- Problem #221 - Posted Saturday, February 12, 2005
- Square, Not a Square . . . . (back to top)
- a) Find the first three squares that are in arithmetic progression.
- b) Find four squares in arithmetic progression, if that's possible.
- c) Find 100 square-free integers in arithmetic progression.
- Here, a "square" is the square of a positive integer. A "square-free" integer is not
- divisible by any square > 1. Show reasoning clearly.
- Solution: a) Many of you blurted out the answer: 1, 25, 49 or {1^2, 5^2, 7^2}. But I like to see some reasons:
- b^2 - a^2 = c^2 - b^2, (b+a)(b-a) = (c+b)(c-b) and deducing stuff. Some of you gave {4, 100, 196} as less odd,
- and Philippe F. offered additional "primitive" sol's {7^2, 13^2, 17^2} and {7^2, 17^2, 23^2}; bien fait, merci!
- b) Thanks for puzzling this out; a couple of holey proofs (not holy), and references to the real proofs that we can't
- have 4 squares in a.p.: first in Dickson's History of the Theory of Numbers, according to Kirk and others:
- Another good page at: http://mcraefamily.com/MathHelp/BasicNumberArithmeticSequenceOfFourSquares.htm
- Denis Borris has a history with this problem and sends http://www.mathpages.com/home/kmath291.htm
- Also Tim Poe offers us http://www.mathpages.com/home/kmath044.htm (has a discussion on Dickson's treatment).
- c) A couple of you mistook 'square-free' to mean 100 non-squares in a row (a 'square-free sequence'); sorry!
- Many of the rest of you realized the common difference must avoid the squares of 2, 3, 5, 7, so d = 210 was good.
- My personal favorite sequence was Kirk's n = 79 + 420k ; a smaller d for Nick's 6463 + 210k , while Denis
- and our old friend Quasi had n = -1847+210k, and so on. (If I named all the numbers we'd be here forever! ;-)
- Square, Not a Square . . . . (back to top)
- WINNERS - Problem 221 . . thanks for entering! (back
to top) . leader
board
- Radu Ionescu . . . . . . . . 9 pts - First correct entry; I would still like to see more details
- Art Morris. . . . . . . . . . . 7 pts - The Morris Express is still going strong; go Art!
- Denis Borris . . . . . . . . . 5 pts - Denis already saw and solved this problem; d=210
- Mark Rickert . . . . . . . . 5 pts - Another good showing; used d=lcm(2^2,3^2,...,11^2)
- Marcello Cammarata. . 4 pts - Search for b) got no results (good); 8513+210k
- Nick McGrath . . . . . . . 4 pts - Counted your 2nd entry as clarif not resub; nice ans
- Mario Roederer.
. . . . . . 3 pts - Eventually complete with a couple of resubs.
d = 7770
- Kirk Bresniker . . . . . . . 3 pts - Dickson saves b); nice seq 79 + 420n, C++ code incl!
- Jeremy Galvagni. . . . . . 3 pts - 'Hip-free to be Square-free!' Huey Lowest and the Nos?
- Quasi-C . . . . . . . . . . . . . 3 pts - Intuition right on b); yes neg ok in c)-1847+210k
- Kirk Bresniker . . . . . . . 3 pts - Dickson saves b); nice seq 79 + 420n, C++ code incl!
- Hermen Jacobs .
. . . . . . 2 pts - I like 1 as a square ok; right not to find
b); c tricky
- Alan O'Donnell. . . . . . . 2 pts - Thanks for entry; good on a), right on b), misread c)
- Zahi Teitelman. . . . . . . 2 pts - Not clear 2r^2 - z^2 = y^2 has no sol'n; d=0 not ok.
- Philippe Fondanaiche. . 2 pts - Thanks for the link on b), for c) I mean sq-free int's
- Allen Druze. . . . . . . . . . 2 pts - Misread def; the numbers are square-free, not the list
- Vince LoCascio. . . . . . . 2 pts - Just in under the wire; yes 1+30030n fails at n=68
- Phil Sayre . . . . . . . . . . . 1 pt - The 3 squares in a.p. don't have to be consecutive...
- Corey Ramsey (new) . . . 1 pt - Correct on a), good intuition on b), and a student of mine!
- LetsGoCubs (new) . . . . 1 pt - You can see from the sol'n what I meant by a.p.of.sq.
