dan's math@home - problem of the week - archives
 
 
Problem Archives page 24
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 . 151-160 . 161-170 . 171-180
181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231-240 . 241-250 . 251+ index
 
231 Obsessed w Squares
232 -- oNly oNe loNe N.
233- Lucas Square Sums
234 - Saturated Numbers
235 FromZeroToInfinity
236 - Gimme an Integer!
237- - - Nearly Integers!
238 - - Cube It Together
239 -- The 3-4-5 Square
240 - - The Pies Have It.
 
problem #231 - posted friday, oct 7, 2005
first problem of this year's contest! (this is my ninth season; I can't believe it!)
Obsessed with Squares (back to top)
Find all integers n for which 2^1994 + 2^1998 + 2^1999 + 2^2000 + 2^2002 + 2^n
is a perfect square. Show your reasoning.
Solution: Ok, everyone factored out the 2^1994. Here's what Phil Sayre had to Say:
"The expression can be factored as 2^1994*(1+2^4+2^5+2^6+2^8+2^d), where d=n-1994.
Since 1994 is even, 2^1994 is a square; hence it only remains to find the number 2^d which makes
the expression in parentheses a square, i.e., 369+2^d=x^2.  Trying d=1, 2, 3, ..., d=8 is a solution:
369+2^8=369+256=625=25^2.  There are no other solutions up thru d=16, after which the distance
between successive squares is greater than 369; hence there can be no other solutions."
 
WINNERS - Problem 231 . . As always, thanks for entering! - Dan (back to top) . leader board
     
Marcello Cammarata . . 10 pts - Off to a fast start, and clever factoring proof!
Ed Wern . (1st entry) . . . . 8 pts - I think you meant 2002 but you wrote 2004...
Nick McGrath . . . . . . . . . 7 pts - Thanks for the thanks, welcome back yourself!
Hermen Jacobs . . . . . . . . 5 pts - I liked your concise Basic prog; y=34 was close!
Mark Rickert . . . . . . . . . 4 pts - Good to find 2^8 + 369 = 25^2; looked for others
Kirk Bresniker . . . . . . . . 4 pts - Nice method of 2-digit endings for squares, n<4000
Phil Sayre . . . . . . . . . . . . 4 pts - Good idea to check 369 + 2^d then squares far apart
Al Nelson . . . . . . . . . . . . 4 pts - Welcome back, hope you make a habit of this site!
Art Morris . . . . . . . . . . . 3 pts - Hi Art, good to notice 2002 = 1994 + 8 worked...
Zahi Teitelman . . . . . . . 3 pts - Right; (2^1994)(369 + 2^8) = (25 * 2^997)^2
Jack Dostal. . . . . . . . . . . 3 pts - Welcome back, Jack; it's been a year; got the 2002
Jeremy Galvagni . . . . . . 3 pts - Cool to try to factor c^2 - 369; J gave 604 digit ans!
David Madfes. . . . . . . . . 3 pts - You were right to believe 369 + 256 was only value
Denis Borris . . . . . . . . . . 2 pts - Basically ok but yes you needed the resub to finish
Vince LoCascio . . . . . . . 2 pts - Ok 2^8 but didn't say 2002, also 2^58 doesn't work
Yakov Macak . . . . . . . . . 2 pts - Sorry, missed your k = 8 entry at first, here you go!
 
 
problem #232 - posted friday, oct 21, 2005 (second problem of this year's contest!)
oNly oNe loNe N (back to top)
Consider the equation below, where x and y are relatively prime positive integers:
Show that there is only one possible value for N. Find it. Show your reasoning.
 
Solution: Yes, it was easier to find the solution x = 5 , y = 4 , N = 2000 than to prove it's unique.
Since gcd(x, y) = 1 it follows that x^3 and y^2 don't have any common factors, so x^3 | (y + 1)^3 ,
meaning x | (y + 1) , x <= y + 1. Similarly y^2 | (3x + 1), so y^2 <= 3x + 1. Combining the two
inequalities gives x - 1 <= y <= sqrt[3x+1] , so that x^2 - 5x <= 0. As x is positive, only
x = 1, 2, 3, 4, 5 are possible. Trying these values in x^3 (3x + 1) = y^2 (y + 1)^3 , we find that
only x = 5, y = 4 work, so N = 4^2 · 5^3 = 2000 is the only solution.
 
