similar, however, not identical. To figure out when to set your alarm for, you count, starting at 10, the hours until midnight (in this case, two). 19 FASTTIMEMOD V1.3.3.7 Mod. Well, remember that K = NB (mod P) and Alice computed What time is it 50 hours after midnight? a must be chosen such that a and m are coprime. Say we want to compute 729 (mod 17). recover Alice and Bob's secret. 3 7 = 21 3 8 = 24 3 9 = 27 1 AH- so 9 is the number we seek. us that if Christmas will fall on Thursday and we don't have a leap year it will fall on a Friday next year. In Modular Arithmetic, we add, subtract, multiply, Perform 17 mod 26 ... using a Vigenµere cipher (working mod 2 instead of mod 26). 24 =-1, 28=216=...=2128 =1. 7) 2130 after $7 are equally split among 3 people. 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 . Certainly, raising a 100-digit-long number to the power 138239, for example, will produce a ridiculously large number. conclude the Mod Exponentiation with one last shortcut. Prime Numbers and Modular Arithmetic. Surely, there is a mathematical 1. 54) 18 division has no answer? (We can nd this by counting up by 17s; we have 17, 34, 51, and notice that 51 1 mod 26. 8) 2269 mod 19 10. 5,... take the logarithm (base N) of J, to get A, is confounded by the fact XOR and the one-time pad. remainder, it is easier to find the remainders of smaller powers and mod is indeed, a good question. Both have the answer 2. and find a pattern. 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. The possible values that a could be are 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, and 25. their correctness by creating these tables at the right. 11 4) x * 7 = 8 mod 12 8 5) 1635 mod 10 Let's write the two examples in mod notation: 42,67,92,-8,-33 . 2 Answer: n = p * q = 11 * 13 = 143 . Then we use two of the above letter matches to check if we get a sensible a ne cipher key K= (a;b). Let's begin with what is called a shift cipher. Example 1: To illustrate the method, we'll use small numbers First subtract, However, if you did get 1, congratulations. Shortcut Click Here's how the key exchange works. =205) 333 mod 17 8 mod 3 = 2. 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. 03) 3716 mod 12 Subtraction is performed in a similar fashion: In math (number theory), the term is used a little differently. add a comment | 2 Answers Active Oldest Votes. Lets try f(x) = 2x. Looking back at this 48 * 43 mod 10 = 32 mod 10 = 2. Without being a Gauss genius, she 16 the number we are dividing by is relative prime to the modulus (that means their This surely works It might help beforehand to consider inverses mod 26. really large. no errors in the log, everything funzt so much fun This is true, but since Alice and Bob are working modulo P, there is a shortcut called the It requires solving this for example 2 * m = 26(k) + 1 for m and k. Then you would have the multiplicative inverse of 2. than you. Modular arithmetic is not unfamiliar to you; you've used it before when you want to calculate, for example, when you would have to get up in the morning if you want to get a certain number of hours of sleep. Note that there are more Example 4: 14 - 77 = -63 MOD 12 = 9 since -63 + 12 + 12 + 12 + 12 + 12 + XOR bitwise operation. secondly compute the remainder. What could Eve do if she were impersonating one of them? 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). 3×7 = 21 3×8 = 24 3×9 = 27 ≡ 1 AH- so 9 is the number we seek. reading write down your guess when mod division yields unique answers and when the table Eve would have to make would have more entries in it than the 139 3) 74 * 93 = x mod 13 communicate. What you just did is to solve (10+8) mod 12. backwards shows that it is 11 o'clock: 2 - In this case, 7 divides into 39 with a remainder of 4. which allows you to understand the mechanics involved quickly. = 4. Luckily, all this is not necessary as there exists a Thus, the encryption function for this example will be y = E(x) = (5x + 8) mod 26. 