- Problem #201 - Posted Sunday, May 9, 2004
- Couples Buy Books (back to top)
- Two men, David and Clifton, and their wives, Kim and Allison, go out shopping
- for books. Each person paid for each book a number of dollars equal to their
- number of books. David bought 1 more book than Kim, while Allison bought
- only 1 book. If each couple spent the same two-digit sum, who is Kim's husband?
- Show reasoning carefully.
- Couples Buy Books (back to top)
- is a perfect square (of the number
of books). So we seek a 2-digit number that's the sum of two
- squares in two different ways. This has come up before if you're a longtime fan of this site; the
- nos. are 50 = 1 + 49 = 25 + 25 ; 65 = 1 + 64 = 16 + 49 ; 85 = 4 + 81 = 36 + 49. Of these, the
- last one doesn't allow for Allison's 1 book, and the first one doesn't involve 2 consecutive squares.
- Thus a^2 + d^2 = c^2 + k^2; 1^2 + 8^2 = 4^2 + 7^2; a=1, d=8, c=4, k=7. Kim is with Clifton.
- Without the 2-digit sum requirement there would be no definitive answer...
- There are Diophantine equations attached to this: a=1, d=k+1; Is Kim married to David or Clifton?
- (1) if K is with D then S = (k+1)^2 + k^2 = 1^2 + c^2 ; 2k^2 + 2k = c^2 ; 2k(k+1) = c^2 ;
- squares in two different ways. This has come up before if you're a longtime fan of this site; the
- sols are k=1, c=2, S=5, too small; k=8, c=12, S=145, too big for 2-digit req'm't.
- (2) if K is with C then S = 1^2 + (k+1)^2 = c^2 + k^2 ; 2(k+1) = c^2 ;
- sol's are k=1, c=2, S=5; c=4, k=7,
S = 65 THIS WORKS! ; c=6, k=17, S=325, too big.
- The only one that gives a 2-digit sum is S = 65, in which Kim is married to Clifton.
- Drew
. . . . . . . . . . . . . . . 10 pts - Another first-place first-timer;
movin' up fast on the list
- Nick McGrath . . . . . . . . 7 pts - This year's leader with another great answer, keep it up!
- Denis Borris. . . . . . . . . . 5 pts - Good steps to reach '2 a^2 + 1 must be a perfect square.'
- Fanfan . . . . . . . . . . . . . . 5 pts - I like the connection (& link) w/ square triangular numbers
- Radu Ionescu . . . . . . . . 4 pts - Woulda been 5 or more pts without the resub alternative
- Art Morris . . . . . . . . . . . 4 pts - Four squares, incr. M until M^2 - 1 - diff of sqs; works!
- Tim Poe . . . . . . . . . . . . . 4 pts - Yes, the three ways listed above will exhaust the squares.
- Jeremy Galvagni . . . . . . 4 pts - Clear two pairings approach; liked the "Pell's Eqn' tag in there.
- Quasi-C. . . . . . . . . . . . . . 3 pts - Could pe paired up the other way if they'd had more money!
- Hermen Jacobs. . . . . . . . 3 pts - Good way to put it; if k > 2 and D with K then C =/= square.
- Ed Wern. . . . . . . . . . . . . 3 pts - K is with C because of a 2-digit bingo game? Oh I see.
- (S.G.) Roy . . . . . . . . . . . . 3 pts - Nice job noting that k=8, d=9 works but is out of range.
- Marcello Cammarata . . 3 pts - If Dave and Kim then $5 or $145, so Kim with Clifton.
- Mike Kosciuk . . . . . . . . 3 pts - Thanks for the vote of approval, and good job yourself!
- Phillipe Fondanaiche . . 3 pts - Oui, la morale est sauve, mais le monde n'est pas sauve'!
- Jack Dostal . . . . . . . . . . 3 pts - Two possible arrangements but there's only one Clifton
- Phil Sayre . . . . . . . . . . . 3 pts - That's it, the difference is $80 between possible couples
- Prachai K. . . . . . . . . . . . 3 pts - I'll try to put in some tougher problems from time to time!
- Vince LoCascio . . . . . . . 3 pts - Right; k^2 + (k+1)^2 = 1^2 + c^2 is "2-limited" a total.
- Ken Duisenberg. . . . . . . 2 pts - Wouldn'ta been conclusive without the 2-digit restric'n.
- Ravi Raja. . . . . . . . . . . . 2 pts - Need d=k+1 to be paired with 1 not c^2 to make same total
- Ajit Athle. . . . . . . . . . . . 2 pts - Your tabular formulas needed motivation; conclu. was ok
- Albert Alvarez . . . . . . . 2 pts - Good solution, AA, just chronologically challenged...
- Alan O'Donnell . . . . . . 2 pts - The $145 solution is partly ok but restricted by 2-digit req't.
- Monika Reynolds . . . . 2 pts - Nice solution, well laid out, just arrived later than most.