Longtime fan Jack D just substituted values into the expressions x^3(3x + 1) and y^2(y + 1)^3 :
   x   x^3(3x + 1) . . . y   y^2(y + 1)3
   1         4   . . . . .   1         8
   2        56    . . . . .   2       108
   3       270    . . . . .   3       576
   4       832     . . . . . 4      2000
   5      2000  . . . . .   . . . . .   . . . . .   . . . . . (There's some hard evidence! - Dan)
WINNERS - Problem 232 . . (back to top) . leader board
These comments aren't meant as critical, just helpful I hope - Dan
Tim Poe . . . . . . . . . . . . . 10 pts - The first entry with a holes-free proof; go poe!
Marcello Cammarata . . 8 pts - I agree with the 2000 but I needed more reasons
Zahi Teitelman . . . . . . . 6 pts - Early entry gets the big points; still near-proof
Nick McGrath. . . . . . . . . 5 pts - I followed (y-4)(y+1)=0 but not why y^2 > 3x+1
Zeke Moore . . . . . . . . . . . 5 pts - Good use of proportions not being integers, Zeke.
Art Morris . . . . . . . . . . . 4 pts - True x^3 =/= y^2 but doesn't make x^3 = (y+1)^3
Hermen Jacobs . . . . . . . . 4 pts - Did thorough search up to x, y <= 1000, not a proof
Kirk Bresniker . . . . . . . . 4 pts - The 5-pointed star wasn't a hint even though x = 5
Ed Wern . . . . . . . . . . . . . 4 pts - Nice that w = y+1 ; prove w = ax/b implies a/b = 1
Mark Rickert . . . . . . . . . 3 pts - Assuming x = y+1 does give y=4 but it's a big ass.
Yakov Macak . . . . . . . . . 3 pts - Yes x = y+1 simplifies it; y and y+1 always rel prime
Phil Sayre . . . . . . . . . . . . 3 pts - Docked 1/2 point for resub; proof still not flawless
Denis Borris . . . . . . . . . . 3 pts - If x = y+1 then y^2 - 3y - 4 = 0 ; thanks for 2nd try
Al Nelson . . . . . . . . . . . . 3 pts - Good, if x =/= y+1 then x | y+1 bec of cube factors
Vince LoCascio . . . . . . . 3 pts - Interesting x = k(y+1); y^2 = k^3(3x+1) prove k=1
Jack Dostal. . . . . . . . . . . 3 pts - Good show; see solution quoted above for details
Akifumi . . . . . . . . . . . . . 3 pts - Yes, if x=1 then L=3, R=8; y^2 | 3x+1 | y^2 => =.
Ravi Raja . . . . . . . . . . . . 3 pts - Nice use of quadratic formula, note discrim = square
Radu Ionescu . . . . . . . . . 2 pts - Unsure whether 3x+1 is a line unless y = 3x+1?
Ajit Athle . . . . . . . . . . . . 2 pts - Halloween eve brings out the fact that x=5 and y=4
     
