Using p=5 and q=7, generate a public and a private key for RSA as shown in class. Use the keys to encode the m
450183
this might prove to be useful:\nhttp://islab.oregonstate.edu/koc/ece575/02Project/Mor/
406361
well the theorem says that a^p ≡ a(modp) where p is prime.\nequivalent to this is a^(p-1) ≡ 1(modp) by just taking the a to the other side. \n\nso we know 5^6≡1(mod7)\n\nso\n5^74 = (5^6)^12 x 5^2 ≡ 1 x 5^2 ≡ 25 ≡ 4 (mod 7)
444950
2P = (2,4)\n4P = (4,sqrt(72))
13208
I assume you talk about symmetric encryption (same key to encrypt and decrypt).\n\nCheck out the 3DES implementation (System.Security.Cryptography.TripleDES). It's really easy:\n1) call TripleDES.Create()\n2) call GenerateKey() and GenerateIV()\n3) configure a CryptoStream with the result of CreateEncryptor()\n4) write to it\n5) save Key and IV somewhere\n\nThen to decrypt:\n1) call TripleDES.Create()\n2) change Key and IV to the values you saved previously\n3) configure a CryptoStream with the result of CreateDecryptor()\n4) read from it
851855
You can use the "CrypTool" for starting. It is a tool for encrypting/decrypting or signing documents with all high end ciphers/coders. the Web address is "http://www.cryptool.com/". If you realy want to know the concepts then you read the book,\n"Applied Cryptography" by Bruce Scheneir. Or you can use the book "Hand book of applied cryptography" , it is free downloadable.\nIf you are looking for some software development then you can use the "cryptlib", it is also free downloadable.
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tennredbear is exactly right. They want to know if you have an apptitude for cryptology (code breaking/writing). Tough but excellent rating.\n\nGood luck!
487838
1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
7923
There isn't one. Mathematicians are working on it, but they have yet to find a pattern that covers all prime numbers (except for the fact they are divisible only by themselves and one.) There are patterns that cover some prime numbers, such as paired primes (where the difference between the two prime numbers is 2, such as 5 and 7), but there is no currently known way to quickly produce or check primes, which has important implications for the RSA algorithm (used in PGP for example), which relies on the complexity of primes for its security.
916413
as both are public key encryptions, both should be equally secure, but ecc is being adopted more often simply because it uses smaller key lengths viz a 163 bit ecc key would be equivalent to a 1024 bit rsa key. different algo\nlol
178588
Rearrange as (-7)(-11) m^(-6) m^(-4) u^(-4). Then combine the two coefficients into 77, and add exponents for the same base m, to get m^(-10). The final result is \n77 m^(-10) u^(-4). By the way, you should ask your teacher or consult the book to see if exponents should be made positive. In that case write the answer as 77 / [m^10 u^4].
55122
Better way is to creat an instance of Random class.\nIt has methods, nextInteger(), nextdouble(), and others.
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The first five are simple\nP(5,1) + P(5,2) + P(5,3) + P(5,4) + P(5,5) \n5 + 20 + 60 + 120 + 120 = 325\n\nSix and seven are a little more complicated\n6!/2! + 7!/2!2!\n360 + 1260 = 1620\n\nEight is special because you limit the maximum.\n\nList of bad values\n9x,xxx,xxx\n7x,xxx,xxx\n69,xxx,xxx\n67,xxx,xxx\n66,xxx,xxx\n65,9xx,xxx\n65,7xx,xxx\n65,6xx,xxx\n\n8!/3!2! - (2 * 7!/3!2!) - (3 * 6!/3!) - (3 * 5!/3!)\n3360 - 840 - 360 - 60 = 2100\n\nNow add the results.\n325 + 1620 + 2100 = 4045
393385
This link will solve your problem, just download the utitlity.  \nNow you better rate my answer the best one since it does not include calling any one :)\n\nhttp://www.softpedia.com/get/Security/Decrypting-Decoding/Asterisk-Key.shtml
749939
Nobody can help you because it's false.\n\nLet a1 = 1, a2 = 4, b1 = 2, b2 = 3, and n = 7\nThen a1 + a2 = 5, b1 + b2 = 5\nBut a1*a2 = 4, while b1*b2 = 6.\n\nYou are talking about modular arithmetic, right?
537744
take a look at this page:\n\n\nsimple encryption using table :\nhttp://www.planet-source-code.com/vb/scripts/ShowCode.asp?txtCodeId=5963&lngWId=3
52716
I hope you can understand the pseudocode, it's pretty loose.\n\nConstruct two arrays P and Q.\nLet P be empty, and Q contain the n numbers you want to permute.\nTo start, call the function PrintPermute below passing in the arrays P and Q.\n\nPrintPermute( P, Q)\n{\n  if Q is empty, print the values in P in order and return\n  else\n\n  for each value q in Q,\n  {\n       Create the strings P' and Q' where P' = P + q and Q' = Q - q\n       Call PrintPermute(P', Q')\n }\nreturn\n}
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learn at\nhttp://iisme.org/etp/HS_Math_-_Intro_to_Cryptography.pdf