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 message 12, then decode your result. Show all intermediate results.
202627
this might prove to be useful:\nhttp://islab.oregonstate.edu/koc/ece575/02Project/Mor/
466315
5| 2 -9 -7 13\n..... 10 ..5 -10\n-----------------\n....2 1 ..-2 3\n\nSo P(5) = 3
15479
When i type the site i get a safe that asks for a combination to open.\n\nif you want the combination, just open the source code and find the combination is 37697 or you can just go to the link below that takes you directly to to the open safe\n\nAny rupees you get you must wire me half j/k
472119
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
601027
Technically there are only 4 answers that will work, and they all rely on parentheses to modify the natural precedence rules:\n\n(5+5/5)*5\n(5/5+5)*5\n5*(5+5/5)\n5*(5/5+5)\n\nAs you can see they are all variations on (5 + 1) * 5, where you use 5/5 = 1.  So there is really just one answer.\n\nYou can easily solve problems like this yourself using the link below...\n\nFor example, can you make 24 using 3, 3, 8, 8?  Try it before you cheat... also, note that the solver allows % (mod) as an operator, so you may have to filter out those results.
427163
(12x^2 +16x) - (12x-7)- (4x-5)\n12x^2 +16x -12x +7 -4x +5\n\n12x^2+12\n\nthat is what I got....\nI think you might be able to put 12 on the other side of the = sign and turn it into 12x^2 = -12 and then since both sides have 12 in common you can divide both sides by 12 and you will have x^2 = -1 then you can add 1 to both sides and you will have x^2 +1 but I think I'm getting off track by that point....it's been a while since I had to do this stuff.
250464
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
659308
0000, 0001, 0002, 0003, 0004, 0005, 0006, 0007, 0008, 0009, 0010, 0011... etc, etc  untill u get to 9998, 9999\n\ntry it out, might take a long time but one combination will work :))
7438
RSA Laboratories usually has good information,\n\nhttp://www.rsasecurity.com/rsalabs/node.asp?id=2247\n\nhttp://www.rsasecurity.com/rsalabs/node.asp?id=2254
293574
the first step is basically to factor 120175.  the obvious factor to do first is 5.  this gives us 5*24035.  then factoring another 5 gives us 5*5*4807.  since 4807 obviously isn't divisible by 5 or 2, try 3.  doesn't work - so keep on trying the next biggest prime number (1,2,3,5,7,11,13,17...)  turns out that 4807 is divisible by 11.  this gives us 5*5*11*437.  keep on going and you get 5*5*11*19*23.  just checking, yes - 5*5*11*19*23=120175.\n\nnow converting these factors into letters, you get the available letters EEKSW.  this can be rearranged to be WEEKS.  however, remember that 5*5*11*19*23 can also be written as 25*11*19*23, since this also equals 120175.  this means the word could also have KSWY.  but i don't think you can rearrange those letters to make a word.\n\nso therefore the english word encoded as 120175 is WEEKS.  basically, you factored the number, converted them into letters, and solved the anagram.\n\n[[edit]]\n\ninuyasha's right - there could be an infinite number of "a"s in there too since 5*5*11*19*23 = 5*5*11*19*23*1*1*1*...\n\nhowever, this gives us WEEKS+AAA... so using the online anagram solver, found at http://wordsmith.org/anagram/ , there are no words that have "WEEKS" in them and "A"s.  therefore, the answer is still WEEKS.
159682
Below you will find the specifications for the AES (Rijndael) and Three-DES algorithms.  \n\nAs for the writing of the programs themselves, I'm almost done with my CSE degree--sounds like your pain is just beginning :-D
394399
P = Ae^rt\nP = 100e^(.2*52)\nP = 100 * 32824.21\nP = 3282421
311681
class publication\n{\n   private :\n      char title[20];\n      float price;\n   public :\n      void getdata()\n      {\n          cin>>title;\n          cin>>price;\n      }\n};\nclass books :: public publication\n{\n   private :\n      int pagecount;\n   public :\n      void getdata()\n      {\n          cin>>pagecount;\n      }\n\n};\nclass tape :: public publication\n{\n   private :\n      int playing_time;\n   public :\n      void getdata()\n      {\n          cin>>playing_time;\n      }\n};
417388
Just expand the right hand side:\n2x^2 - 12x + p = q(x^2 - 2xr + r^2) + 10\n2x^2 - 12x + p = qx^2 - 2qrx + qr^2 + 10.\nNow, for that to be always true, the coefficients have to match up.\nSo q = 2, 2qr = 12, qr^2 + 10 = p.\nSo q = 2, r = 3, p = 28.
436803
First you will have to generat your key:\n\ngpg --gen-key\n\nYou will have to answer a few questions to make the key you want. Defaults will be OK. But if you want more security try a 4096K key.\n\nOnce you have your key you will only need to execute the next command:\n\ngpg -e file_you_want_to_encrypt\n\nYou will need to introduce the name you used in your key, leave the destinatary as the default since you only want to encrypt and that will be all\n\nIn case you want to exchange files with friends you will need a public key to receive files from them and their public key to encrypt files for them.
393984
I don't follow this formula.  I tried it for 7.\n\n((7-1)/2)! (mod 7) is 3! mod 7 = 6 mod 7 = 6.\n\nNow,6 mod 4 = 2 and is not =1.\n\nAlso, it doesn't work for p=2.  I think it should be odd primes.
283351
V=UG >> a code word = input which is of 2^k different values\nG=[P I]>> G =generating matrix \nH=[I P']>> H=Parity check matrix\n\nso using some row operations \n\n1. switch places of R1 and R4\n2. add R3+R4 place it in R3\n3. add R2+R1 place it in R2\n4. add R1+R3 place it in R1\n5. add R1+R2 place it in R1\n\nyou'll get \n\nG = 1 1 1 1 1 0 0 0 \n    1 1 1 0 0 1 0 0 \n    1 0 0 1 0 0 1 0\n    1 1 1 0 0 0 0 1\n\nthus \n\nH=  1 0 0 0 1 1 1 1\n    0 1 0 0 1 1 0 1 \n    0 0 1 0 1 1 0 1 \n    0 0 0 1 1 0 1 0 \n\n\nRegarding the required to prove:\n\nit's simple \n\nyou have k=4, then you'll have 2^4 inputs which are multiplied by your generating matrix ,for example if u=0011\nyou'll sum last two rows of G to get V\n\nthen any V(codeword) is only a summation of a number of rows of G\n\nand since that the weight of any row of G is even that's why the output V's will always have an even weight ...!!\n\nsee i told u its simple... :)\n\nhope i ve been of help