     
problem #233 - posted monday, nov 7, 2005
Lucas Square Sums (back to top)
The Lucas numbers are defined as L0 = 2, L1 = 1, Ln+1 = Ln + Ln-1 for n > 1
Find a closed form for the sum of the squares of Lk , from k = 0 to n ,
in terms of the Ln's. Verify your result numerically up to n = 10.
"Closed form" is an algebraic formula without "sum of" or ". . . " Show your reasoning.
Solution: Opinions varied as to what "closed form" meant. Here are three or more answers:
(1) Let Sn = L0^2 + L1^2 + . . . + Ln^2. Make a table of values in Excel and have it do the sum
of squares, and notice a pattern: Sn = Ln Ln+1 + 2 or also Sn = L(2n+1) + 3 + (-1)^n.
Note: MathWorld had this answer listed as Ln Ln+1 - 2 but this was an error. - Dan
(2) Let a = (1+\/5)/2 and b = (1-\/5)/2 ; then Ln = a^n + b^n is well known (MathWorld e.g.)
Lk^2 = (a^k + b^k)^2 = a^(2k) + 2(ab)^k + b^(2k); add these up and form three geom series.
Sn = (1 + a^2 + . . . + a^(2n)) + 2(1+ ab + . . . + (ab)^n) + (1 + b^2 + . . . + b^(2n))
Sn = (1 - a^(2n+1))/(1 - a^2) + 2(1 - (ab)^(n+1))/(1 - ab) + (1 - b^(2n+1))/(1 - b^2)
(3) Phil noticed that the logs of the Ln get more evenly spaced, making Ln appox geom with ratio
a = (1 + \/5)/2. Thus Ln is approx [a^n] the greatest (really nearest) integer to a^n, for n > 2.
Thus Sn = 2^2 + 1^2 + a^2 + a^3 + . . . + a^n = 5 + (a^2 - a^(n+1))/(1 - a) ;
but I think this is off by 5 for the 10th term; maybe without the 5 it's correct-er!
(4) Use the golden rectangle pattern as sent in by Yakov M, Tile squares : 2x2 then 1x1, then
3x3 below, 4x4 to the right, 7x7 below, etc. You get a Ln by Ln+1 rectangle, with 2 square
units hanging off the top from the first 2x2. Thus visually Sn = Ln Ln+1 + 2 as desired.
(5) And new contestant Claudio gave Ln = x^n + y^n, x and y are roots of z^2 - z - 1 = 0
An appropriate table for this problem: 
 n . . . Ln. . . . .Ln^2 . . . .Sn . . . Ln*Ln+1 
 0 . . . 2 . . . . . 4 . . . . . 4 . . . . 2
 1 . . . 1 . . . . . 1 . . . . . 5  . . .  3
 2 . . . 3 . . . . . 9 . . . . .14 . . . .12
 3 . . . 4 . . . . .16 . . . . .30 . . . .28
 4 . . . 7 . . . . .49 . . . . .79 . . . .77
 5 . . .11 . . . . 121 . . . . 200 . . . 198
 6 . . .18 . . . . 324  . . . .524 . . . 522
 7 . . .29 . . . . 841 . . . .1365 . . .1363
 8 . . .47 . . . .2209 . . . .3574 . . .3572
 9 . . .76 . . . .5776 . . . .9350 . . .9348
10. . .123  . . .15129  . . .24479 . . 24477
11. . .199  . .  . . . . .
Tim Poe . . . . . . . . . . . . . 10 pts - Phrased it as Ln(Ln+1 + (2/Ln)) which works
Ed Wern . . . . . . . . . . . . . 8 pts - Your spreadsheet was good but shoulda had +2
Marcello Cammarata . . 7 pts - Second fully correct answer; I like use of phi...
Nick McGrath . . . . . . . . 6 pts - Caught the error of MathWorld; should we send bug?
Mark Rickert. . . . . . . . . 5 pts - Also looked up incorrect formula, used it as hint ;-}
Radu Ionescu. . . . . . . . . 4 pts - Left out the in between ones but had ends covered
Kirk Bresniker . . . . . . . 4 pts - Pointed out that Ln = a^n + b^n as above, thanks!
Akifumi . . . . . . . . . . . . . 4 pts - Thanks for the induction proof and staying up late!
Denis Borris. . . . . . . . . . 3 pts - No deduct for 2nd entry; didn't catch orig question sorry
Joe Alvord . . . . . . . . . . . 3 pts - Got 2nd entry 1st, thanx for gen result, good geom series
Hermen Jacobs. . . . . . . . 3 pts - Nice approach with the triple geometric series Hermen
Phil Sayre . . . . . . . . . . . . 3 pts - Cool idea taking logs of Ln and finding differences
Claudio Baiocchi (new) . 3 pts - Welcome to my contest, thanks for generating function
Jeremy Galvagni . . . . . . 3 pts - Nice table, verified Ln = phi^n + (1-phi)^n on TI-83
Yakov Macak . . . . . . . . . 3 pts - I really like the clear geometric approach and picture
Vince LoCascio . . . . . . . 2 pts - Correct general formula for the individual Ln's.
 
 
problem #234 - posted saturday, nov 19, 2005
Saturated Numbers? (back to top)
Ok, bear with me on this . . . Every integer n > 1 has a prime factorization.
If no primes are skipped; 2^a 3^b 5^c . . . the n is called saturated (sat).
If also exponents in order; a >= b >= c . . . n is saturated ordered (s.o.).
If n has more divisors than any smaller number, it's supercomposite (s.c.).
a) For n <= 20, 40, 60, 80, 100, 120: how many composite, sat, s.o.,
. . and s.c. nos are there? (ex: for n <= 10 there are 5, 4, 4, and 3 of each.)
b) List all the n <= 200 that are sat but not s.o.
c) List all the n <= 200 that are s.o. but not s.c.
d) And finally, list all the n <= 200 that are s.c.
Show your reasoning.
     
Solution: A couple of you rightly needled me for calling 2 "supercomposite" when it's
not composite. (And is 1 supercomposite, because no smaller number has a divisor?)
So let's count 2 as saturated, ordered, and supercomp, but not composite.
(Listen to my podcast dansmathcast_002 for details on these.) Then a) is:
n<= . .com. . . . .sat . . . .sa.or. . su.co 
 10 . .  5 . . .  . 4 . . . . . 4 . . . . 3
 20 . . 11   . . .  7 . . . . . 6 . . . . 4
 40 . . 27 . . . . 11 . . . . .10  . . .  6
 60. . .42 . . . . 14 . . . . .12 . . .   8
 80. . .57 . . . . 16 . . . . .14 . . . . 8
100. . .74 . . . . 18 . . . . .15 . . . . 8
120     89 . . 11 .20. . . . . 16 . . . . 9
200. . 153 . . . . 26  . . . . 20  . . . 10
b) Sat but not Ord: n with <exps> of prime fac. of n
18=<1,2>, 54=<1,2>, 90=<1,2,1>, 108=<2,3>, 150=<1,1,2>,162=<1,4>
c) Sat Ord but not Supercomp:
8=<3>, 16=<4>, 30=<1,1,1>, 32=<5>, 64=<6>, 72=<3,2>, 96=<5,1>,
128=<7>, 144=<4,2>, 192=<6,1>
c) Sat, Ord, and Supercomp (all s.c. are s.o.)
2=<1>, 4=<2>, 6=<1,1>, 12=<2,1>, 24=<3,1>, 36=<2,2>, 48=<4,1>,
60=<2,1,1>, 120=<3,1,1>, 180=<2,2,1>
     
