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CTK Exchange
Paul_B
guest
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Dec-23-07, 06:46 PM (EST) |
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"Virginia Tech Regional Mathematics Contest"
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I have been struggling unsuccessfully with problem 5 (below) from the
29th Annual Virginia Tech Regional Mathematics Contest October 27, 2007. Not sure my physics degree maths is up to it!https://www.math.vt.edu/people/plinnell/Vtregional/E07/index.html Any hints/pointers gratefully received. Many thanks, Paul.
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Find the third digit after the decimal point of (2 + sqrt(5))^100((1 + sqrt(2))^100 + (1 + sqrt(2))^-100) For example, the third digit after the decimal point of pi = 3.14159... is 1. |
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alexb
Charter Member
2155 posts |
Dec-23-07, 08:51 PM (EST) |
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1. "RE: Virginia Tech Regional Mathematics Contest"
In response to message #0
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>Find the third digit after the decimal point of
>
>(2 + sqrt(5))^100((1 + sqrt(2))^100 + (1 + sqrt(2))^-100)
>First note that the second factor is an integer. It's the same as (sqrt(2) + 1)^100 + (sqrt(2) - 1)^100. Denote it S. You can estimate S: S < 2*10^38. The first factor is clear in the form n + m*sqrt(5), where n and m are integers. Its conjugate, viz., (2 - sqrt(5))^100 equals n - m*sqrt(5). So that the two add up to an integer. You can estimate the conjugate to be 2*10^(-63). It follows that the product (2 - sqrt(5))^100 * ((sqrt(2) + 1)^100 + (sqrt(2) - 1)^100) has at least 20 zeros after the point. This means that the given expression has as many 9's after the point. In particular, the third digit is 9.
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alexb
Charter Member
2155 posts |
Dec-24-07, 09:50 AM (EST) |
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3. "RE: Virginia Tech Regional Mathematics Contest"
In response to message #1
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> In particular, the third digit is 9. I just ran an infinite precision calculator bc in unix setting the scale parameter (the precision) to 1000 (places after the point). The result of (2 + sqrt(5))^100 * ((sqrt(2) + 1)^100 + (sqrt(2) - 1)^100) was this 94158733601034420664808450657998303298219601745567527892456021922994\
873597395955752869490271254871747.9999999999999999999999996186915243\
50724296156402933296675021218116222226597721314268654625211899960519\
42368418009208191432319697804575525158667593134273807734411639388949\
32307648957195371189623886453935449571231143314936542103641377974274\
57373848409994346615470477246474706794413372086958553952574754666142\
29151114932075632574364865209856433169780322945448733389644181892441\
27420973432139212328150618338564776159770105288608103158800096011223\
72506941768760367348138758305106586004736218238712569915542585026230\
90330479907851266703159940179345128325678400225654712399487081639016\
84261320793017096568967024222159685302092319797120240474004627295974\
79740246314514910447171527980998956002895372283771020400439508108455\
74371548252312823314171379553283665051681156641376209779187966659681\
94531615433196352583116152196656759946748691099745060576940076512147\
50792255686835245432440450203344714912238473742309557459413566080518\
10248051261191700248203749909991013069100325677059884894391440358506\
72718850849642 with 24 9's after the point. Recall 63 - 38 = 25, give or take. |
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alexb
Charter Member
2155 posts |
Dec-24-07, 09:55 AM (EST) |
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4. "RE: Virginia Tech Regional Mathematics Contest"
In response to message #3
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Ah, for (2 - sqrt(5))^100 * ((sqrt(2) + 1)^100 + (sqrt(2) - 1)^100) I got .0000000000000000000000003813084756492757038435970667033249787818837\
77773402278685731345374788100039480576315819907918085676803021954244\
74841332406865726192265588360611050676923510428046288103761135460645\
50428768856685063457896358622025725426261515900056533845295227535252\
93205586627913041446047425245333857708488850679243674256351347901435\
66830219677054551266610355818107558725790265678607876718493816614352\
23840229894711391896841199903988776274930582312396326518612416948934\
13995263781761287430084457414973769096695200921487332968400598206548\
71674321599774345287600512918360983157386792069829034310329757778403\
14697907680202879759525995372704025202597536854850895528284720190010\
43997104627716228979599560491891544256284517476871766858286204467163\
34948318843358623790220812033340318054683845668036474168838478033432\
40053251308900254939423059923487852492077443131647545675595497966552\
85087761526257607592788312756316597036348867940950977200762216898101\
9809954995385479261942702746357095042592317611233 |
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mr_homm
Member since May-22-05
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Dec-24-07, 01:37 AM (EST) |
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2. "RE: Virginia Tech Regional Mathematics Contest"
In response to message #0
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These numbers of the form (a + b*sqrt(c)) always show recursion when raised to integer powers. (a + b*sqrt(c))^2 = a^2 + b^2*c + 2ab*sqrt(b), so the coefficients collect into a part with sqrt(c) and a part without it, so the number retains the same form as you raise it to higher powers. (2 + sqrt(5))*(2 + sqrt(5)) = 9 + 2sqrt(5)
(9 + 2sqrt(5))*(2 + sqrt(5)) = 28 + 13sqrt(5)
and so on. Let's say that the current value is x + y*sqrt(c). Then the next value is
(x + y*sqrt(c))*(a + b*sqrt(c)) = ax + bcy + (ay + bx)sqrt(c) = x' + y'*sqrt(c).
Now you can treat these as vectors with basis {1, sqrt(c)}, so that at each stage you are multiplying by a matrix:
x' a bc x
y' = b a * y
Let's call this matrix M. Then you are computing M^100 applied to the vector (x_0, y_0) = (1, 0). Computing powers of a matrix should be within the type of math you learned for physics, since it is part of the theory of exponentiating a matrix, which occurs in solving simultaneous linear differential equations. Note that you will only want the y value, since the x value corresponds to an integer, so it does not contribute anything to the third decimal place. You can probably take it from there. Hope that helps ! --Stuart Anderson |
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