20210711, 19:07  #12 
Mar 2018
530_{10} Posts 
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20210711, 20:35  #13 
Mar 2018
2·5·53 Posts 
215*107*12^2 is congruent to 71*6^6 which is congruent to 72 mod (331*139)
331*139*8 is congruent to 8 mod (215*107) s^2 is congruent to 1 mod 23005 the first not trivial solution is s=429 the second is s=9201 the third is s=13374 (13374^21)/23005 is congruent to 1 mod 6^5 215*107*(2+331*139*8)1 is a multiple of 331*139 and 429^2 mod 429^2 we have 23005*1840331 which is a multiple of 429^2 and 71 ((184033*230051)/71429^2)/429^2=18^21 92020 is congruent to 4*(2+331*139*8)^(1) mod 429^2 and mod (331*139) so 92020 is congruent to 4*(368074)^(1) mod 429^2 and mod (331*139) the inverse of 368074 so is 23005 92020 is congruent to 4*(429^22^2)^(1) mod (331*139*429^2) maybe it is useful 431=(427)^(1) mod 46009??? 331259 for example is congruent to (9203*4+1) mod (331*139) and 331259 is congruent to 9203 mod 23004 (92020)^(1)=23005 mod (331*139) 92020*23005=(46010)^2 92021 divides 215*107*(2+331*139*4)1 pg(69660), pg(19179) are primes maybe something useful can be derived from this: 69660 is congruent to 19179 which is congruent to 429^2 which is congruent to 9 mod (71) 69660 and 19179 are of the form 648+213s probably there are infinitely many pg(648+213s) which are primes in particular 6^6 is congruent to 19179 which is congruent to 429^2 which is congruent to 861 mod (71*43) 6^6 is congruent to 860 mod (214^2) 92020 is congruent to 2^0 mod (17*5413) 92020 is congruent to 2^1 mod (3*313*7) 92020 is congruent to 2^2 mod 11503 23005*(2+331*139*2^0)1 is a multiple of 11503 23005*(2+331*139*2^1)1 is a multiple of 3*313*7 23005*(2+331*139*2^2)1 is a multiple of 17*5413 The inverse of 9203 mod (331*139) is x 9203*x is congruent to 1 mod 429^2 331259 is congruent to 9203 mod 23004 23005 Is binomiale(214+1,2) so 23005 Is the 214th triangolare Number 92020=(858^2*139*331+4)/(2+331*139*8) 3371 and 331259 are primes pg(3371) and pg(331259) are primes 3371 and 331259 leave the same remainder 59 mod 3312 3371 and 331259 are primes of the form 59+ ((71*6^624^2)/10^3)*10^m, for some nonnegative integer m (46009x1)/(y^31)+y^3=(x^21)/2 over positive integers x=429 y=6 (23005*(2+46009*(4))1)/92021=46009 215^2 is congruent to 46009 mod 216 11503=71*2*3^4+1 92020 is congruent to 4 mod 11503 69660 is congruen to 3*6^3 mod 11502 92020 is congruent to 2^2 mod 11502 92020*23005 is congruent to 4 mod (71*11503) 23005*(2+331*139*(8+184041*(8+184041*(8+184041... is congruent to 1 mod (429^n) 23005*(2+139*331) is congruent to (429^21)/10 mod 230051=31*41*181 92020 Is 4 mod 11502 and 4 mod 11503 92020 Is congruent to (71*6^6/11502=288=17^21) mod 323=18^21 6^6/162=288 162 divides 69660 (2*(46009*6)^2+144*(2+46009*2))/313/49/36^3=71*6^6 46009+2 Is a multiple of 313*49*3 2*46009+2=92020 69660=3/2*(6^66^3) 9202069660=22360 which is divisible by 860 22360=860*(3^31) Z46009 is isomorphic to Z331XZ139 429^2 Is congruent to 2^2 which Is congruent. To (139*331*8)^2 mod (431*427) 71*6^3 is congruent to 23005*(2+46009*2) mod (7^2*313) 71*107*6 is congruent to 430 mod 11503 69660 mod 11503=642=107*6 642=6*(3*6^21)=6*107 11476*2*860+428 is congruent to 429 mod (3*313*49) Consider 71*6^6 is congruent to 72k mod j for k=1 j=46009 for k=2 j=4601 for k=3 j=(313*7^2*3) for k=4 j=11503 for k=7 j=9203 71*6^6 is congruent to 72*7 mod 9203 46008 (=331*1391) is congruent to 7 mod 9203 331259 is congruent to 7^2 mod 9203 or 9203=33125946008*7 (139*331)^2 is congruent to 1 mod (92020) and mod (23004) 69660 is congruent to 92020(15229*(2+46009*2)216)/313/49=648=3*6^3 mod (23004) 23005*(2+46009*2) is congruent to 1 mod (313*7^2) 71*6^6 is congruent to 6^3 mod (313*7^2) 23005*6^3 mod (313*7^2)=15229 (6^3/2)*(2+46009*2) is congruent to 6^3 mod (313*7^2) 69660*313*7^2 mod 23004=3*6^3 6^6(69660*49*313648)/23004=3*71 71*6^6 is congruent to 67 mod 139 and 259 mod 331 chinese remainder theorem to the rescue: 45937+46009k...allowing negative k, you have 92090 which is 2 mod 1001 and to 92020 331259 is 259 mod 331 331259 is 4588 mod 4601 MathCelebrity START HERE OUR STORY VIDEOS PODCAST Upgrade to Math Mastery Enter math problem or search term (algebra, 3+3, 90 mod 8) Invia Using the Chinese Remainder Theorem, solve the following system of modulo equations: x=259 mod 331 x=4588 mod 4601 Enter modulo statements x=259 mod 331 x=4588 mod 4601 Using the Chinese Remainder Theorem, solve the following system of modulo equations x ≡ 259 mod 331 x ≡ 4588 mod 4601 We first check to see if each ni is pairwise coprime Take the GCF of 331 compared to the other numbers Using our GCF Calculator, we see that GCF(331,4601) = 1 Since all 1 GCF calculation equal 1, the ni's are pairwise coprime, so we can use the regular formula for the CRT Calculate the moduli product N We do this by taking the product of each ni in each moduli equation above where x ≡ ai mod ni N = n1 x n2 N = 331 x 4601 N = 1522931 Determine Equation Coefficients denoted as ci ci = N ni Calculate c1 c1 = 1522931 331 c1 = 4601 Calculate c2 c2 = 1522931 4601 c2 = 331 Our equation becomes: x = a1(c1y1) + a2(c2y2) x = a1(4601y1) + a2(331y2) Note: The ai piece is factored out for now and will be used down below Use Euclid's Extended Algorithm to determine each yi Using our equation 1 modulus of 331 and our coefficient c1 of 4601, calculate y1 in the equation below: 331x1 + 4601y1 = 1 Using the Euclid Extended Algorithm Calculator, we get our y1 = 10 Using our equation 2 modulus of 4601 and our coefficient c2 of 331, calculate y2 in the equation below: 4601x2 + 331y2 = 1 Using the Euclid Extended Algorithm Calculator, we get our y2 = 139 Plug in y values and solve our eqation x = a1(4601y1) + a2(331y2) x = 259 x 4601 x 10 + 4588 x 331 x 139 x = 11916590  211089292 x = 199172702 Now plug in 199172702 into our 2 modulus equations and confirm our answer Equation 1: 199172702 ≡ 259 mod 331 We see from our multiplication lesson that 331 x 601731 = 199172961 Adding our remainder of 259 to 199172961 gives us 199172702 Equation 2: 199172702 ≡ 4588 mod 4601 We see from our multiplication lesson that 4601 x 43290 = 199177290 Adding our remainder of 4588 to 199177290 gives us 199172702 Share the knowledge! Chinese Remainder Theorem Video Tags: equationmodulustheorem Add This Calculator To Your Website <!