**dan's math@home - problem of the week - archives****Problem Archives**page 14**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 index131 - 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
- Back at the 7-11! (back to top)
- 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!!

**WINNERS - Problem 131 .**(back to top) . leader board .- @ 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)**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**. . . . . .**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**.**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
- The Top Five (back to top)
- 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)."

**WINNERS - Problem 132 .**(back to top) . leader board . LOTS of answers; I think it's a record response!**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**. .**Sudipta Das**. .**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**.**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
- Walk, Eat, Bike (back to top)
- 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!

**WINNERS - Problem 133 .**(back to top) . leader board .**Steve Lawrie**. . . . . . . 10 pts - Good approach ; thanks for the 7.11 answer as well! (see above)**Sudipta Das**. .**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**.**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**. .**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
- A Peachy Quintet (back to top)
- 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.

**WINNERS - Problem 134 .**(back to top) . leader board .**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**. .**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**.**Hermen Jacobs**. . . . . . 4 pts - Thanks for carrying & thinking about my problem throughout Holland!**Michelle Lee**. .**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
- Reciprocal Pairs (back to top)
- 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

**WINNERS - Problem 135 .**(back to top) . leader board .**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**. .**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**. .**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**.**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)**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
- How Low Can You Go ? (back to top)
- 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! ;-)
- Here's Derek's answer:
- "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

**WINNERS - Problem 136 .**(back to top) . leader board .**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**.**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**. .**Nikita Kuznetsov**. . . . 2 pts - Right, answer is 1.9 for 019; you can't go lower.**Sudipta Das**. .**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
- 2000 Shuffles ! (back to top)
- 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!**;-}

**WINNERS - Problem 137 .**(back to top) . leader board .**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**. .**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**.**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
- The Twelve Coins (back to top)
- 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

**WINNERS - Problem 138 .**(back to top) . leader board . Regulars & other contestants - thanks for your patience! - Dan**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**. .**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**.**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
- A Funny Inverse ! (back to top)
- 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!

- 3)

**WINNERS - Problem 139 .**(back to top) . leader board .**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**Lisa Schechner**. . . . . . 3 pts - Cool: {(1/3,2),(1/2,1/3),(2,3),(3,1/2)} inverse = recip indeed.**Quasi-C**.**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)**Derek Chin**. . . . . . . . . 1 pt - Those are equal to their inverse, not reciprocal, but thanks for entering!- Problem #140 - Posted Monday, January 28, 2002
- Swimmer Loses Cap ! (back to top)
- 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 !)

**WINNERS - Problem 140 .**(back to top) . leader board .**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**.**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**. .**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 **- Dan the Man Bach**- 2*7*11*13 A.D.

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