie . . . . . . . . 2 pts - Right, the 12 reverts to its orig pos. Oh, you said that.
Allen Druze . . . . . . . . 2 pts - Yes, Briana bragged and Heather shuffled; 2 in pos 5.
Drew . . . . . . . . . . . . . . 1 pt - Right about Briana, but the 7th shuffle same as first is 6-cycle
Anirban Bhattacharyya 1 pt - Good try! Heather ok but Briana takes 10 shuffles not 5

Problem #138 - Posted Wednesday, December 26, 2001
I just got twelve rare coins as a gift, all identical in apperance. I know one is a fake; it's
too light or too heavy (I'm not sure which). I do have a balance scale with two pans.
How can I tell which is the fake coin (and whether it's light or heavy) in only three weighings?

Solution: The best advice on these weighing problems is to divide the possibilities as equally as you can
depending on outcomes: there there are three things that can happen: balanced, heavy left, and heavy right.
There are 24 possible solutions: coin 1 heavy, coin 1 light, coin 2 heavy, . . . , coin 12 light.

The first step is to weigh four coins on each pan: 1,2,3,4 vs. 5,6,7,8. a) If they balance, then the fake coin
is 9,10,11 or 12. b) If it's heavy left then either one of 1,2,3,4 is heavy, or one of 5,6,7,8 is light.
c) If it's heavy right, then either one of 1,2,3,4 is heavy or one of 5,6,7,8 is light. Down to 8 possibilities!

The second step is to weigh three known coins against three suspect coins: in cases a,b,or c:
a) weigh 9,10,11 vs 1, 2, 3. If balance, then 12 is fake; a 3rd weighing vs coin 1 tells if 12 is light or heavy.
if heavy left, then 9,10 or 11 is heavy; weigh 9 vs 10 in 3rd weighing to tell which (or 11 if neither)
if heavy right then 9,10 or 11 is light; proceed as in heavy case. Now I'll use part of Steve's answer:

b) weigh 1,2,3,5 vs 4,10,11,12. If left is heavy then since 10,11,12 are normal, one of 1, 2, 3 is heavy,
We proceed as 9,10,11 above. If right is heavy then either 4 is heavy or 5 is light. Weigh 4 against any
normal coin for the 3rd weighing. If they balance, then one of 6,7,8 is light; proceed as above; done.
c) is just like b) only in reverse; you figure it out or read on for more on this story!
Here's Anirban's different approach: (See comment below)
We first divide the 12 coins into 4 groups A,B,C and D each containing 3 coins. Now we place two
groups A and B on one side of the scale pan and the two groups C and D on the other side of the scale
pan. Either : i > Side containing A and B is heavier. ii > Side containing C and D is heavier.
Now we place the two groups A and C on one side of the scale pan and B and D on the other side
of the scale pan. On weighing we might observe ; i> Side containing A and C is heavier.
ii>Side containing B and D is heavier. Based on the observations abtained from these two experiments
it is possible to deduce* which group contains the defective coin and whether the coin is heavier or lighter.
Let us suppose that A contains a heavier coin. Now we isolate group A containing 3 coins. Now we keep
away one coin from group A and check equilibrium between the two remaining coins. If there is equilibrium
then the coin we kept away is the heavier coin. Otherwise the pan weighing heavier contains the heavier coin.

Ok; Tim Poe has written in pointing out *Anirban's argument doesn't work. Tim is correct, and gets a bonus pt.
I won't take any points away from Anirban; it's my fault for not catching the error. Good try though, A.B.! - Dan
"The first part of the approach described is flawed. The results of the two weighings as described
would be the same if one of the A coins were heavy or if one of the D coins were light (or vice versa).
Similarly, if one of the B coins were heavy or one of the C coins were light (or vice versa), you would
get the same results through two weighings. The rest would be correct if the first part weren't flawed.
However, because you don't know which of the two possible scenarios you have, you can't be sure of
selecting the correct group to apply the final weighing to." - Tim

