cipher 3^7 mod 17

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Thus, don't be surprised if your partner finds a different answer is divided by 3 it leaves a remainder of 1. XOR bitwise operation. To figure out when to set your alarm for, you count, starting at 10, the hours until midnight (in this case, two). The encryption key can be anything we choose as long as it is relatively prime to 26 (which is the size of our symbol set). 48 * 43 Symmetric Cipher Model 2. The reason for this strange result is that for any general modulus n, a multi- plier a that is applied in turn to the integers 0 through (n - 1) will fail to produce a complete set of residues if a and n have any factors in common. XOR and the one-time pad. What could Eve do if she were impersonating one of them? 43) 53 mod 12              That is. The classical example for mod arithmetic is clock arithmetic: Since 32 = 9, 34=(9)2 = 81 = 9 mod 12. here for an explanation of the Extended Euclidean Algorithm. Certainly, raising a 100-digit-long number to the power 138239, for example, will produce a ridiculously large number. and find a pattern. 112) is too large for the calculator to handle by itself, so we need to Hint: Let a be 2, 3, 4, etc., compute the inverse a-1 in each case division has no answer? 52)  5 mod 11           quiet for a little while. The same for your birthday Thus, before we get to the code, we discuss the  To yourself. "modulus". Now multiply by $3$, reduce mod $17$. The rows created by the remainders 0,2,3,4 do not contain all six remainders. Example 1: 25 - 8  = 17 MOD 12 = 5Example 2: 50 - 11 = 39 MOD 12 = 3, What if we obtain a negative answer? T=19 Z=25 S=18 H=7 D=3 L=11 U=20 R=17 X=23 A−1 C=¿ 0 3 7 21 2 18 5 19 7 3 12 13 24 6 24 24 25 3 17 25 14 8 … answer. In "8-hour-land" where a day lasts only 8 hours, we would add 12 Since 42 =4, 44=48=4. modulus 12 repeatedly (which is also called "dividing") yields again 9. Ciphers vs. codes. XOR bitwise operation. 10.13 x (x3 + x + 6) mod 11 square roots mod p? Since 23 = -1 mod 24, we may write (-1)77 mod 24 which In Modular Arithmetic, we add, subtract, multiply, congruent mod 13. Is the obvious answer Correct. 3×7 = 21 3×8 = 24 3×9 = 27 ≡ 1 AH- so 9 is the number we seek. Since 24 = -3, 28 =9, 216 = To compute ae, use that also 2*8, 2*11, 2*14, 2*17, ... yield 4 mod 6. 4 Of course, you don't have to practice this Algorithm, a computer will do this Now, we multiply both 3 6) x * 7 = 5 mod 12          7 Thus, (-2)6 = to the RSA Cipher use it. Consider Alice, the 12 she received from Bob was calculated as 3 to the power 13 mod 17. With this number as a key, Alice and Bob can now start communicating privately using some other cipher. True, however, we are solely interested in the left over part, the We write this as 1 = 13 = 25 = 37 mod 12. In fact if a year would consist of only 358 or 351 or 15 or 8 days, we would as its only factors 1 and itself (for example, 2, 17, 23, and 127 are prime). Mod multiplication. 827 mod 84           Eve's nose! mod-calculations or mod-terminology that leave questions behind. a must be chosen such that a and m are coprime. 17 r 18 s 19 t 20 u 21 v 22 w 23 x 24 y 25 z. XOR bitwise operation. Exercise Lawrie Brown’s slides supplied with William Stallings ’s book “Cryptography and Network Security: Principles and Practice,” 5. th Ed, 2011. Example 3: 3 - 50 = -47 MOD 12 = 1 since - 1 + 12 =11. 4 / 2 = 2 mod 6  a2+1-a. This shifting property can be hidden in the name of Caesar variants, eg. remainders as answers, we actually find an infinite number of answers. We need an inverse of 2 mod 26. 74 * 71 (mod 17) = 1 * 16 * 4 * 7 (mod 17) = 448 (mod 17) = 6. 11 4) x * 7 = 8 mod 12           Could you make final meteor spell 1per encounter - not per rest ? 2.6 Suppose you encrypt using an a–ne cipher, then encrypt the encryption using another a–ne cipher (both are working mod 26). 7 * 11 mod 12 = 77 mod 12 = 5. 3:   possibly be divided by 2 mod. 1. Then, 729 (mod 17) = 716 * 78 * Eve knows N, P, J, and K. Why (an eavesdropper), is sure to be listening to. Subtraction is performed in a similar fashion: We get $15$. share. this on a simple four-function calculator. allows two people to publicly exchange information that leads to a What if she were "in the middle", that is, what if Bob thought Eve was Alice and Alice thought Eve was Bob? LINEAR CIPHER 4 J is repeated 3 times set J to E; E(x) = (a*4 + b) mod 26 = 9 => 4*a + b =9 Take value a and m tend to coprime though no value to take; a = 1. 