WINNERS - Problem 234 . . (back to top) . leader board
 
Mark Rickert. . . . . . . . . 9 pts - Wonder how to use the gap-system.org for this
Marcello Cammarata . . 8 pts - Lists made answer easy to read, good use of exps
Nick McGrath . . . . . . . . 6 pts - Thanks for extra table and results up to 200 sauf 100
Vince LoCascio . . . . . . . 5 pts - Good to pin down the rules before listing data
Tim Poe . . . . . . . . . . . . . 4 pts - T\Yes the resub was worth a couple, minus one
Jeremy Galvagni. . . . . . 4 pts - Yes 2 is super-comp and also prime, even even.
Kirk Bresniker . . . . . . . 4 pts - Your full table and clear answer scored a bonus pt
Denis Borris. . . . . . . . . . 3 pts - Bonus pt for list of n: s(n) and d(n) primes (all sqs)
Zahi Teitelman . . . . . . . 3 pts - Gave an accurate (and uncaptioned) spreadsheet
Jack Dostal. . . . . . . . . . . 2 pts - Wrote C++ prog ; it missed some and gave others
Hermen Jacobs. . . . . . . . 2 pts - Some good table values but 2^k not s.c. for k>2
       

 

problem #235 - posted sunday, dec 4, 2005
From Zero to Infinity, Rationally! (back to top)
Prove there are no rational numbers x, y such that
but there are infinitely many rational x, y such that
Show your reasoning.
Solution: A more selective group this week / fortnight... Keep 'em coming! - Dan
(1) Some of you used mod 3 or mod 4 arguments to prove the first eqn had no integer solutions,
or x = m/d, y = n/d gets to 2m^2 - 3n^2 = d^2 or -d^2 ; only squares are 0 and 1 mod 3 or 4.
 
(2) Hermen and Nick made reference to the hyperbola for this equation, and Nick found a paper
on Elliptic Curves: http://mat.uab.es/~xarles/elliptic.html  follow this argument:
Draw a line of rational slope r through (1,1) on 2x^2 - 3y^2 = -1 ; using algebra (!) gets
the coords of the second point the line hits the hyperbola as (x, y) where
x = (-3r^2 + 6r - 2) / (2 - 3r^2) and y = (3r^2 - 4r + 2) / (2 - 3r^2).
This may be the same as Mark and Phil's recursions xn+1 = 10 xn - xn-1 ; same for y's.
Some solutions are (1, 1), (11,9), (109, 89), (1079, 881), etc. Also (1/5, 3/5) gives more rats.
Kudos to Marcello for using 2x^2 - 3y^2 = (x\/2 - y\/3)(x\/2 + y\/3) and raising to powers;
odd powers of this give inf no of solutions of form (a, b) ; (a\/2 - b\/3)(a\/2 + b\/3) = (-1)^n.
       
WINNERS - Problem 235 . . (back to top) . leader board
 
Marcello Cammarata . 10 pts - Yes, solve 2x^2 - 3y^2 = d^2 in integers, use conjug!
Vince LoCascio . . . . . . . 6 pts - Great use of mod 3 and sqrts, need pf of inf of sols
Mark Rickert. . . . . . . . . 5 pts - First ans submitted, nice recursion, lacking 1st part
Nick McGrath . . . . . . . . 5 pts - Good 1st ans, great 2nd ans, see above curves link.
Hermen Jacobs. . . . . . . . 4 pts - Good argt for 1st part, needed proof of inf, typo 119
Phil Sayre . . . . . . . . . . . 4 pts - I like using the 'last digit' & u proved the recursion
Denis Borris. . . . . . . . . . 1 pt - Good ans, using mod 3 to disprove, nice list of sol'ns!
 
 
problem #236 - posted wednesday, dec 14, 2005
Gimme an Integer! (back to top)
Find all positive real numbers c such that the following is an integer.
Show your reasoning.
 