- Shanthy Krishnamurthy (new) 1 pt - Welcome! Totals are squared bec of price per book.
- Yakov Macak . . . . . . . . 1 pt - Your nos. work but couples got mixed up - could be serious!
- Divyanshu Mishra (new) 1 pt - Welcome to my contest; remember to explain reasoning!
- Allen Druze. . . . . . . . . . 1 pt - Your eqn. cases seem appropriate but the spouse swap isn't.
- Nick McGrath . . . . . . . . 7 pts - This year's leader with another great answer, keep it up!
-
- Problem #202 - Posted Thursday, May 20, 2004
- Magic Square Primes (back to top)
- Choosing nine distinct integers from 1-25, make a magic
- square with as many primes as possible. (The example has 4
- primes but you can do better. Note there are 9 primes from 1-25: 2,3,... ,23)
- The winner will be determined using the following comparisons:
- 1) The most primes, 2) Smallest magic constant , 3) order received.
- A magic square has rows, columns, and main diagonals that add
- up to the same thing, the "magic constant."
- Show reasoning carefully.
- Problem #202 - Posted Thursday, May 20, 2004
-
- "The following square contains
7 primes
with a constant 33: (17,
7, 9) (3,11,19) (13,15,5). Remembering
that
- the sum of the nine numbers must be divisible by 9, that the magic constant C is one third the sum, that the central value
- a22 is one-third the magic constant, that a11 (or any other corner element) must be less than 2 a22 and that either a13 and
- a31 summed to a11 must be less than C, all magic squares with a given central value can be built without too much effort.
Moreover, no 9 primes square can exist, because 2 should be included, and a 3x3 magic square either cannot contain- even numbers or it must contain at lest three (in order to have all the sums of the same parity). Any 8 primes square
- should contain all eight odd primes; now, the sum of these primes is 98, and the only ninth number that makes the sum
- divisible by 9 is 1 (10 is even and 19 would be repeated). Such a square would have a constant C = 33, a central value
- a22=11, and cannot therefore contain the prime 23. It follows that the required square cannot contain more than 7 primes.
I built magic squares starting with a22=5, and then checking for a22 = 7, 9, 11 ? (since I was looking for 7 primes magic- squares, central even values could be skipped). The first one I encountered was (17, 7, 9) (3,11,19) (13,15,5) as above,
- and since any 7 primes magic square with a22 > 11 would have a constant greater than 33, no further search was needed."
- Tim Poe and Philippe F. also found the smallest known totally-prime square: (17,113,47) (89,59,29) (71,5,101); S = 177
- Everyone should know that 1 is never considered a prime by a mathematician, only by Carl Sagan (in the book 'Contact')
- the sum of the nine numbers must be divisible by 9, that the magic constant C is one third the sum, that the central value
- Yakov Macak
. . . . . . . . 10 pts - Third first-time winner in a row. Go
Aussies; good on ya!
- Marcello Cammarata . . 8 pts - A bonus point for one of the only complete proofs of bestness!
- Quasi-C. . . . . . . . . . . . . . 6 pts - Good reasoning about 9 and 8 primes being impossible.
- Denis Borris. . . . . . . . . . 5 pts - Got the above solution S=33; threw in 2 more with S = 39
- Vince LoCascio . . . . . . . 5 pts - Yes, the 9 primes add to 100, which is not a mult of 3 (or 9)
- Fanfan . . . . . . . . . . . . . . 4 pts - Seven primes do constitute a magic-nearly-prime square!
- Nick McGrath . . . . . . . . 4 pts - I also was (just) in the slide rule and log-table generation.
- S Khoo . . . . . . . . . . . . . . 4 pts - Correct ans; bonus pt for incl all 180 w/7 pr and one with 0.
- Prachai K. . . . . . . . . . . . 3 pts - Used 4 variables in key locations and others in terms of em
- Shanthy Krishnamurthy 3 pts - Nice to see 'median prime is 11' and build around that!
- Ajit Athle. . . . . . . . . . . . 3 pts - Your answer was short and sweet and I enjoyed reading it
- Phil Sayre . . . . . . . . . . . 3 pts - Thanks for liking this problem; 8 eqs in 10 unks for 9 nos.
- Ed Wern. . . . . . . . . . . . . 3 pts - The winner (first in) among the S = 39 crowd; right about 2
- Tim Poe . . . . . . . . . . . . . 3 pts - Woulda been 2 with resub but I liked the 9-prime square
- Art Morris . . . . . . . . . . . 2 pts - 6 primes + 1 resub + 1st entry = 2 pts; the 'new math'!
- Drew . . . . . . . . . . . . . . . 2 pts - Forcing primes in corners may not have been smallest S
- Jeremy Galvagni . . . . . 2 pts - My favorite mini-computer too - TI-83; several w/5,6,7 primes
- Hermen Jacobs . . . . . . . 2 pts - A long-time magic-sqare playa - also mentioned S=15 basic 1-9
- Phillipe Fondanaiche. . 2 pts - L'unite n'est pas un premier (1 is not prime) but resub was good
- Allen Druze. . . . . . . . . . 2 pts - Good reasoning starting with 1-9 then expanding to arith prog
- Carl Chenard. . . . . . . . 2 pts - Welcome back - got 7 primes with sums 39 or greater...