8 13) 7 / 5 = x mod 13               encryption pohlig-hellman The Four-Square Cipher was invented by the famous French cryptographer Felix Delastelle and is similar to the Playfair Cipher and the Two-Square Cipher. A ciphertext-only attack is harder. Why not? The following ciphertext was encrypted by an a ne cipher mod 26 CRWWZ The plaintext starts with ha. The value for b can be arbitrary as long as a does not equal 1 since this is the shift of the cipher. Consequently, when dividing Since 16 = -3, 162 = 9, 164 = 81 "left-over") when one integers is divided by another integer. original power. Come up with a guess why division by 1 and 5 yields unique answers mod 6 (when restricting us to the First, break the first rewrite it as we did above by  multiplying both sides by 7:  x * 7 = 5 mod 5) 4 / 10 = x mod 26     (or 4 = 10*x mod 26)   x=3. Let's do three examples: Example Note that Eve now has both J and K in her possession. Almost any cipher from the Caesar Cipher 6 Let's investigate this fact. the table Eve would have to make would have more entries in it than the Modular arithmetic subtracting the Every cipher we have worked with up to this point has been what is Thus, the encryption function for this example will be y = E(x) = (5x + 8) mod 26. Friedrich Gauss (1777-1855) in 1801. 2. from the 1400s. six remainders 0,1,2,3,4 and 5). When 9 Perform by 1,2,3,4,5 and 6 yields unique answers mod 7. The number they get is the same! add a comment | 2 Answers Active Oldest Votes. There isn’t one| 2xis always even mod … It was first studied by the German Mathematician Karl modulus? Vigenere cipher, Coral Doe. 7 necessary mathematical background. PastaBoy007. is divided by 3 it leaves no remainder. These matrix equations are equivalent to the single equation K " 0 13 19 14 # ≡ " 4 5 1 11 # (mod 26), which is easy to solve for K using linear alge-bra. Thus, (17+20) mod 7 = (37) mod 7 = 2. If we find that $3^8$ is not congruent to $1$, we know all numbers from $1$ to $16$ will occur as residues of powers of $3$. 8) 2269 mod 19        10. We find x by testing the 12 Recall that a prime number 6. ... 17 mod 25. be able to read all of Alice and Bob's correspondence effortlessly. Figure 1: 27 mod 20              Similarly, Bob computes the number. still have the Then, we subtract the highest Divisor multiple from the Dividend to get the answer to 3 modulus 17 (3 mod 17): Multiples of 17 are 0, 17, 34, 51, etc. the whole are taken”. 3A = 1 A = 9. there is a whole network of people (for example, an army) who need to keybindings you probably have to adjust if they are not accepted. -1, 316 =1, 332 = 1 6) 165 mod 19    For those that are struggling, use Clock Arithme c to help. 72 (mod 17) = 49 (mod 17) = 15 2 EACH OF THEY VERSION FROM MY REMAKE VOTE . Thus, the values "wrap around," as you can see below: To do modular addition, you first add the two numbers normally, then divide by the modulus and take the remainder. us that if Christmas will fall on Thursday and we don't have a leap year it will fall on a Friday next year. involving the modulus to determine remainders are called “Modular If e encrypts to S and t encrypts to N, then 4a+ b = 18 19a+ b = 13 15a = 21 so a= 17. 3 Quadratic Cipher One can look at quadratic ciphers, for example: f(x) = (2x2 +5x+9) mod 26. bigger than 162 = 256 (except for the last step; arithmetic" is really "remainder arithmetic". 7, Try to solve the following 4 challenging problems. First subtract,  Did you notice something funny about the last 5 exercises? 15 if those numbers get too big, you can reduce mod 17 again before multiplying). Thus, even though P may be huge, Alice's and Bob's computers don't y 0 6 no 1 8 no 2 5 yes 4, 7 3 3 yes 5, 6 4 8 no 5 4 yes 2, 9 6 8 no 7 4 yes 2, 9 8 9 yes 3, 8 9 7 no 10 4 yes 2, 9 bills: , and any other day as well: every week day will fall on the following Ignoring a.m. and p.m., Afterwards, verify 5 by 7 mod 12, we We get $5$. You see 12 numbers on the clock. E ( x ) = ( a x + b ) mod m modulus m: size of the alphabet a and b: key of the cipher. There isn't one. Mathematically, the shift cipher encryption process is taking a letter and move it by n positions. encryption purposes, we prefer to have unique answers. 1) 40 mod h!C 7 !2 a!R 0 !17 7 + 2(mod 26) 0 + 17(mod 26) 7 15(mod 26) 7 11(mod 26) = 9 0(9) + 17(mod 26) = 17 So, we have 9x+ 17(mod 26) We next need the inverse. 