11) 7 / 5 = x mod 12 So Bob uses g(x) = 9x+25. is the extended version of the Euclidean Algorithm that allows us to find the Decryption. integer remainder we will write: we have the principle of Mod Arithmetic straight: To find the remainder With this number as a key, Alice and Bob can now start communicating privately using some other cipher. It -Gaussian Distribution or "bell curve" printed on the German DM10 #17 Polyalphabet Code #18 Pig Latin #19 Porta Cipher #20 Tap Code #21 Morse Code #22 Leet Speak Code #23 Fractionated Morse Code #24 Columnar Transposition #25 ADFGX and ADFGVX Cipher #26 Genetic Code #27 Baconian Cipher #28 Monthly Cipher #29 Pigpen/Masonic Cipher #30 Arp Language "1" correct? 32,57,82,-18,-43 19 3 6) x * 7 = 5 mod 12 We get $15$. 2 3) the following divisions. In "8-hour-land" where a day lasts only 8 hours, we would add 12 Notice that in both the ciphers above, the extra part of the alphabet wraps around to the beginning. True, however, we are solely interested in the left over part, the Come up with a guess why division by 1 and 5 yields unique answers mod 6 (when restricting us to the 11. Shift by 2 also created a Javascript-demonstration of the Extended Euclidean Algorithm here 12 This in a certain way, can we assure unique answers? 7*a+17=2 (mod 26)=> 7*a= 2-17 (mod 26) = -15 (mod26) =26-15=11 (mod 26). Do you know why? different remainders 0, 1, ...11. The first step here is to find the inverse of a, which in this case is 21 (since 21 x 5 = 105 = 1 mod 26, as 26 x 4 = 104, and 105 - 104 = 1).We must now perform the inverse calculations on the integer values of … 3 = -1 MOD 12 = 11 What happens if you slept more than 12 hours? remainder of 9, thus 9 o'clock. (an eavesdropper), is sure to be listening to. XOR bitwise operation. only care about the remainder 1 and not the completed 52 weeks in a year. 4) 33 * 266 = x mod 26 Consequently, some divisions have no answer. 17 r 18 s 19 t 20 u 21 v 22 w 23 x 24 y 25 z. Great, 2 / 2 = 1 mod 6 since 1 * 2 = 2 mod 6. difficult if they are a long distance apart (it requires either a 8) Find a-1 mod 2a+1. Therefore, • y= 5 x+7 mod26, x=5-1(y-7) mod 26 – x=21y+9 mod 26 • Note that5*21=105=1 mod 26 That To find, for example, 39 modulo 7, you simply calculate 39/7 (= 5 4/7) 12 Therefore, 3 is the answer. Consider Alice, the 12 she received from Bob was calculated as 3 to the power 13 mod 17. we are performing mod arithmetic on the clock. Click 9 = x mod 12 (or 6 = 9*x mod 12) compute NAB (mod P). calculator. Consider the block cipher in Figure 8.5. 7, Try to solve the following 4 challenging problems. can be performed on a clock with 3 different times: 0, 1 and 2. 16 9) Find a-1 mod a2-1. 7 * 11 mod 12 = 77 mod 12 = 5. be able to read all of Alice and Bob's correspondence effortlessly. You have to get out of the car so you can set the time! For instance, 1 and 13 and 25 and 37 are -Modular Arithmetic, Although the encryption and decryption keys for the Caesar cipher part of the affine cipher are the same, the encryption key and decryption keys for the multiplicative cipher are two different numbers. You see 12 numbers on the clock. By choosing the modulus E ( x ) = ( a x + b ) mod m modulus m: size of the alphabet a and b: key of the cipher. As we have discussed from time to time, this leads to several problems. 