- Problem #222 - Posted Monday, March 7, 2005
- Spare Sum Squares? (back to top)
- Find three distinct positive integers, whose sum is a perfect square, and the
- sum of any pair is a square. a) Find all such solutions with total sum less than 1000.
b) Can a solution exist so that the original three numbers are also squares? (Bonus pt this part)- Here, a "square" is the square of a positive integer. Show reasoning clearly.
- Solution: These 'rightly' reminded many of you of Pythagorean Triples, espec part b.
- a) There are 352 triplets with the pair sum square property, according to Ed, but only
- three triples have a square sum: (41, 80, 320), (57, 112, 672), (88, 168, 273).
- For example, 41 + 80 = 121 = 11^2 ; 41 + 320 = 361 = 19^2 ; 80+320 = 400 = 20^2 ,
- and the total = 41 + 80 + 320 = 441 = 21^2. How about that!
- Alan O'Donnell. . . . . . . 2 pts - Thanks for entry; good on a), right on b), misread c)
Similarly b and c can be written in terms of L, M, N so if we can find three
squares less than K^2 which sum to 2 * K^2, we can find the corresponding values of a, b, c.
a b c K L M N
41, 80, 320 from the squares of 21, 11, 19, 20
88, 168, 279 from the squares of 23, 16, 19, 21
57, 112, 672 from the squares of 29, 13, 27, 28
-
- WINNERS - Problem 222 . . thanks for entering! (back to top) . leader board
- To my contestants - I'm still struggling to catch up and I apologize; thanks again for your patience!
- Mark Rickert . . . . . . . 10 pts - First answer with some reasons or code behind it!
- Marcello Cammarata. . 8 pts - Holding on in all-time top ten!
Also (65, 464, 560), etc.
- Kirk Bresniker . . . . . . . 6 pts - Nice presentation, correct idea on part b), thanks.
- Radu Ionescu . . . . . . . . 5 pts - Got the right three; I like your Pythag reference in b.
- Philippe Fondanaiche . 5 pts - Hey fans, check out Philippe's www.diophante.fr
- Mario Roederer. . . . . . . 4 pts - Thanks for sending ref for #221 too, also 5 w/dupes
- Jeremy Galvagni. . . . . . 4 pts - Cracked the all-time top 20, nice ans 17, 32, 32 etc.
- Quasi-C. . . . . . . . . . . . . 4 pts - I'll fix up your cyl in cone score, good job on a).
- Kirk Bresniker . . . . . . . 6 pts - Nice presentation, correct idea on part b), thanks.
- Nick McGrath
. . . . . . . 4 pts - Took your time and got a chance to 'think
deeply.'
- Gordon Berg (new) . . . . 4 pts - Nice entrance into my contest, thanks 4 yr expertise!
- Art Morris. . . . . . . . . . . 3 pts - Good answer after one false start, resub one pt off
- Hermen Jacobs . . . . . . . 3 pts - Earlyish entry, 2 of 3 ans, and good intuition part b.
- Denis Borris . . . . . . . . . 3 pts - Similar theme to last week's, but different problem!
- Larry Corrado (new) . . . 3 pts - Welcome to my contest, thanks 4 Java pseudo-code
- Phil Sayre . . . . . . . . . . . 3 pts - Nice method (Python of course) with triple nested loop
- Vince LoCascio. . . . . . . 3 pts - You weren't sure there weren't more than 3 but ok!
- Zahi Teitelman. . . . . . . 2 pts - Nice expln finding sol'n 273,168,88; see the 2 others
- Problem #223 - Posted Friday, April 1, 2005
- Divisor Take-Away (back to top)
- Here's the game: the first player is given an integer n > 1 , then the second player subtracts a proper
- divisor d < n of that number, telling the first player the difference n - d , then the first player subtracts
- a proper divisor of that new number, etc. The player who announces a difference of 1 is the winner.
- Does either player have a winning strategy ? If not explain why not; if so, which player is it, does it
- depend on n, and what's the strategy? Show reasoning clearly.
- Solution: MathWorld defines 'proper divisor' as possibly 1 but not n. From longtime contestant Tim Poe:
"In order to have a difference of 1, d must be 1 less than n. Proper divisors are all 1/2 of n or less, so n in- this situation must be 2, and d must be 1. If your opponent gives you n=2, then you use d=1 and win.
If a player is given an odd number, they cannot win on that turn. If a player is given a prime number as n,- they must select d=1 and give n-1 back to the other player.
Since all divisors of an odd number are odd, If you give your opponent an odd number, they must return- an even one. Because you can use 1 as the proper divisor, you return an odd number for an even one that
- they give you. The victory then goes to the first person who can force their opponent to give them an even
- number, and they win by never using an even divisor from that point forward.
