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wu :: forums    riddles    medium (Moderators: ThudnBlunder, william wu, SMQ, Icarus, Eigenray, towr, Grimbal)    Concatenated Squares « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Concatenated Squares  (Read 423 times) ThudnBlunder wu::riddles Moderator Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Concatenated Squares   « on: Jan 13th, 2004, 7:32am » Quote Modify Find two four-digit numbers A,B such that when they are concatenated they form an eight-digit number N equal to A2 + B2, as in the simpler example below.     122 + 332 = 1233   (Numbers with leading zeros are inadmissible.)   « Last Edit: Jan 13th, 2004, 9:10am by ThudnBlunder » IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: Concatenated Squares   « Reply #1 on: Jan 13th, 2004, 11:16am » Quote Modify :: first analyze the problem a bit a^2+b^2=10000*a + b a^2 -10000a + b^2-b = 0 a = (5000+-sqrt(25000000-b^2+b))))   and by trying out b's between 1000 and 5000 we find a=9412 b=2353 :: IP Logged Wikipedia, Google, Mathworld, Integer sequence DB Quetzycoatl Newbie     Gender: Posts: 46 Re: Concatenated Squares   « Reply #2 on: Jan 13th, 2004, 11:23am » Quote Modify :: 9412, 2353 ::   Dagnabit Towr, I just noticed you snuck your answer in before mine. Anyway I got mine the same way you did. « Last Edit: Jan 13th, 2004, 11:28am by Quetzycoatl » IP Logged ThudnBlunder wu::riddles Moderator Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Re: Concatenated Squares   « Reply #3 on: Jan 13th, 2004, 5:25pm » Quote Modify on Jan 13th, 2004, 11:16am, towr wrote: we find..... In fact, brute force is not required on this one.   Let 10000A + B = A2 + B2   Multiplying by 4 and completing the square,   (2A - 10000)2 + (2B - 1)2 = 1 + 100002 = 17*5882353   As both prime factors are of the form 4n+1, we can express them as the sum of two squares. 17 = 12 + 42 = x2 + y2 (say) 5882353 = 5882 + 23532 = z2 + w2 (say)   (x2 + y2)(z2 + w2) = (xz - yw)2 + (xw + yz)2 or (x2 + y2)(z2 + w2) = (xz + yw)2 + (xw - yz)2   The first identity yields nothing interesting. The second gives 47052 + 88242     So 2A - 10000 =  [smiley=pm.gif]8824 and 2B - 1 =  [smiley=pm.gif]4705   Hence A = 9412 (588 is inadmissible) B = 2353 N = 94122353   « Last Edit: Jan 15th, 2004, 1:53am by ThudnBlunder » IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: Concatenated Squares   « Reply #4 on: Jan 14th, 2004, 12:21am » Quote Modify on Jan 13th, 2004, 5:25pm, THUDandBLUNDER wrote: In fact, brute force is not required on this one. It's not like I tried to match any 1000<=B<=9999 to any 1000<=A<=9999 , So it's not that brute..   Yours is a nice solution. But I don't really see how I could have found it, unless I knew about it beforehand.. At least not without using a lot more time, or some luck (which I never have).   And how did you find Quote: 5882353 = 5882 + 23532 = z2 + w2 (say) ? (interesting how that second square is B squared..) « Last Edit: Jan 14th, 2004, 12:29am by towr » IP Logged Wikipedia, Google, Mathworld, Integer sequence DB ThudnBlunder wu::riddles Moderator Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Re: Concatenated Squares   « Reply #5 on: Jan 15th, 2004, 3:28am » Quote Modify Quote: And how did you find... I used the function FactorInteger in Mathematica. FactorInteger[5882353, GaussianIntegers -> True]   (I guess that's also brute force.) I don't think there's an analytical method, as opposed to an algorithm, for finding two squares which when added equal a given prime of the form 4n+1.     Quote: (interesting how that second square is B squared..) Do you mean it's interesting that one of the factors (5882353) of 1000001 happens to feature in the answer? Yes, it is.     I looked at some smaller numbers.   10A + B = A2 + B2 (2A - 100)2 - (2B - 1)2 = 1 + 102 = 101 As 101 is prime we can conclude that there are no numbers such that 10A + B = A2 + B2   (except A = 10, B = 1)   ===========================================   100A + B = A2 + B2   (2A - 100)2 + (2B - 1)2 = 1 + 1002 = 11*13*17   13 = 22 + 32 17 = 12 + 42   We get A = 12 or 88 B = 33 N = 1233 or 8833   =========================================   1000A + B = A2 + B2   (2A - 1000)2 + (2B - 1)2 = 1 + 10002 = 101*9901   101 = 12 = 102   9901 = 102 + 992   We get A = 990   B = 100 N = 990100     « Last Edit: Jan 15th, 2004, 7:36am by ThudnBlunder » IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. 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