- Alan O'Donnell . . . . . . 1 pt - 6 primes was better than my demo but not maximal or early
- Marcello Cammarata . . 8 pts - A bonus point for one of the only complete proofs of bestness!
-
- Problem #203 - Posted Wednesday, June 2, 2004
- Where's The Crowd? (back to top)
- My friend's music fan website started out fast: lots of visitors the first week.
- But every week, there were fewer and fewer; the second week there was one
- more visitor than 2/3 of the first week, the 3rd week there were 2 more than
- 2/3 of the second week, the 4th week there were 3 more than 2/3 of the 3rd
- week, and so on, until the final week, when there were just 533. How many
- weeks was the website up, how many visitors did it get the 1st week, and how
- Problem #203 - Posted Wednesday, June 2, 2004
- If there are n weeks and at the end
of the first week there were x visitors then we have at the end
of each week:
Week 1 : x . . . . . . Week 2 : (2x+3) / 3 . . . . . . Week 3 :(4x+ 24) / 9 . . . . . . Week 4 : (8x+129) / 27
Week 5:(16x+582) / 81 . . . . . . Week 6: (32x + 2379) / 243 . . . . . . Week 7: (64x+ 9132) / 729
Week 8 : (128x + 33573) / 2187 . Week 9: (256x + 119634) / 6561 ..Week 10: (512x + 416415) / 19683
. . . Week n: (2/3)^(n-1)* x + 1*(2/3)^(n - 2) +2*(2/3)^(n 3) +...+ (n -1)*(2/3)^0
It is clear that solutions are possible only at the end of even number of weeks since 533*3^(n -1) will- produce an odd number from which an odd number should be subtracted so that the resultant will be
- divisible by 2^(n-1). Thus, we set 2x/3 + 1 = 533, which gives us our first solution x = 798.
If there are 798 visitors at the end of 1st week. then there will be 533 visitors at the end of
second week and in all there would be 1331 visitors with the site surviving only for 2 weeks..
The second solution occurs with x = 19677,and the site is up for 10 weeks with the following position:- Week 1: 19677 visitors . .. Week 2: 13119 . . . . . Week 3: 8748 . . . . . Week 4: 5835 . . . . Week 5: 3894
- Week 6: 2601 . . . . . Week 7: 1740 . . . . . Week 8: 1167 . . . . . Week 9: 786 . . . . . Week 10: 533
- Total Number of Visitors: 58100 - Solution 1: Two Weeks, 798, 533, total 1331 (goes against problem stmt- Dan)
Solution 2: Ten Weeks, 19677 first week and 58100 total.- (Alan O'D and Drew both kept going past 533 and noted that visitors bottomed out at week 20, then slowly
- gained in popularity, 3 hits per week, until reaching 19677 again in about 126 years! - Dan)
- produce an odd number from which an odd number should be subtracted so that the resultant will be
- Ajit Athle.
. . . . . . . . . . . 9 pts - Woulda been 10 but a late resub
and equal trmt for 2 wks
- Alan O'Donnell . . . . . . 6 pts - Again thought of n=2 wks and x1 = 798 as allowable
- Tim Poe . . . . . . . . . . . . . 5 pts - Thanks for direct approach to our tens of thous of visitors
- Fanfan . . . . . . . . . . . . . . 5 pts - Good answer and proof that there are no larger solutions
- Hermen Jacobs . . . . . . . 4 pts - Later resub was no help so not needed, orig ans fine!
- Art Morris . . . . . . . . . . . 4 pts - That's a good way to look- backwards! Or is it backward?
- Marcello Cammarata . . 4 pts - Keep on sending those solutions in, all the way from Italy!
- Quasi-C. . . . . . . . . . . . . . 3 pts - Let N = number of contestants this week... then N > 4!(!)
- Radu Ionescu . . . . . . . . . 3 pts - Welcome back to my little contest; good answer!
- Denis Borris. . . . . . . . . . 3 pts - Yes I love monkey and coconut problems. Answer: -4.
- Ed Wern. . . . . . . . . . . . . 3 pts - Excel rules again; 9 weeks correct if you start with 0.
- Nick McGrath . . . . . . . . 3 pts - Good recursion: proved #wks is even, then 2 mod 4, etc.
- Carl Chenard. . . . . . . . . 3 pts - I liked yr comment abt fractional people being not all there
- Allen Druze. . . . . . . . . . 3 pts - Was that 3n - 3 + 9(2/3)^n+2 +((2/3)^n+1)x = 533?
- Jeremy Galvagni . . . . . . 3 pts - A new TI-83 func: sum(seq(2/3 y + k)) or something
- Drew . . . . . . . . . . . . . . . . 3 pts - You & Alan figured another 19677 hits the 6562nd week!