The phrasing of your question reads 'the first player is given an integer', but that the 1st subtraction is done- by the 2nd player. I don't understand this, but there it is. If the n used in the 2nd player's first subtraction
- is an odd number, then the 1st player has already won if they avoid even divisors. If the n used in the 2nd
- player's first subtraction is an even number, the 2nd player has already won if they avoid even divisors."
- [Dan's nutshell: a player hearing an even number wins, a player hearing an odd number loses.
- If the given number is odd the first player wins; if it is even the second player can always win.]
- Akifumi had a tight formal proof by induction using the P, N concepts from game theory, cool !
- WINNERS - Problem 223 . . thanks for entering! (back to top) . leader board
- Nick McGrath . . . . . . 10 pts - Got to the key issue: odd vs even, and odd divisors.
- Akifumi . . . . . . . . . . . . 7 pts - Nice to see you back in my contest, great answer!
- Mark Rickert . . . . . . . 6 pts - Good recovery from my original April Fools wording
- Gordon Berg (new) . . . . 4 pts - Nice entrance into my contest, thanks 4 yr expertise!
- Marcello Cammarata. . 4 pts - You recognized it as the famous
Aliquot Game.
- Denis Borris . . . . . . . . . 4 pts - The resub and good timing rounded you up 1/2 pt.
- Art Morris. . . . . . . . . . . 3 pts - Concise answer had all the important stuff in it.
- Mario Roederer. . . . . . . 3 pts - "A player who's given an odd number will alw. win"
- Denis Borris . . . . . . . . . 4 pts - The resub and good timing rounded you up 1/2 pt.
- Prachai K. .
. . . . . . . . . . 3 pts - Nice idea to work backwards; reverse
induction!
- Jeremy Galvagni. . . . . . 3 pts - Good distinction first player picking vs outside source
- Zeke Moore . . . . . . . . . . 3 pts - Fine expl'n; first player to get stuck with odd loses
- Phil Sayre . . . . . . . . . . . 3 pts - I agree, the second player is really the first player.
- Philippe Fondanaiche . 3 pts - Bonne explication avec demonstr'n d'induction
- Yakov Macek. . . . . . . . . 3 pts - It's been a while, from the dead of winter down under!
- Hermen Jacobs . . . . . . . 2 pts - Need to count 1 as proper div; p.s. my book is done!
- Vince LoCascio. . . . . . . 2 pts - Correct; each player has 50% inherent chance to win
- Keith Anker (new) . . . . . 2 pts - Welcome to my contest; I like the 'move' viewpoint
- Zahi Teitelman. . . . . . . 1 pt - Wanted a general ans dep on n even or odd; part ok.
- Bob Seegmiller. . . . . . . 1 pt - Welcome back after a long siesta; 3 would work if..
- Problem #224 - Posted Wednesday, May 11, 2005
- Fractionacci Numbers? (back to top)
- Let's define a Fibonacci-esque sequence of fractions: ao = 1 ; an+1 = (an) / (1 + n an)
- i) List the first ten 'Fractionacci Numbers' ii) Give an explicit formula (and proof) for
- the exact value of an (the nth Fractionacci). Show reasoning
- Solution: Most of you noticed these were all unit fractions (numerator 1), and that the
- denoms progressed as bn+1 = n + bn, making bn = sum of 1, 0, 1, 2, 3, ... = n(n-1)/2 + 1
- [not n(n+1)/2 + 1]. A popular answer format was an = 2 / (n^2 - n + 2) .
- This can (and should) be proved by induction on n: a0 = 2/(0-0+2) = 1 yes; now assume
- true for n; then an+1 = an / (1 + n an) = (2/(n^2 - n + 2)) / (1 + (2n/(n^2 - n + 2)) =
- = 2 / (n^2 + n + 2) = 2 / ((n+1)^2 - (n+1) + 2) ; TIJ ('That's It Jack'; Californian for QED)
- There was an issue with notation; some of you said a1 = 1/2; others (correctly) said
- a1 = 1 or 1/1. (note n=0 in a1 = 1/(1+0*1)); this made a10 = 1/46 not 1/56. The first 10 are
- a0 = 1, a1 = 1, a2 = 1/2, a3 = 1/4, a4 = 1/7, a5 = 1/11, a6 = 1/16, a7 = 1/22, a8 = 1/29, a9 = 1/37.