Solution: Here's Vince's systematic nine-step approach!
Step 1: let A=3+sqrt(c) and B=3-sqrt(c)
Step 2: we are interested in all values for c, for which A^(1/3) + B^(1/3) = n, where n is an integer.
Step 3: cubing step 2, A +3(A^2B)^1/3 +3(AB^2)^1/3 + B = n^3 / / Step 4: recognizing that
. . A+ B =6, regrouping the middle terms of step 3 yields: 6 + 3(AB)^1/3(A^1/3 +B^1/3) = n^3
Step 5: recognizing from 2 that A^1/3 + B^(1/3) = n, rewrite as: 6 + (3n)(AB)^1/3 = n^3
Step 6: bringing all but AB to the RHS, (AB)^1/3 = (n^2)/3 - 6/n
Step 7: cubing each side, AB = n^6/27 - 8/(n)^3 - 6n^3/9 + 4
Step 8: AB = 9 - c, substituting above and solving for c:  c = 5 + 8/(n)^3 + [18(n)^3 - (n)^6] / 27
Step 9: the only values for n that produce positive rational number for c are
. . n=1, giving c = 368/27 and n=2, giving c = 242/27
       
WINNERS - Problem 236 . . (back to top) . leader board
 
Marcello Cammarata . 10 pts - Wrote as x + y = n, c = 9 + ((6 - n^3)/(3n))^3. Nice!
Tim Poe . . . . . . . . . . . . . 7 pts - Good, work bkwds and put in the n first; then Excel it
Hermen Jacobs . . . . . . . 5 pts - Nice work cubing the binomial x+y ; solving n=1 and 2
Kirk Bresniker . . . . . . . 4 pts - Yes, f(c) has n-int at (0, 2.884), lim = 0, but prove decr
Denis Borris . . . . . . . . . 4 pts - That's it; c=242/27 for n=2 and 368/27 for n=1, inv func!
Nick McGrath. . . . . . . . 4 pts - I think this inverse function idea is catching on! Go Nick!
Mark Rickert. . . . . . . . . 4 pts - Yes, 9 + (5/3)^3 or 9 - (1/3)^3. I left out gory details too
Radu Ionescu. . . . . . . . . 3 pts - Your resub had you covered; x^3 = 6 + 3x*(9-c)^(1/3)
Jeremy Galvagni. . . . . . 3 pts - That's right, deriv of f(c) is alw neg.; decimals are close
David Madfes . . . . . . . . 3 pts - Nice report-style ans with formulas, graphs, 242 not 235.
Vince LoCascio. . . . . . . 3 pts - A fine quotable answer, also the most 'recent' submitted!
Ed Wern. . . . . . . . . . . . . 3 pts - I like the k = x+y, x = (3-\/c)^(1/3), \/c = 3 - x^3, s.b.s.
Nemesis Enforcer (new) . 3 pts - Welcome N.E.; recommends WalterZorn.com grapher
Phil Sayre . . . . . . . . . . . 2 pts - Right to work inside out, resub had exact integer value
Rich Futrell (new). . . . . . 2 pts - Thanks for listening to my podcast and then entering!
V Balakrishnan (new) . . 1 pt - Hello, u got in just in time, good ansr, welcome2dmc
Claudio Baiocchi (new) . . 1 pt - I got your answer along with #238; welcome to dansmath!

 

 
problem #237 - posted tuesday, dec 27, 2005
Nearly Integers! (back to top)
Show that for any positive integer n, the number (1 + \/2)^n is
less than 1/ 2^n away from the nearest integer. \/2 means square root of 2.
Show your reasoning. (Time's up on this one)

 

Solution: Here's new fan V Balakrishnan's clear and concise answer:
Note that (1+\/2)^n+(1-\/2)^n is an integer , since the kth term is C(n,k).(\/2)^k in (\/2+1)^n and
C(n,k).(-\/2)^k in (\/2-1)^n. When k is odd the terms cancel; when k is even it is rational as it is of
the form 2^m. Hence (1+\/2)^n+(1-\/2)^n is an integer ; also |1-\/2| < 0.5
So the distance from nearest integer is |1-2|^n=(2-1)^n as (1+\/2)^n+(1-\/2)^n is the nearest one
Since 2 < 9/4 => \/2 < 3/2 or \/2-1 < 1/2 ; so (\/2 - 1)^n < (1/2)^n
       
From Phil:  (1+\/2)^n = An + Bn \/2, with A1=1, B1=1, A(n+1)=An+2Bn, B(n+1)=An+Bn.  (n's are subscripts)
multiplying [An+Bn\/2](1+\/2) = (An+2Bn) + (An+Bn)\/2 = A(n+1) + B(n+1) \/2 = (1+\/2)^(n+1) QED
The first few coefficients are ... [also included actual integer deviations showing < 1/2^n]
n=  1   2   3     4     5    6       7      8
An=1   3   7   17   41   99   239   577
Bn=1   2   5   12   29   70   169   408
       