- Prachai K. . . . . . . . . . . . 2 pts - Figured out that 512 = 2^9 might lead to 3^9 - 6 right ans
- Mark Rickert (new) . . . . 2 pts - Spreadsht isn't really cheating, it's the weapon of choice for some.
- S Khoo . . . . . . . . . . . . . . 2 pts - That's a good way to describe the inverse func: reverse engrg.
- Vince LoCascio . . . . . . 2 pts - Correct that wording implied n>3 then brute force takes over
- Sumita Das (new) . . . . . 2 pts - Welcome! Yes 3^9 - 6 (=?) is best answer since k > 2.
- Phillipe Fondanaiche. . 2 pts - Merci pour me donner la solution en francais! Bien fait, pas tot.
- T Miller (new) . . . . . . . . 2 pts - You are correct sir. Welcome to my contest, see you again.
- Lisa Schechner . . . . . . . 2 pts - Welcome back Lisa! The difference equation seemed to work.
- Phil Sayre . . . . . . . . . . . 2 pts - Yep, you're right that 798 'works' but wasn't intended.
- Ken Duisenberg . . . . . . 2 pts - Good idea; check all Nk-1 = 3/2(Nk - k)+1 are integers
- Jack Dostal . . . . . . . . . . 2 pts - Start with Vn and work backwards . . . is right!
- Charleen Yen . . . . . . . . 1 pt - I followed your program but didn't see it run or give output.
- Zahi Teitelman . . . . . . 1 pt - Only n=2 solution , n < 3 was discouraged by wording
- Daniel Baczkowski (new) 1 pt - Thanks for entering; some entries ahead of you but good start!
- Alan O'Donnell . . . . . . 6 pts - Again thought of n=2 wks and x1 = 798 as allowable
- Being a perfect square is
a relative thing. If these
are, find their square
roots in their
- domains (m, n are integers); if not, prove they can't be squares of the given form.
- Is a = 205879238043281 the square of a regular integer? What about b = 205889238043281 ?
- Is c = 1357 + 874 i the square of a Gaussian integer m + n i ? What about d = 573833 + 595944 i ?
- Is e = 275643 + 187110 \/ 2 the square of any m + n \/ 2 ? What about f = 41289 + 43589 \/ 2 ?
- Is g = 71257 + 40976 \/ 3 the square of any m + n \/3 ? What about h = 1967856 - 1136159 \/ 3 ?
- Show reasoning carefully. One point penalty for resubmissions.
- domains (m, n are integers); if not, prove they can't be squares of the given form.
- Fanfan:
Notice that a is a multiple of 11 but not of 11^2 (No,
a is not a square)
- 14348841^2 = 205889238043281 = b is an integer square
- (answer can also be found by factoring b = 3^2 x 13^2 x 127^2 x 2897^2)
- Quasi-C: Is c = 1357 + 874 i the square of a Gaussian integer m + n i ? No.
- If c = (a + bi)^2 = a^2 b^2 + 2*a*b * i we have a*b = 437 (437 = 19*23)
- a and b must be both odd (1 and 437 or 19 and 23) so a^2 b^2 is even therefore <> 1357.
- Nick McG: m^2 - n^2 = 573833 and 2mn = 595944 Writing a little looper in BASIC we find:
d = 573833 + 595944 i = ( 837 + 356 i ) ^ 2- Ajit: e = (315 + 297 \/2)^2 is a perfect square found by calculating at www.1728.com
- As before, f =(m+n\/2)^2 = m^2 + 2n^2 + 2mn \/2 then 2mn = 43589 ... (odd) impossible
- Sumita: The square of any (m + n sqrt(3)) is given by m^2 + 3n^2 + 2mn sqrt(3).
g = 71257 + 40976 sqrt(3) = 71257 + 2*(2^3)*13*197 sqrt(3) Possible (m,n) can be:- (1,20488), (2,10244), (4,5122), (8,2561), (13,1576), (26,788), (52,394), (104,197), (197,104), (394,52),
- (788,26), (1576,13), (2561,8), (5122,4), (10244,2), (20488,1) m^2+ 3n^2 = 71257 only for (m,n) = (197,104)
- So sqrt(g) = 197 + 104 sqrt(3). h = 1967856 - 1136159 sqrt(3) cannot be a perfect square
- for integer m and n, as 43589 is not divisible by 2. (Also the real number h is negative - Dan)
- 14348841^2 = 205889238043281 = b is an integer square
- Marcello Cammarata . 10 pts - Also gave quotable results, good
job and prompt!
- Art Morris . . . . . . . . . . . 7 pts - All-time leader finds limits of MATLAB and value of pencil
- Fanfan . . . . . . . . . . . . . . 6 pts - Impressive methods including rings Z[\/2], norms, etc.!
- Alan O'Donnell . . . . . . 5 pts - Good answers and thanks for summarizing; hello back.