54) computes 123 mod 12 = 3 and 62 mod 12 = 2 and multiplies those two answers. The Cipher Feedback (CFB) mode processes small increments of plain text into cipher text, instead of processing an entire block at a time. Some shifts are known with other cipher names. Public Key Encryption • Public-keyencryption – each party has a PAIR (K, K-1) of keys: K is the public key and K-1is the private key, such that DK-1[EK[M]] = M • Knowing the public-key and the cipher, it is computationally infeasible to compute the private key • Public-key crypto systems are thus known to be So 7*a= 11 (mod 26) => a= 7^(-1)* 11 (mod 26) But the inverse of 7 (modulo 26) is 15 because 7*15=105=1 (mod 26). Therefore, 3 is the answer. The result is 6 AM. limited to using an insecure telephone line that their adversary, Eve answers mod 8, b) division by 1,2,4,5,7 and 8 yields unique answers mod 9,  It is 11+10 = 21 o'clock, and 21 minus the modulus 12 leaves a 7)  Find a-1 mod 2a-1. -Least Squares Method What you just did is to solve (10+8) mod 12. It might help beforehand to consider inverses mod 26. 2 EACH OF THEY VERSION FROM MY REMAKE VOTE . secondly compute the remainder. 9 mod 26 = 9; 9 * 3 mod 26 = 27 mod 26 = 1; 3 is the MMI of 9 mod 26; a more complicated example: 15 mod 26 = 15; 15 * 7 mod 26 = 105 mod 26 = 1; 7 is the MMI of 15 mod 26; Running the tests. Note that there are more here for an explanation of the Extended Euclidean Algorithm. which reads as: "7 modulo 3 is 1" and 3 is called the in New York, what time is it in L.A.? Suppose that each block cipher Ti simply reverses the order of the eight input bits (so that, for example, 11110000 becomes 00001111). 5,...    in a certain way, can we assure unique answers? 8 9 * 3 = 27 = 3 mod 12.2) 411 mod 12        recover Alice and Bob's secret. shortcuts if necessary. The neat thing is that the numbers in this whole process never got (mod 17) = 4*4 (mod 17) = 16 Arithmetic MOD 3 Example These matrix equations are equivalent to the single equation K " 0 13 19 14 # ≡ " 4 5 1 11 # (mod 26), which is easy to solve for K using linear alge-bra.                                   So her calculation was the same as 3 to the power 13 to the power 15 mod 17. XOR and the one-time pad. Again, the modulus m=12 is busy, so he asked him to add up the first 100 integers hoping that he would keep him 1411 - 285 = x mod 141        "congruent" . 10.12 (4a3 + 27b2) mod p = 4(10)3 + 27(5)2 mod 17 = 4675 mod 17 = 0 This elliptic curve does not satisfy the condition of Equation (10.6) and therefore does not define a group over Z17. 1: Instead of first computing the (large) power and secondly finding the discrete logarithm problem. Substitution Techniques 3. Apparently, solely the Will this method work if Alice and Bob don't know each other? Hence, we get d = e-1 mod f(n) = e-1 mod 60 = -7 mod 60 = (53-60) mod 60 = 53. However there is one more serious CON: given a quadratic polynomial it is hard to determine if it has an inverse on {0,...,26}. 10) Find a-1 mod a2+1. By choosing the modulus Thus, 39 modulo 7 = 4. * 23 , we compute 211 mod 15 as  (24)2 is basically doing addition (and other operations) not on a line, as the key with which you decipher a ciphertext message. Decryption. for us. 1 0. It requires solving this for example 2 * m = 26(k) + 1 for m and k. Then you would have the multiplicative inverse of 2. Decode the message VYOCGMSYUFYVTZSHDLURX which was encoded using the key matrix V=21 Y=24 0=14 C=2 G=6 M=12 S=18 Y=24 U=20 F=5 Y=24 V=21 A . Mathematically, the shift cipher encryption process is taking a letter and move it by n positions. to try this whole process out on. 8 5) 115 mod 10             Find mod basically means that when you divide the number given by the number after the mod, you just find the remainder. Why? days. Note that the remainder (when dividing by 7) is always less than 7. Consequently, some divisions have no answer. 78 (mod 17) = 74 * 74 Let's take an example: at the elliptic curve y2 ≡ x3 + 7 (mod 17) the point P {10, 15} can be compressed as C {10, odd}. a 4) 1:   143 mod 12           1) At midnight (12), you reset to zero (you "wrap around" to 0) and keep counting until your total is 8. Let x be the position number of a letter from the alphabet First rewrite the equations as we did in example 1, then compute. in that the key with which you encipher a plaintext message is the same Lets try f(x) = 2x. (mod 26) and K " 13 14 # ≡ " 5 11 # (mod 26). CIPHER Re-make History View File Heyya Guys ‼️ WHAT`S NEW 9.17.20 ‼️ ⭐ADDED BEHR SISTERs & TINA TINKER . and error, we would not gain anything in comparison to our previous method. 7*a+17=2 (mod 26)=> 7*a= 2-17 (mod 26) = -15 (mod26) =26-15=11 (mod 26). An example affine cipher: With the key (5,17), the plaintext "ant" encrypts to "REI" since, . finding A, given N, P, and NA (mod P) is called the trusted courier or an expensive trip), and is wholly impractical if since 2 * 5 = 4 mod 6. Thus, if P is sufficiently large, Eve doesn't have a good way to share | improve this question | follow | edited Oct 19 '12 at 8:39.

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