WINNERS - Problem 237 . . (back to top) . leader board
 
Marcello Cammarata . 10 pts - Great solution involving convergents of continued fractions!
Nick McGrath. . . . . . . . 7 pts - The complement 1 - \/2 when expanded, cancels the odd terms
Mohamed Omar . . . . . . 6 pts - Welcome back, it's been a couple of years! Good answer too.
Phil Sayre . . . . . . . . . . . 5 pts - Ok, your lemmas combined to prove diff < (1-\/2)^n < (1/2)^n
V Balakrishnan . . . . . . 4 pts - Second week, right on the mark, remember show your steps...
Akifumi. . . . . . . . . . . . . 4 pts - Taking a break from studying by doing contest problems!
Denis Borris . . . . . . . . . 3 pts - Might have proved odd terms drop out, and maybe Kidd > Nash
Kirk Bresniker . . . . . . . 3 pts - That's right; | b\/2 - nint | < |1 - \/2|^n ; also cont frac bn/an.
Vince LoCascio. . . . . . . 2 pts - Right idea to get recursion ; I missed the integer part of 5\/2 etc
Hermen Jacobs . . . . . . . 2 pts - Good idea bounding then taking log of powers; but 1+\/2 /</ 2
Thad (new) . . . . . . . . . . . 1 pt - Welcome new podcast fan! See the above solution to problem.
Claudio Baiocchi . . . . . 1 pt - I like the recursion leading to the continued fractions. Nice!;-}
 

 

 
problem #238 - posted monday, jan 9, 2006
Cube It Together! (back to top)
a) (warmup) Start with a 2" by 6" rectangle. Say how to cut it up in the fewest number
of pieces and rearrange into a 3" by 4" rect. b) Start with a block 8cm by 8cm by 27cm.
Find the fewest number of pieces to cut it into to rearrange into a 12 by 12 by 12 cube.
Explain precisely how to do it. (Dan's note: I actually did this, with cheese!)
Partial answers (e.g. part a) accepted. Show your reasoning. Simple GIFs ok if necessary
Solution: There were mostly two camps : a large "7 pieces" component, and a small "4 pieces" group.
Yes, the first part was supposed to be a hint, but I overlooked the obvious two 2x3s solution; sorry!
Please see the attached artwork page with art from Phil, Tim, and
And Jeremy G, not content with three dimensions, asks:
"Is there a 4-d rectangular hyper-prism that must be cut into 15 pieces to
fit in a hyper-cube? Probably. Maybe I'll try cutting a 16*16*16*81,
rearranging the pieces and rearranging them to fit in a 24x24x24x24."
       
WINNERS - Problem 238 . . (back to top) . leader board
 
Nick McGrath . . . . . . . 10 pts - Seventh entry received, but the first to get b) in 4 pieces!
Akifumi. . . . . . . . . . . . . 7 pts - The eleventh entry and second with just 4 pieces. 7-11!
V Balakrishnan . . . . . . 5 pts - There you go, and thanks for the simple clear pictures!
Tim Poe . . . . . . . . . . . . . 4 pts - Nice graphics for first part, I put 'em on the artwork page
Yakov Macak . . . . . . . . 3 pts - The mental puzzle gymnastics are almost always worth it!
Marcello Cammarata. . 3 pts - Ok for first in 2 rectangles, then 7 blocks for part b)...
Ed Wern. . . . . . . . . . . . . 3 pts - Good starting point to check the sum of volumes works
Mark Rickert. . . . . . . . . 3 pts - The 7 pieces worked; did you find a better way later?
Claudio Baiocchi . . . . . 3 pts - I got your 236 too; thanks for the 7-piece description
Zeke Moore . . . . . . . . . . 3 pts - Your seven piece scheme was pretty effective and popular
Jeremy Galvagni. . . . . . 3 pts - Yes a) was easier than I realized, but b) was trickier, eh
Phil Sayre . . . . . . . . . . . 3 pts - That was a cool sketch for 7 pcs, I put it on artwork page
Kirk Bresniker . . . . . . . 3 pts - Managed part b) in 8 pieces including a cool L-shape
Hermen Jacobs . . . . . . . 3 pts - The second part with 9 pieces was different than others
Radu Ionescu . . . . . . . . 2 pts - a) was ok, I didn't see how many pieces yr 5 cuts made
Thad . . . . . . . . . . . . . . . 2 pts - The 7 pcs were ok, but I didn't get your diagram for a).
David Madfes . . . . . . . . 2 pts - Ok for 7 pieces of cheese, but no part a) was included
Denis Borris . . . . . . . . . 2 pts - Eventually got 2 pcs and 7 pcs after 2 resubs received
Ajit Athle . . . . . . . . . . . 2 pts - The 7 pcs were ok on b) but 3 pcs on part a) can be improved
Zahi Teitelman. . . . . . . 2 pts - Ok on a); I see the 7 pcs in b) but not how to assemble
John Smith (new). . . . . . 2 pts - Welcome! 10 pcs for part b), some 4 x 4 x 12, I think.
problem #239 - posted thursday, jan 26, 2006
The 3, 4, 5 . . . Square ? (back to top)
I recently got this e-mail question sent to me: You are sitting inside a big square
painted on the floor. One corner is three meters (3m) from you, another corner
is 4m away from you, and another is 5m away from you. How big is the square?
a) Is the problem solvable? . b) Is the solution unique? . c) How many solutions are
there, and what are they? . Show your reasoning.
     