- Ed Wern. . . . . . . . . . . . . 4 pts - Yes, "squares have pairs" (of prime factors). Nice slogan!
- Quasi-C . . . . . . . . . . . . . 4 pts - Quotable Quasi strikes again. Nice I gave UFDs, eh?
- Nick McGrath . . . . . . . . 4 pts - Nice way to put it: 11 divides a but not 11^2. Et cetera.
- Tim Poe . . . . . . . . . . . . . 3 pts - Short on details but nailed all the answers fair & 'square'
- Sumita Das . . . . . . . . . . 3 pts - Good discussion, thanks. Managed to use norms for Z[i]
- Phil Sayre . . . . . . . . . . . 3 pts - Good on e: use quad formula; disc=209522032, not a sq.
- Ajit Athle . . . . . . . . . . . 3 pts - Wonderful site for any kind of calculation: www.1728.com
- Zahi Teitelman . . . . . . . 3 pts - Nice program that works on c-h. Good idea Zahi.
- Prachai K. . . . . . . . . . . . 3 pts - Solving m2-n2=A, 2mn=B was popular and effective.
- Jeremy Galvagni . . . . . . 3 pts - We'll check out MIRACL calc. and see if we like it!
- Albert Alvarez . . . . . . . 3 pts - Welcm bak AA. Good plan, take real sq root or complex
- Jack Dostal . . . . . . . . . . 3 pts - Used a Pyth Triple idea on c: one is mult of 3, other of 4
- Denis Borris. . . . . . . . . . 3 pts - This was an example of an easy one? Not for some of us!
- Vince LoCascio . . . . . . . 2 pts - Came closer in resub; still could have m2-n2 = odd-odd.
- Drew . . . . . . . . . . . . . . . . 2 pts - Missed out on your initial problem, D. Otherwise good.
- Hermen Jacobs . . . . . . . 2 pts - Finally found a computer in your travels. Go Rabobank!
- Allen Druze. . . . . . . . . . 1 pt - Some of the roots were right, some were left (out).
- Art Morris . . . . . . . . . . . 7 pts - All-time leader finds limits of MATLAB and value of pencil
- Problem #205 - Posted Monday, June 28, 2004
- Orange Pyramids (back to top)
- The grocery person likes to stack the oranges
- in tetrahedral piles; triangular-based pyramids.
- Each orange is exactly 10 centimeters in diameter.
- a) Exactly how tall (in cm) is a pile of 10 oranges ?
- b) How many oranges are in a pyramid of n layers ?
- c) Exactly how tall is that pile of n layers of oranges ?
- Show reasoning carefully.
- Orange Pyramids (back to top)
- Problem #205 - Posted Monday, June 28, 2004
- The height of a regular tetrahedron
with side lengths a is a sqrt(6)/3: this follows from
- applying the Pythagorean theorem twice : once to find the height f of a triangular face,
- which gives f^2 + (a/2)^2 = a^2 => f = sqrt(3)/2 a ; once to find the height h of the
- tetrahedron, which gives (f/3)^2 + h^2 = f^2 => h = sqrt(8)/3 f = sqrt(6)/3 ~ 0.8165
- (continued by Radu Ionescu) A tetrahedral with edge L has height H=L*sqr(2/3) ;
- [total is] 2R+4Rsqr(2/3) = 10(1+2sqr(2/3)) ~ 26.33
- b) How many oranges are in a pyramid of n layers ? (also from Radu)
- 1 + (1+2) + (1+2+3) + (1+2+3+4) +... + (1+2+3+......n) =
1*2/2 + 2*3/2 + 3*4/2 + 4*5/2 +... + n(n+1)/2 =
1*2*3/6 + (2*3*4-1*2*3)/6 + (3*4*5-2*3*4) + ... + (n(n+1)(n+2)-(n-1)n(n+1)) = n(n+1)(n+2)/6- c) Exactly how tall is that pile of n layers of oranges ? (See part a for method)
- 2R+2R(n-1)sqr(2/3)=10(1+(n-1)sqr(2/3)) (Also 10 + (10/3)(n-1)\/ 6 works.)
- applying the Pythagorean theorem twice : once to find the height f of a triangular face,
- Jeremy Galvagni . . . . . 10 pts - Yes,10+(n-1)h; Player-of-the-Week!
How's it feel, JG?
- Tim Poe . . . . . . . . . . . . . 7 pts - You coulda proved the height of a tetra if you'd wanted to
- Art Morris. . . . . . . . . . . 5 pts - Using 3D co-ords was refreshing to calc ht of center ball
- Radu Ionescu . . . . . . . . 5 pts - Nice answer, and early on in the contestant rush hours!
- Nick McGrath. . . . . . . . 4 pts - Nice form 10((2/3)\/6 + 1); answer seemed ok w/o typo fix
- Marcello Cammarata. . 4 pts - Yes, 26,33 cm; I'm trying to get the US to go metric too.