Tim Poe . . . . . . . . . . . . 11 pts - Nice to have old problems explained by a new generation!
Anirban Bhattacharyya 7 pts - I liked your idea of dividing into 4 groups of 3; see above.
Steve Lawrie . . . . . . . . 5 pts - You started out with the same notation as me so I borrowed!
Sudipta Das . . . . . . . . 4 pts - Good solution: '... second weighing HHLL vs GGHL ...'
Arthur Morris . . . . . . 3 pts - Great structure: If 1+2+3+4<5+6+7+8 then if 1+2+5<3+4+10...
Lisa Schechner . . . . . . 3 pts - You bet, I sprinkle in classic problems shamelessly!
Omar Gymboski . . . . . 3 pts - Nicely organized; if A=B and C>B then fake is in C, etc.
Zahi Yeitelman . . . . . . 2 pts - Right you are; 3 groups of 4 (or as we saw 4 grps of 3 also works)
Quasi-C . . . . . . . . . . . . 2 pts - Divide them 4 and 4, right, then you can find the fake.
Tim Bock . . . . . . . . . . . 2 pts - I like the technical terms such as 'depth-first method'...
Hermen Jacobs . . . . . . 2 pts - Four weighings isn't that easy either; bonus pt for new solution.
Allen Druze . . . . . . . . 1 pt - Slipped in 2 answers a few days post-Hermen, but ok!
Phil Sayre . . . . . . . . . . 1 pt - Bonus point for the timely reference; not sure Freeman X. Dyson

Problem #139 - Posted Tuesday, January 8, 2002
Many of my algebra and precalculus students think the 'inverse function' of f(x), often
written f^(-1)(x), is the same as the reciprocal 1/f(x) (mistaking the -1 for an exponent).
This (as I am obliged to remind them) is almost always false. But can you find at least
one function whose inverse is also its reciprocal? Tiebreaker: Find as many as you can!

Solution: There were three distinct types of correct answers on this one; I had to use my judgement!
The function f(x) = 1/x doesn't work, since f^(-1)(x) = 1/x while 1/(f(x)) = x. . . I had thought of the
power function, and I decided Art's answer fills the bill but loses the tiebreaker. -- Dan

1) Art Morris submitted this 'single point' function: f(1) = 1 ; at all other x , f(x) is undefined.

2) Hermen Jacobs found that if f(x) = x^a , then f^(-1)(x) = x^(1/a) while 1/(f(x)) = x^(-a) ,
so we solve 1/a = -a to get a^2 = -1 , so a = i or -i , where i = Sqrt[-1]. Thus f(x) = x^i or x^(-i).
Wait patiently for Dan's new lesson on complex number polar geometry so you can figure out
what the heck (a + b i)^(c + d i) means!
 3) Jack Dostal astounded me with examples of real-valued functions defined on discontinuous intervals; I made a little picture at right -->   Here's the function (in black; slight correction on Jack's first interval) f(x) = {(2x+1)/2 if 0 < x < 1/3 , 2/(2x-1) if 1/2 < x < 5/6 , . . . . . {(2-x)/2x if 6/5 < x < 2 , 2x/(x+2) if x > 3}   Then any input x=a, where a is in any of the four intervals, will follow a 4-cycle path : a --> f(a) --> f(f(a)) --> f(f(f(a))) --> f(f(f(f(a)))) = a ; (shown from blue ---> red): additionally f(f(x)) = 1/x (in gray). Ref: Euler/Foran, Math Magaz Sep1981; Cheng/Dasgupta, Amer Math Monthly Oct1998   4) Lisa Schechner sent in a 4-point solution (see below) that has a similar cycle!