Solution: I give you Marcello's answer, this year's contest leader! My first impulse was to use
the law of cosines, as some of you did, but for me it produced a nasty octic equation! - Dan
 
"Refer the square to an orthogonal coordinate system ...Let s be the side of the square;
then, the three given corners have coordinates (0,0), (s,0) and (0,s) respectively.
Let a, b, c be the three corresponding distances, and (x,y) my coordinates; then
x^2 + y^2 = a^2; (s-x)^2 + y^2 = b^2; (s-y)^2 + x^2 = c^2. Subtracting the first
eqn from the other two, we find x = (s^2 + a^2 - b^2)/2s; y = (s^2 + a^2 - c^2)/2s,
and by substitution into the first one the biquadratic equation
2s^4 - 2(b^2+c^2)s^2 + (a^2-b^2) + (a^2-c^2) = 0 is obtained. By substituting all the
permut'ns of 3, 4, 5 for a, b, c, we find that only a=3 and a=4 give real roots (b and c are
obv. interchangeable), and in both cases only the higher s^2 root gives positive x, y values.
There are therefore only two distinct solutions, which are
s = sqrt((41+sqrt(1071))/2) = 6.7015, x = (s^2 - 7)/2s = 2.4593, y = (s^2 - 16)/2s = 1.7181; and
s = sqrt((34+sqrt(896))/2) = 5.6539, x = (s^2 + 7)/2s = 3.44460, y = (s^2 - 9)/2s = 2.0310; plus
two mirror solutions with interchanged x, y. " -- (Sorry this wasn't as "very soon" as I'd hoped! -- Dan)
       
WINNERS - Problem 239 . . (back to top) . leader board
 
Jeremy Galvagni. . . . . . 9 pts - Algebra good, radical expressions ok, but had 6.0715 twice
Marcello Cammarata. . 8 pts - Nice answer ^ with simultaneous coord equations, buono!
Ed Wern . . . . . . . . . . . . 6 pts - Thanks for the re-send, and good showing of your work!
Kirk Bresniker . . . . . . . 5 pts - Got your re-send; another one slipped by; sorry!
Zeke Moore . . . . . . . . . . 5 pts - Three eqns for 3 dist, expand, solve for x, and off you go!
V Balakrishnan . . . . . . 4 pts - Making a habit of good early answers,VB. Imag soln too!
Mark Rickert . . . . . . . . 4 pts - Good answer; I didn't get it originally (or I stupidly deleted it?)
Denis Borris . . . . . . . . . 4 pts - Thanks for solving my problem and rewriting it 'correctly'!
Nick McGrath. . . . . . . . 3 pts - Thanks for listening, your 3rd solution was short of 6...
David Madfes . . . . . . . . 3 pts - Used law of cos like i tried; one ans good, 6.007 a bit short
Ajit Athle . . . . . . . . . . . 3 pts - Some cosine equations led to 5.65 but also 6.16 too wide
Joe Alvord. . . . . . . . . . . 3 pts - Used the old Alaskan Geometry Sketchpad, rotate perps!
Tim Poe . . . . . . . . . . . . . 2 pts - Cool story, reduce the square until 3 circles concurrent!
Phil Sayre . . . . . . . . . . . 2 pts - Serious equations; got the 3-4-5-case fine; no other orders.
Radu Ionescu . . . . . . . . 2 pts - Used law of cos ok for 3-4-5 ; other orders not corn-sidered
Luis de Brito Camacho . . 2 pts - Welcome back! Nice graphics, 3 simul quadratics, cool.
Zahi Teitelman. . . . . . . 2 pts - I agree with 2z^2 - 82z + 305 = 0, but order 345 only
Barry Thompson (new) . 2 pts - Hi! Defined a func f(a,b,c,d) a^2 + b^2 = c^2 + d^2...
Hermen Jacobs . . . . . . . 1 pt - Hi, HJ. Sort of misinterp the prob using concentric circles
John Smith . . . . . . . . . . 1 pt - One alias per customer... 4 is a reasonable guess, I guess
Pee Wee . . . . . . . . . . . . . 1 pt - Welcome, and you're welcome! 6m was close (not unique)
Thad. . . . . . . . . . . . . . . . 1 pt - Your intuition was right about law of cos; doable =/= done

 

 
problem #240 - posted sunday, feb 5, 2006
: The Pies Have It : (back to top)
Using only the number pi (), the symbols for addition, multiplication, square root (),
parentheses ( ), and greatest integer brackets [ ], but no other symbols or operations,
construct each of the integers 1 through 10, using as few total 's as you can.
[n] is the largest integer less than or equal to n. Show your reasoning. (Time's up on this one.)
 