- Fanfan . . . . . . . . . . . . . . 4 pts - Thanks for recurring the Pythm and the formula for T(n) too
- Ed Wern. . . . . . . . . . . . . 3 pts - Early entry partly ok but ht involves more than \/ 3 .
- Quasi-C . . . . . . . . . . . . . 3 pts - No problem with height formulas but 'closed form' too open
- Denis Borris. . . . . . . . . . 3 pts - Height needs \/ 2 as well; thanks for space-filling percent
- Phillipe Fondanaiche . . 3 pts - Great answer all parts mais un peu plus en retard qu'les autres.
- Yakov Macak . . . . . . . . . 3 pts - Glad you enjoyed a bit of first-place glory; good answer
- Phil Sayre. . . . . . . . . . . . 3 pts - Nice thorough answer; flowed well incl. induction proof
- Mark Rickert . . . . . . . . . 3 pts - That's it; 3 layers 10 + 20\/(2/3) ; n layers n(n+1)(n+2)/6
- Ajit Athle. . . . . . . . . . . . 2 pts - Height of 10 + 10 \/ 3 isn't that far off as it turns out.
- Hermen Jacobs. . . . . . . . 2 pts - Almost correct height of (20/3) \/ 6 ; just left off 2 fives.
- Drew . . . . . . . . . . . . . . . . 2 pts - Good try; first part may have been 10 layers not 10 orngs
- Albert Alvarez. . . . . . . . 2 pts - Again, 10 + 10 \/ 3 isn't half bad, in fact it's within 5%.
- Prachai K. . . . . . . . . . . . 2 pts - I'm interested in the factor of \/ 2 / 2 usually for 45deg ang.
- Zahi Teitelman . . . . . . . 1 pt - May have been trying to fit a grape under the 3 oranges?
- Tim Poe . . . . . . . . . . . . . 7 pts - You coulda proved the height of a tetra if you'd wanted to
-
- Problem #206 - Posted Tuesday, July 13, 2004
- Given the GCD and LCM (back to top)
- A while ago, I gave my Prealgebra class a puzzle problem: Given two natural numbers,
- m and n. If their GCD is G=6 and their LCM is L=72, what are the numbers ?
- I was surprised when more than one correct answer came in! a) What were all possible
- (m, n) for G=6 and L=72 (m<=n) ? b) What's the smallest sum, m+n, for any (m, n) pair
- that share the same G=gcd and L=lcm (with another m<=n; G>=2)? c) Find the (G, L) pair
- with the most solutions (m, n) for the same G=gcd(m,n) and L=lcm(m,n) (G>1, L < 1001).
- GCD: greatest common divisor ; LCM: least common multiple. Show reasoning carefully. (Answer/winners here soon)
- Problem #206 - Posted Tuesday, July 13, 2004
- a) G=6 and L=72 if (m, n) = (6,
72) or (18, 24) Note 72/6 =
12 has 2 distinct prime factors.
- b) G = 2 and L = 12 for (m, n) = (2, 12) ~ (4, 6) ; the smallest sum m+n is 4+6 = 10.
- c) G I had no idea I didn't have the best solution with (L, G) = (6, 720) (4 solutions) - Dan
- As Fanfan states, "Let m = Ga, n = Gb. For L/G with k different prime factors, the (m, n) can
- be done in 2^k different ways (simply choose which pi^ai go to a, and the others go to b, none
- can be split between a and b). If we assume m <= n, hence a <= b, we have to remove half of the
- solutions, and we get 2^(k-1) pairs (m,n). To get many solutions, L/G must have many different
- prime factors (exponents do not play any role). Since G>=2 & L<=1000, we need L/G <= 500.
- It is then obvious that the best L/G is equal to 2x3x5x7 = 210 (a fifth prime factor would lead
- to 2x3x5x7x11 = 2310), with [k=4] and thus 2^3= 8 solutions : we can choose for example
- (G, L)=(2, 420) and get (2,420), (4,210), (6,140), (10,84), (12,70), (14,60), (20,42), (28,30).
- Tim P. found all (G,L) = (4,840), (3,990), (3,630), (2,924), (2,840), (2,780), (2,660), and (2,420)
- each of the 8 (G, L) pairs has 8 (m, n) solutions with G = gcd(m, n) ; L = lcm(m, n) ; m <= n .
- b) G = 2 and L = 12 for (m, n) = (2, 12) ~ (4, 6) ; the smallest sum m+n is 4+6 = 10.
- Fanfan
. . . . . . . . . . . . . . 10 pts - That was an impressive rollout
of number theory.
- Ed Wern. . . . . . . . . . . . . 7 pts - Yes 96 was a boo-boo, and b was unclearly worded too
- Tim Poe . . . . . . . . . . . . . 6 pts - Nice job, extra point for being first to lisy all 8 with 8.
- Nick McGrath. . . . . . . . 5 pts - Right; (6, 72) and (18, 24) ok but (12, 36) fails G=6.