Hermen Jacobs . . . . . . 9 pts - Nice answer on second try; your wife is wise! (think first write later)
Jack Dostal . . . . . . . . . 8 pts - Great answer! I tried making a GIF animation but it lost color.
Arthur Morris . . . . . . 6 pts - Good "pinpoint solution" {(1,1)} is its own inverse and its recip!
Tim Boomer . . . . . . . . 5 pts - Yes, try various types of funcs; logs, trigs, exps, then powers!
Lisa Schechner . . . . . . 3 pts - Cool: {(1/3,2),(1/2,1/3),(2,3),(3,1/2)} inverse = recip indeed.
Quasi-C . . . . . . . . . . . . 2 pts - A little more time might have paid off; 1/x and -1/x don't work
Allen Druze . . . . . . . . 2 pts - Nice try y = (x+a)/(x-a) but no consistent a. Try y = (ax+b)/(cx+d).
Phil Sayre . . . . . . . . . . 2 pts - Interesting thought; (x,y)->(x,y)/(x^2+y^2) inversion but inv func?
Anirban Bhattacharyya 1 pt - You want f(x)*f^(-1)(x) = 1 but y = 1/x doesn't make that true.
Uwe Buenting (new) . . . 1 pt - Welcome to my contest; see above solutions, I'm not always this slow!
Derek Chin . . . . . . . . . 1 pt - Those are equal to their inverse, not reciprocal, but thanks for entering!

Problem #140 - Posted Monday, January 28, 2002
As a swimmer jumps off a small bridge and begins to swim upstream,
her swim cap comes off and floats downstream. Ten minutes later she turns
around, swimming downstream with the same effort, past her original bridge.
At the next bridge, 1000 meters away from the first, she catches the cap.
What was the speed of the current? Of the swimmer?

Solution: Many of you took the point of reference as the cap (or the water).
This is a good approach, as Hermen points out, with a decidedly European train analogy:
"The speed of the current is (1000 meter in 20 minutes) 50 m/min. There are not enough data to resolve the
second question about the speed of the swimmer. (Dan's note: The speed of swimmer must be more than 50 to pass orig bridge.)
"As far as the reasoning is concerned: let me show by the following exemple. My wife and I are sitting in a runing train.
I walk to the WC, that takes me 10 minutes and back to my wife again 10 minutes. In the meantime (20 minutes) the
train has covered 1000 m (not so fast!). The speed of the train is 1000/20= 50 m/min. It is irrelevant whether I walked
or ran to the toilet. You can not say anything about my speed." (Dan's note: But I bet I know what you had in the dining car !)

Arthur Morris . . . . . . 10 pts - See the cap, be the cap, think like the river, win the week!
Steve Lawrie . . . . . . . . 8 pts - Good to notice S drops out of eqns. Bonus pt: near same time as Art.
Hermen Jacobs . . . . . . 6 pts - Love the train example, also bonus pt for ternary new answer to #138.
Anirban Bhattacharyya 5 pts- Yes, v > u so u = current = 50 --> swmr goes any speed faster.
Tim Poe . . . . . . . . . . . . 4 pts - Right, 'same effort' means 10 mn out, 10 mn back, x = any > 50.
Quasi-C . . . . . . . . . . . . 3 pts - You 'seconded the motion' I'll accept 5/6 m/sec and any s > 5/6.
Nikita Kuznetsov . . . . 2 pts - Asked "If she wanted to swim why not go to the beach?" Hmm!
Allen Druze . . . . . . . . 2 pts - Nice set of equations and solution, n = 10 mn, y = 50 m/mn
Jack Dostal . . . . . . . . . 2 pts - All ok except the swimmer can't float up and back under orig bridge.
Phil Sayre . . . . . . . . . . 2 pts - Correct, dist=veloc*time holds the key, the Vs cancels obligingly.
Myung Hwa Yang (new) 2 pts - Welcome to the contest; You are right not to know swimmer's dist.
Lisa Schechner . . . . . . 1 pt - I was with you up to T=10, but U=50 not 20, true any v but > 50.
Sudipta Das . . . . . . . . 1 pt - A few days after others, but largely right c = 50 but need s > 50.
Zahi Yeitelman . . . . . 1 pt - Yes, current is 50, good use of algebra; also need s > 50 to work
Uri Andrews . . . . . . . 1 pt - Welcome back! c(t+10)=1000 is compact and leads to c=50.

 THANKS to all of you who have entered, or even just clicked and looked. My site is now in its fifth season - OVER 28,000 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.

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