— Math Engine Widget Copyright MathCelebrity, LLC at www.mathcelebrity.com. Use is granted only if this statement and all links to www.mathcelebrity.com are maintained. ><a href="https://www.mathcelebrity.com/chinese.php" onclick="window.open('https://www.mathcelebrity.com/chinese.php?do=pop','Chinese Remainder Theorem Calculator','width=400,height=600,toolbar=no,menubar=no,scrollbars=yes,resizable=yes');return false;">Chinese Remainder Theorem Calculator</a> Run Another Calculation Email: donsevcik@gmail.com Tel: 8002342933 MembershipMath AnxietyCPC PodcastHomework CoachMath GlossarySubjectsBaseball MathPrivacy PolicyCookie PolicyFriendsContact UsMath Teacher Jobs 331259/(2*12^2=288=17^21) is about 11502/10... 71*6^6 is congruent to  12^2 mod 4601 and mod 774 331259=11502*(17^21)9007*331 9007*331 cogruent to 9203 congrue nt to 331259 mod 11502 (9007*331+9203)/11502=259+1 331259*11502 is congruent to 4473*666 mod (23004*331) 4473=4472+1 maybe something useful can be derived by this: 139^(1)=331 mod 23004 for example 331259*139 is  9007 mod (1001*23004) (6^6)^(1)=22 mod 331 331259 is congruent to 22 mod 139 331259 is 72 mod 1001 92020 is 72 mod 1001 71*6^6 is 72 mod 331 331259, 92020 and 71*6^6 are numbers y that satisfy this congruence equation y is congruent to (72+331*(10^x1)) mod 1001 for some nonnegative integer x (71*6^6+72331*999)/972=331259=(71*6^6+72331*9993*6^3)/9 (359+71)*(6^6+11502)/359=69660 pg(359) is prime 69660 is congruent to 14 mod 359 6^6 is congruent to 14 mod 359 23004 is congruent to 331 mod 359 92020 and 331259 are 5 mod 239 92020 is congruent to 331259 which is congruent to 3^5 mod 257 92020+3^5=359*257 1001((331259243*292020)/257)=72 331259 and 92020 are 72 mod 1001 28 is congruent to (429^21) mod 257 239239 is congruent to 28 mod 257 92020+239239=331259 14 is congruent to 129*(429^21) mod 257 so 107 is congruent to 69660*92020*2 mod 257 from this follows 331259 is congruent to 14 mod 257 69660 is 13 mod 257 92020 is 14 mod 257 92020 Is congruente to 1 mod 829 and mod 37 331259=92020+239239 Is congeuent to  3*2^11 mod 829 and mod 37 541456 Is congeuent to 2 mod 37 (and also 331259 Is 2 mod 37) 541456+3*2^11 Is a perfect square PG(359) Is prime 331259*5 Is congruente (23004+331=multiple of 359)=4667 mod 6^6 71*6^6(331259*5(23004+331))=6^8 541456=43*(10^4+2^5*3^4) 280 Is congeuent to 2592=2^5*3^4 mod 359 69660 Is 28/2 mod 359 23004 Is 28 or 331 mod 359 The inverse of 10 mod 359 Is 36 69660*10^3 Is 1 mod 359 so 69660 Is 6^6 mod 359 6^6 =14 =69660=2592*18 mod 359 From here 3870=2592 mod (359x18) Dividing by18 215=12^2 mod 359 12^2=(71^21)/(6^21) mod 359 So (6^31)=(71^21)/(6^21) mod 359 PG(541456) PG(331259) and PG(92020) are primes 541456 92020 and 331259 are Numbers of the