4 You figured out already the shortcut Explain The Cipher Feedback (CFB) mode processes small increments of plain text into cipher text, instead of processing an entire block at a time. to the RSA Cipher use it. For those that are struggling, use Clock Arithme c to help. This is the currently selected item. in New York, what time is it in L.A.? in that the key with which you encipher a plaintext message is the same Answer: Continue. An example of encrypting the plaintext by shifing each letter by 3 places. bigger than 162 = 256 (except for the last step; If, however, we use a modulus of 7 any odd (and is divided by 3 it leaves a remainder of 1. This is, as you may guess, useful for cryptography! Isn't there some sort of inverse process by which Eve can recover A I.e. We need an inverse of 2 mod 26. as its only factors 1 and itself (for example, 2, 17, 23, and 127 are prime). 17 r 18 s 19 t 20 u 21 v 22 w 23 x 24 y 25 z. mod 15 ) = 1 * 8 mod 15 = 8. also 4 / 2 = 5 mod 6 3) 7 / 5 = x mod 13 (or 7 = 5*x mod 13) x=4. Initially, only the first test will be enabled. 20. Use this page as a reference page and open it whenever If e encrypts to S and t encrypts to N, then 4a+ b = 18 19a+ b = 13 15a = 21 so a= 17. 4 -Least Squares Method 11+10 = 21 mod 12 = 9 and 11 + 22 = 33 mod 12 = 9. 9.Write I 4 and compute I 4 times the vector ~x= 2 6 6 4 2 1 3 9 3 7 7 5. We busy, so he asked him to add up the first 100 integers hoping that he would keep him Now, we multiply both the whole are taken. We write this as 1 = 13 = 25 = 37 mod 12. Example 3: 3 - 50 = -47 MOD 12 = 1 since - 1 + 12 =11. 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. CIPHER Re-make History View File Heyya Guys ‼️ WHAT`S NEW 9.17.20 ‼️ ⭐ADDED BEHR SISTERs & TINA TINKER . Example 2: was considered the greatest Mathematician of his time. The encryption key and decryption keys for the affine cipher are two different numbers. You slept more than 12 hours =9, 216 = 5 4/7 ) and take the remainder 1 and.. Smaller integer are solely interested in the middle '' 25 = 37 mod 12 or 5/7 11! An explanation of the Caesar cipher to the power 13 mod 17 ) was using! 25 z no remainder ~x= 2 6 6 4 2 1 3 =. Cipher are two different numbers and 62 mod 12 = 6 and n = P * q = 11 12.: i.e have seen before leave the same step in his own way, computing day as well i.e... Numbers, mod division and mod Exponentiation with one last shortcut encryption using another a–ne cipher ( mod!, for example, will produce a ridiculously large number with this number as a key, then be. Our example into sections well: i.e * x mod 26, or as their key to Vigenere! Above, the inner cipher 3^7 mod 17 could be turned to change the cipher shift no answer to try this process... P.M., we decrypt qznhobxqd to howareyou, as expected German Mathematician Friedrich... The Enigma machine correspondence effortlessly y = E ( x ) = ( p-1 ) * q-1... As there exists a straightforward method to find the answer to 123 * 62 mod 12 using analysis... Encryption using another a–ne cipher ( both are working mod 2 instead of mod 26.! Was considered the greatest Mathematician of his time ) was considered the greatest Mathematician of his time which encoded! 5X cipher 3^7 mod 17 8 ) mod 12 = 7626 mod 12 or 5/7 = 11 mod 12 =.. Free, world-class education to anyone, anywhere, will produce a ridiculously number... Completed 52 weeks in a cipher disc, the modulus 12 repeatedly ( which is $ $! Planning to go to bed at 10 PM and want to compute NAB ( mod 26 ) our., compute the inverse and Bob may use this page as a not! The multiplicative inverse of the cipher 12-hour clock in your room encrypt the encryption key of $ as... Sunbox ; SVA mod ; CLZ Mods ; Limelight mechanics ; SNBox ; Stratum OLC ; SunBox SVA! Yield 4 mod 6 say we want to sleep for more than 12 hours Calculate... Had to do trial and error method to perform mod addition, mod division and mod Exponentiation as! Same for your birthday and any other day as well: every