- Marcello Cammarata. . 4 pts - Yup m close to n has smallest sum, and 2^(k-1) sol'ns
- Denis Borris . . . . . . . . . 4 pts - Mentions as well that G=2, L=60060 has 32 solutions
- Prachai K. . . . . . . . . . . . 3 pts - Nice work on a and c; got 7 of 8. My b was hard to interpret
- Jeremy Galvagni . . . . . 3 pts - The next biggerst: G=2, L=4620 first with 16 solutions.
- Drew. . . . . . . . . . . . . . . . 3 pts - Good descrip. of role of prime factors in no. of sol'ns
- Hermen Jacobs. . . . . . . . 2 pts - Correct in general way abt number of divisors; L=mn/G
- Vince LoCascio . . . . . . . 2 pts - Welcome back from vacation; (30, 900) has 4 like (6, 720)
- Quasi-C . . . . . . . . . . . . . 2 pts - There are smaller sums around than 12+18 < 6+36.
- Allen Druze. . . . . . . . . . 2 pts - (18, 24) yes; (2, 12) and (4, 6) right idea but sum = ?
- Zahi Teitelman . . . . . . . 1 pt - (24, 18) was right; good try on part b but m=n won't do
- Ajit Athle. . . . . . . . . . . . 1 pt - Right on part a) and true that min m+n if m,n ~\/(mn).
- Ed Wern. . . . . . . . . . . . . 7 pts - Yes 96 was a boo-boo, and b was unclearly worded too
-
- Problem #207 - Posted Thursday, July 22, 2004
- Lost Bill of Sale (back to top)
- There were six prices for various TV sets sold at the store: $231, $273, $429, $600.60,
- $1001, and $1501.50. One day, a motel owner came in and bought a bunch of TVs.
- The total came to $13519.90 but the bill of sale was lost. How many of each TV
- type did the motel guy buy? Show reasoning carefully.
- Problem #207 - Posted Thursday, July 22, 2004
- Let a, b, c, d, e, and f be the number
of TV's bought at the prices in their listed order. Thus, after
- rewriting the prices, 3003a/13 + 3003b/11 + 3003c/7 + 3003d/5 + 3003e/3 + 3003f/2 = 135199/10
- Multiplying by 3003 gives a/13 + b/11 + c/7 + d/5 + e/3 + f/2 = 135199/30030 = 135199/(2*3*5*7*11*13).
- Thus, by clearing fractions, 2310a + 2730b + 4290c + 6006d + 10010e + 15015f = 135199.
- Now, consider the equation mod 13, 11, 7, 5, 3, and 2. Each element on the left of the equation
- is nonzero mod 13, 11, 7, 5, 3 and 2, respectively, and is zero modulo each of the other primes.
- Thus, the equation yields the following 6 equations: .9a = 12 = 9*10 mod 13,
- 2b = 9 = 2*10 mod 11,. 6c = 1 = 6*6 mod 7, ..d = 4 mod 5, . 2e = 1 = 2*2 mod 3, . f = 1 mod 2.
- Thus, a >= 10, b >= 10, c >= 6, d >= 4, e >= 2, and f >= 1. (... all these can be more strictly defined ...)
- However, 231*10 + 273*10 + 429*6 + 600.60*4 + 1001*2 + 1501.50*1 = $13519.90.
- Therefore, ( 10, 10, 6, 4, 2, 1 ) is the only solution.
- Skeptics included Nick who figures the expensive one was for the motel owner himself,
- and Ed, who figured since there was no sales tax it must be a non-profit motel!
- rewriting the prices, 3003a/13 + 3003b/11 + 3003c/7 + 3003d/5 + 3003e/3 + 3003f/2 = 135199/10
- Marcello Cammarata. . 9 pts - Answer was 1st and right (but you said d was mult of 4...)
- Art Morris. . . . . . . . . . . 6 pts - True, can eliminate even val of f etc; MATLAB mysterious
- Tim Poe . . . . . . . . . . . . . 5 pts - You suspect _me_ of having a non-unique solution? ;-}
- Vince LoCascio . . . . . . . 5 pts - Good way of doing 4 of d and 1 of f then totaling the rest
- Ed Wern. . . . . . . . . . . . . 4 pts - Excel does encourage thinking in a certain structured way
- Quasi-C . . . . . . . . . . . . . 4 pts - True, 135199 is rel prime to all values, Diophantine eqn.
- Joel Haywood (new) . . . 4 pts - Very nice (and quotable) work; welcome to my contest!
- Hermen Jacobs. . . . . . . 3 pts - Good use of BASIC, was made for this type of looping.
- Jeremy Galvagni . . . . . 3 pts - All are factors of 3003 so use as many as poss then back off
- Denis Borris . . . . . . . . . 3 pts - Assuming at least one of each simp the prob but risky
- Nick McGrath. . . . . . . . 3 pts - True, mostly cheap TV's at this guy's motel, hee hee
- Fanfan . . . . . . . . . . . . . . 3 pts - Used d and f get rem is 9616, then used congruences!