- I didn't mind if you included a10 = 1/46, but including a11 = 1/56 or calling a10 = 1/56 was
- counted as (slightly) incorrect and wasn't given my highest ranking.
- Dan's Note: I feel bad about the delay between problems this year; thanks for staying!
- I haven't forgotten about all of you; I have just been burned out on book revisions, my
- marathon training, and my upcoming podcasting career; stay tuned for developments!
- WINNERS - Problem 224 . . thanks for entering! (back to top) . leader board
- Art Morris. . . . . . . . . . . 9 pts - I thought n(n-1) was wrong but I was wrong. Steps?
- Jeremy Galvagni. . . . . . 3 pts - Good distinction first player picking vs outside source
- Nick McGrath
. . . . . . . 6 pts - Moved up a bit with your n(n-1). 26 miles this Sunday!
- Marcello Cammarata . 5 pts - Nice early ans and good proof but a1 = 1 not 1/2, etc
- Radu Ionescu
. . . . . . . . 4 pts - Good but steps lacking and n(n+1) rather than n(n-1)
- Jeremy Galvagni. . . . . . 4 pts - I like your doing this on a TI-83 - I can't find mine!
- Kirk Bresniker . . . . . . . 4 pts - That's correct that 0 + 1 + . . . + n-1 = n(n-1)/2.
- Prachai K. . . . . . . . . . . . 4 pts - Induc step ak+1 = 2 / ((k+1)^2 - (k+1) + 2), yes!
- Mario Roederer. . . . . . . 4 pts - Good induction proof, easy to follow, thanks.
- Denis Borris . . . . . . . . . 3 pts - Glad you ironed out the question then got it right!
- Phil Sayre . . . . . . . . . . . 3 pts - Unknown coeffs yay! I'd like to see that Fib-o-quilt.
- Arijit Bhattacharyya (new) 3 pts - Welcome to my contest; nice use of telescoping series!
- Ravi Raja . . . . . . . . . . . 3 pts - Nice to see you back; thanks for 10 extra terms!
- Mark Rickert . . . . . . . . 3 pts - Correct list but I think 1/56 is 11th or 12th term...
- Juan Carlos Carrara . . 3 pts - Welcome back; good method; guess & prove!
- Charlene Bong (new) . . . 3 pts - Good to have a contestant from Brunei, Borneo!
- Ken Duisenberg . . . . . . 2 pts - Approx ok but for + sign in n^2 - n + 2; proof?
- Ajit Athle . . . . . . . . . . . 2 pts - Careful when saying 'clearly'; also was n^2 - n + 2
- Zahi Teitelman. . . . . . . 2 pts - Minus sign, no proof, 11th; otherwise good ans!
- Lisa Schechner . . . . . . . 2 pts - Hi Nancy, yr Erdos #4 is very cool! (I met a #1.)
- Hermen Jacobs . . . . . . . 2 pts - Hello Hermen; sorry had to charge for the resub.
- Philippe Fondanaiche . 2 pts - Need the induction step for full 3 pts on later ans.
- Pascal Picard (new) . . . . 2 pts - Bienvenue! Bonne explication avec dem. d'induction!
- David Madfes (new) . . . . 2 pts - Nice to have you abord; induc'n ok but minus sign
- Chong Zhang (new) . . . . 2 pts - Glad you like my site, thanks 4 entering; good ans.
- Dawn Snyder (new) . . . . 2 pts - Thanks for good ans; joint effort w/Damon Haught
- Vince LoCascio. . . . . . . 2 pts - Just under the wire with most of a complete proof
- Damon Haught (new) . . 1 pt - I'm giving you a point here for helping Dawn. ;-}
- Quasi-C. . . . . . . . . . . . . 1 pt - The last three of you respondents
- Jeff Olson (new) . . . . . . . 1 pt - all came up with regular Fibonacci
- Polly Lau (new) . . . . . . . 1 pt - number fractions. Keep entering!
- Problem #225 - Posted Wednesday, June 8, 2005
- 3 Sums of 2 Nums (back to top) Answer these questions three; send the sums to me.
- a) The product, the quotient, and the difference of two real numbers are all the same.
- Find the sum of the two numbers.
- b) If [ x + \/(x^2 + 1) ] [ y + \/(y^2 + 1) ] = 1, find the value of the sum of x and y.
- c) The equation log3x(3) + log27(3x) = - 4/3 has two positive solutions. Find their sum.
- Show reasoning clearly ; \/(a) is sqrt[a]; time's up on this one; solutions and winners posted very soon.