Solution: The best (tied by many) was 21 total pi's (or pies?)
Ed debated plural of pi; pies too deserty, pis too distasteful, how bout just pi, like deer?
Firsts on lists from new entrant Doug; nexts from Denis, Nick, and others
 
1 = [sqrt(pi)].
2 = [pi - sqrt(sqrt(sqrt(sqrt(pi))))]. (Dan's note: illegal minus sign; replace with [sqrt(pi)] + [sqrt(pi)].)
3 = [pi].
4 = [pi + sqrt(pi)].
5 = [pi * sqrt(pi)].
6 = [pi + pi].
7 = [pi + pi + sqrt(sqrt(pi))].
8 = [pi + pi + sqrt(pi)]. (= [(pi + pi) + sqrt(sqrt(pi))] = [(pi + pi) * sqrt(sqrt(pi))].)
9 = [pi] * [pi]. (= [pi * pi] , fewer symbols)
10 = [pi] * [pi] + [sqrt(pi)].
 
Doug, who's going to start grad school next year, says, "I greatly enjoy mathematical problems and
puzzles, and having looked through the archives, I can say that it appears that your site offers the sort
of puzzles I particularly enjoy." - Thanks Doug, you have 240+ to browse; pace yourself! - Dan
WINNERS - Problem 240 . wow, a quarter of a hundred entries this time!. (back to top) . leader board
 
Denis Borris. . . . . . . . . 10 pts - Denis's style was to separate the integers as in [pi]+[pi]=6
Nick McGrath. . . . . . . . 7 pts - Set the trend with [pi+pi]+[\/pi]=7 and [pi*pi]+[\/pi]=10
Zeke Moore . . . . . . . . . . 5 pts - Set clean expressions for all ten; nice job in pi contest
Ed Wern . . . . . . . . . . . . 5 pts - Your .gif \/ pics got mixed around by e-mail but i deciphered
Ajit Athle . . . . . . . . . . . 4 pts - Moved up on the list by not making any errors, 21 pi's.
Tim Poe . . . . . . . . . . . . . 4 pts - Used [\/(pi+pi)] = 2 good but your 10 worked out to 12
V Balakrishnan . . . . . . 4 pts - A new force on the list; first ans but minus was illegal for 2
Juan Carlos Carrara . . 3 pts - Welcome back, nice expressions for 7 & 9, needed \/ in 10
Kirk Bresniker . . . . . . . 3 pts - Good, optimal on 9 of 10, used 3 pi's for 5, extra 3.1416
Zahi Teitelman. . . . . . . 3 pts - Thanks for your longtime entries; here your 10 was 11 w/o [ ]
Akifumi. . . . . . . . . . . . . 3 pts - Nice try with 7 = [pi^(\/pi)] ; alternate (legal) expres'n too!
Phil Sayre . . . . . . . . . . . 3 pts - First 9 were letter perfekt; the last may have needed [ ] .
Jeremy Galvagni. . . . . . 3 pts - Thanks for trying for the minimal symbols, got the min 21
Marcello Cammarata. . 3 pts - MCam tried to use each symbol just once; got up to 8!
Mark Rickert . . . . . . . . 3 pts - I now posted points for your 239, thanks for re-sending.
Yakov Macak . . . . . . . . 3 pts - There was exactly as much to this as met the eye (and the pi)
Mohamed Omar. . . . . . 3 pts - I liked your proofs involving weakly decreasing [ ] and \/.
Al Nelson . . . . . . . . . . . 3 pts - Hey Al, any time! Wow, 7 = [pi*(\/\/\/pi + \/\/\/pi)]. Cool!
Radu Ionescu . . . . . . . . 2 pts - Early entry; expressions seemed ok but way too many pi's
Hermen Jacobs. . . . . . . 2 pts - Thanks for sending them; one minus and a couple extra pi's.
David Madfes. . . . . . . . 2 pts - 5 = [\/(pi*pi*pi)] has nice symmetry but an extra pi.
Joe Skolnik (new) . . . . . 2 pts - Welcome to my contest; enter often! Much good answer.
Brady Wern (new) . . . . . 2 pts - Sorry about the delay; I hadn't rec'd your answer; welcome!
Thad . . . . . . . . . . . . . . . 2 pts - Good to avoid parentheses, extra challenge! Yr 10 ws 11.
Doug Babcock (new) . . . 2 pts - Thanks for the compliment on my site; g'luck in grad skool.
Ravi Raja. . . . . . . . . . . 2 pts - Just under wire; spelled out GreatestIntegerFunction lots ;-}

 

 THANKS to all of you who have entered, or even just clicked and looked.
My site is now in its 9th season - OVER 68,000 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 - 2006 A.D.
 
Problem Archives Index
 
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