- Nikita Kuznetsov . . . .. 3 pts - Hi back, nice idea to put f=2f'+1, d=5d'+4 and subst.
- Prachai K. . . . . . . . . . . . 3 pts - Glad you enjoyed the problem, me too. Div trix yay!
- Drew. . . . . . . . . . . . . . . . 3 pts - Good; divide out by 3003 and the system gets easier!
- Allen Druze. . . . . . . . . . 3 pts - Subtotals are 231a+273b+429c+1001d=9616 or 6613 or 3610
- Zahi Teitelman . . . . . . . 2 pts - Gave a good argument; solution slightly off at $13495.90
- Ajit Athle. . . . . . . . . . . . 2 pts - Trial and error by hand builds character; you have a lot!
- Albert Alvarez. . . . . . . . 2 pts - Nice method; start with fixed value of e then table thru.
- Jack Dostal. . . . . . . . . . . 2 pts - Welcome back; cheap search with restrictions works.
- Mark Rickert. . . . . . . . . 2 pts - Good; odd of 6th and 5x-1 of 4th to get 90 cts, etc.
- Yakov Macak. . . . . . . . . 2 pts - I like your actual bill-of-sale format with subtotals.
- Jon Stearn . . . . . . . . . . . 1 pt - Nice to see you again! Yours worked out to $13159.90.
- Art Morris. . . . . . . . . . . 6 pts - True, can eliminate even val of f etc; MATLAB mysterious
-
- Problem #208 - Posted Monday, August 2, 2004
- The Third Degrees (back to top)
- Can you find a triangle, whose sides are all equal to their opposite angles,
- in degrees? Prove your answer is unique or prove it's impossible.
- Show reasoning clearly; one point penalty for resubmissions.
- Problem #208 - Posted Monday, August 2, 2004
- Most of you had to endure my e-mail
alert and resubmit a proof that this was unique.
- Thanks, and sorry to have put you through that, my poor dear 'dansmath-heads'.
- Quasi-C went further and said the misleading picture inspired another problem
- involving the central angles A+B+C=360, and their near-opposide sides. (Bonus pt!)
- Special kudos for Jeremy G. who responded 2 days before his wedding; excellent math tolerance, fiancee!
- Proof that equilateral triangle is unique: from Fanfan, Quasi-C, Nick, and others:
By Law of Sines, a / sin a = b / sin b = c / sin c. Consider the function f(x) = x / sin x.- This is increasing on (0, 180) so we must have a = b = c ; since a+b+c=180, all=60.
- Thanks, and sorry to have put you through that, my poor dear 'dansmath-heads'.
- Nick McGrath
. . . . . . . 10 pts - Finally got that proof in; first first
part so URD winner!
- Ed Wern. . . . . . . . . . . . . 7 pts - Flash of inspiration a=b=c followed by logical proof.
- Denis Borris . . . . . . . . . 5 pts - Says there are 674 possible triangles with integer sides
- Art Morris. . . . . . . . . . . 5 pts - Decided if units equal it must be spherical tri. a=b=c=90!
- Hermen Jacobs. . . . . . . 4 pts - Checked nearby triangles 59-60-61; they didn't work...
- Fanfan. . . . . . . . . . . . . . 4 pts - Nice proof by monotone function, meaning one-to-one.
- Quasi-C. . . . . . . . . . . . . 4 pts - Bonus point for also constructing point P (pic from #173)
- Zahi Teitelman. . . . . . . 3 pts - Right, it was too easy; good job showing contra if not equi.
- Vince LoCascio. . . . . . . 3 pts - Yep, it wasn't my intent for it to be so simple; caught me!
- Tim Poe . . . . . . . . . . . . . 3 pts - Another using x /(sin x) being an increasing function.
- Nikita Kuznetsov . . . . . 3 pts - Easy part first; I'm glad the proof part was 'more fun'!
- Jon Stearn . . . . . . . . . . . 3 pts - True if you use law of sines the ratios are actually all 1
- Marcello Cammarata. . 3 pts - Inversely, sin x / x is strictly decreasing (Eastern hemisphere!)
- Jeremy Galvagni. . . . . . 3 pts - Congratulations, a bonus point as a wedding present!
- S Khoo . . . . . . . . . . . . . . 2 pts - True answer; no follow-up proof but thanks for entry!
- Radu Ionescu . . . . . . . . 2 pts - Good terminology: x/sin x is 1-1 corresp for 0<x<180.
- Alan O'Donnell . . . . . . 2 pts - By 'sine rule' all sides add to 180- and ratios all = 1.
- Ajit Athle . . . . . . . . . . . 2 pts - I believe your statement about isosceles but not general
- Yakov Macak. . . . . . . . . 2 pts - Your equilateral case can happen, but yr proof of non can't
- Mark Rickert. . . . . . . . . 1 pt - Answer was ok; but after 209 was posted so just 1 point.
- Ed Wern. . . . . . . . . . . . . 7 pts - Flash of inspiration a=b=c followed by logical proof.