- Solution: Prachai K. sends in a nice winning solution:
- a) ab = a/b = a-b ; If a is not 0, b^2=1 ; b=1,-1 ; b=1 doesn't give a soln. ; b=-1 does, giving a= -1/2 ;
- a=0 doesn't work. So a+b = -3/2
b) We know that (x-sqrt(x^2+1))(x+sqrt(x^2+1))(y-sqrt(y^2+1))(y+sqrt(y^2+1)) = 1
So from the given condition, we know that (x-sqrt(x^2+1))(y-sqrt(y^2+1))=1
So x+sqrt(x^2+1) = 1/(y+sqrt(y^2+1)) = -(y-(sqrt(y^2+1)))
and x-sqrt(x^2+1) = 1/(y-sqrt(y^2+1)) = -(y+sqrt(y^2+1)))
Add the last two equations. 2x = -2y ; x+y = 0.
c) We have (log3/log3x)+(log3x/3log3) = -4/3
Let log3/log3x = a, we have a + 1/(3a) = -4/3, which gives a = -1/3, 1.
Then we have x = 1/9, 1/81 . Their sum is 10/81.
- WINNERS - Problem 225 . . thanks for entering! (back to top) . leader board
- Prachai K. . . . . . . . . . . .10 pts - Is this your first weekly win? Congrats, I think so!
- Marcello Cammarata . . 7 pts - Good job explaining all parts. I like mult by \/(x2+1) - x
- Hermen Jacobs . . . . . . . 5 pts - x=y=0 works; so does any y = -x as you said in resub.
- Zeke Moore . . . . . . . . . . 5 pts - if x-y or y-x then x/y or y/x should also be ok...
- Mark Rickert . . . . . . . . 5 pts - Assumed single sum for b, and x=y=0 worked...
- Jeremy Galvagni. . . . . . 4 pts - I like your doing this on a TI-83 - I can't find mine!
- Chong Zhang . . . . . . . . 4 pts
- I like b) (u)(v) = 1; get x and y in terms of u, v!
- Nick McGrath . . . . . . . 4 pts - a/b = b-a is switchy but otherwise great. Go Nick!
- Mario Roederer. . . . . . . 4 pts - All parts well done; good to prove x+y=0 w/no assump.
- Jeremy Galvagni. . . . . . 4 pts - Glad you liked this one, esp first part. Nice ans.
- Art Morris. . . . . . . . . . . 3 pts - y = 1 may not work if same order of diff and quot.
- Denis Borris . . . . . . . . . 3 pts - Yes x=y=0 works in b); is sum unique or nec both 0?
- Kirk Bresniker . . . . . . . 3 pts - Parts a) and c) perfect; b) proved' If x=-y then 1=1'
- Zahi Teitelman. . . . . . . 3 pts - True -3/2 and 10/81; b assumes 0+0; any others?
- Phil Sayre . . . . . . . . . . . 3 pts - Nice alg steps on b) w/o assuming; reason on a)?
- Philippe Fondanaiche . 3 pts - Yes there are 2 answers if order of quot reversed
- Ajit Athle . . . . . . . . . . . 2 pts - Reversed a) to get +3/2; I like 10/81 = 0.12345679
- Vince LoCascio. . . . . . . 2 pts - Nice work; got the resub for 10/81, good recovery
- Quasi-C . . . . . . . . . . . . . 2 pts - I liked proof of b). And the 1/9 and 1/81 were true
- David Madfes . . . . . . . . 2 pts - Yes on a); for part b) there are more than 0, 1, -1.
- Damon Haught. . . . . . . 1 pt - (After 226 was up) a) -3/2; b) just sum is 0; c) good!
- Problem #226 - Posted Wednesday, June 22, 2005
- Con-hex-utive Numbers (back to top)
- These three tasks require you to fill in the integers from 1 to 9 in the hexagons at right,
- using each one exactly once for each part, so that, in turn:
- a) No two adjacent hexagons contain either consecutive numbers nor
- numbers whose names have the same number of letters. . . . b) No two
- adjacent hexagons contain digits whose sum is a multiple of four or five. .
- c) In any hexagon, the total of the numbers in the surrounding hexagons
- is a multiple of the number in the original hexagon.
- Each part has a unique solution up to rotation or reflection. Give each answer in three rows of 3, starting with left-to-top. Sorry, time's up on this one; solutions and winners soon.
- Solution: From longtime dans-math-head Tim Poe:
- Nick McGrath . . . . . . . 4 pts - a/b = b-a is switchy but otherwise great. Go Nick!