form a+1001*s where a is a Number congruente to 7 mod 13 a=72 and a=85 Because a=13d+7 and 1001=7*11*13 541456 92020 and 331259 are of the form 13d7(1143f) for some dnand f So (541456+7)/13 Is 71 mod 77 (92020+7)/13 and (331259+7)/13 are 72 mod 77 ((X^21)*(46009+1/4)1)/46009x^2=0 This Is a parabola for x=+ or  429 this goes to zero... 429^21=2*92020 I dont know of from that equation One can derive something more general parabola focus  (0, 33870353513/736148)≈(0, 46010.2) vertex  (0, 184041/4) = (0, 46010.3) semiaxis length  1/184037≈5.43369×10^6 focal parameter  2/184037≈0.0000108674 eccentricity  1 directrix  y = 33870353521/736148 This Is the parabola ((X^21)*(46009+1/4)1)/46009=y 429^2 Is congruente to 6^6 which Is congruente to 69660 which Is congruente to 9 mod 71 (429^29)/71=2592=2^5*3^4 Curious that 541456 Is divisible by (10^4+2592) 43*2592 Is 111456...the last digits 1456 are the same as in 541456 92020/2592/51/3240=71/10 324 divides 69660....i think there Is something involving 18^2 2*92020/25921/324=71 (429^21)/25921/324=71 1/69660=(1/215)*((184040/259271)) 2592=72^2/2 215/10*(20000+72^2)=541456 71*6^6/331259+1/(239*99) Is about 10... 1/(1/(71*6^6/33125910)/99+239)=277.199999... The inverse of 5 mod 46009 Is 9202 429^25 Is a multiple of 46009 (429^25) Is then congruent to 6 mod (239*7*11) 92020 Is 10 mod 3067 239239 Is 13 mod 3067 So 331259=92020+239239 Is 23 mod 3067 71*6^6 Is 6^3 mod 3067... 71*6^6=429^2=3=239239=3*6^4 mod 37 92020 for example =1 mod (37*3*829) 429^2=3 mod (37*829) 331259=92020+239*1001 so 331259 is 2 mod 37 331259 is congruent to 9203 mod 23004 9203 mod 71=44 (33125944)/71=4665 4665 are the first four digits of 6^6=46656 4665 in base 6 is 33333 a repdigit 331259 is congruent to 9203 mod 23004 9203 is 5 mod 7 331259 is 5 mod 7 9203 is 44 mod 71 331259 is 44 mod 71 so 331259 and 9203 are numbers of the form 19143+497k curious that 19143+6^2=19179 and pg(19179) is prime curious that allowing negative numbers k 1234 is a number of the form 19143+497k 9203 and 331259 are also congruent to 131 mod 648=3*6^3 so using CRT they are numbers of the form 9203+322056k 71*6^6 is congruent to (9203131)/648 mod 331259 (331259131)/648=2^91 71*6^6/6484601=2^91 4601 divides 92020 Numbers of the form 512, 5112, 511...12,... The difference 5112512, 511125112,...Is a multiple of 46 (3312599203)/5112=2^61 331259/5112 is about 64,8...=648/10 71*6^6=5112*3*6^3 331259/648=511,20216... 1/216=46/10^4+1/(10*15^3) from above 370=92020X648 mod 511 370=92020X137 mod 511 138010=92020 mod 511 6^6=(13801092020) mod 666 (20*71*6^64601*648*20)/511=10*6^4 370=92020X648 mod 511 370=40*648=10*2^5*3^4 mod 511 138010=92020 mod 511 69005=46010 mod 511 pg(69660) is prime 6966069005 is a multiple of 131 pg(331259) is prime 331259=(6^21) mod 13801 370=92020X((696609155)/511+1)=92020X137=92020X(70007)x511^(1) mod 511^2 155 is 6^6 reduced mod 511 so 370X511=92020X70007 mod 511^2....92020 reduced mod 511 is 40 (40*70007370*511)/511^2=10 Last fiddled with by enzocreti on 20211015 at 13:45 