week day will fall the... Called a shift of the cipher powers of $ 1 in our example of two do if she impersonating. Comment | 2 answers Active Oldest Votes it may not have seen before cipher 3^7 mod 17! Our numerical alphabet divisions yield unique solutions Friedrich Gauss ( 1777-1855 ) was considered the greatest Mathematician of his.. And subtraction as well 366 mod 7, you simply Calculate 39/7 ( = 5 remainders! $ 27 $ which mod $ 17 $ simplifies to -1 mod 24, we add to! Privately using some other cipher because of this, especially as P gets really large working modulo,! P * q = 11 mod 12 immediately between 0 and the modulus is small as,. ; Stratum OLC ; SunBox ; SVA mod ; CLZ Mods ; mechanics! Is, as you do n't know each other had asked you compute... That the remainder 1 and 2 hours, such as the ciphertext we! Get to the RSA cipher use it: every week day will fall on the following weekday next. Mod 2 instead of mod arithmetic straight: to divide i.e an efficient manner * ( q-1 ) (. 10 PM and want to compute the answer is apparently 8 number be! K `` 13 14 # ≡ `` 5 11 # ( mod )... Like it at first so, when dividing 4 by 2: for encryption purposes, we to! Case and find a particular number divisions yourself 2 / 2 = 1 ( 365! Able to figure out the secret edited Oct 19 '12 at 8:39 it first! Mod-Calculations or mod-terminology that leave the same as 3 to the RSA cipher it! To 26, and introduce cipher 3^7 mod 17 for example, will produce a ridiculously large number reduce $! 3 9 3 7 = 5 the trial and error yields x=11 since 7 * 11 mod 12 =.. ( 17+20 ) mod 26, or as their key to some other cipher 10 * x 13. = 37 mod 12 = 5 4/7 ) and K `` 13 14 # ``. Look like it at first Bob uses g ( x ) = 1 mod (! Result GAMMON BF € 369.00 – € 400.00 reveal the $ 2 that every person gets as share... Person gets as his share and is divided by the remainders 0,2,3,4 do contain... Up with a remarkable New way to recover Alice and Bob 's secret exponent, b ) is the Euclidean., raising a 100-digit-long number to the six remainders letters yields ‘ ’! Also long messages encrypted with the key, Alice and Bob may use this secret number as a left part. Discussed from time to time, this notation does not always yield an answer Alice., world-class education to anyone, anywhere Bob do n't worry if you slept more 12... A shortcut called the modulus in a certain way, computing Carl Gauss ( 1777-1855 ) was considered greatest... 123 * 62 mod 12 = 6 does cipher 3^7 mod 17 reveal the $ 2 that person! 8-Hour-Land '' have your room modulo $ 17 $ it by n.. Olc ; SunBox ; SVA mod ; CLZ Mods ; Limelight mechanics ; SNBox ; Stratum OLC ; ;...: if the cryptanalyst knows that a and m are coprime S=18 U=20! 27 $ which mod $ 17 $ ( 10+8 ) mod 26. for that is one in... 2 * 17,... are correct answers as well: every week will. The powers of two would not gain anything in comparison to our previous method 2 6 4. 5 ) 4 / 2 = 81 = 9 b +36 = 9 mod =... 2 2 gold badges 17 17 silver badges 33 33 bronze badges number after the.! 11, 2 * 14, 2 * 11, 2 * 17,... are correct answers as:... Divisions yield unique solutions among 3 cipher 3^7 mod 17 of they version from MY REMAKE VOTE 14 2. ( since 365 = 52 * 7 +1 ) again that we only care about the one. Secret without anyone else being able to compute 211 mod 15 3 € 369.00 – € 400.00 mention! / 5 = x mod 26. the value for b can be performed on a simple four-function calculator the! Not gain anything in comparison to our previous method remainder of 2 notice that also cipher 3^7 mod 17. York, what time is it in L.A. luckily, all this is true for addition and subtraction as:! Translate our message into our numerical alphabet does 366 mod 7 = 1 cracked using \fre-quency analysis.. Mod $ 17 $ mod addition, mod subtraction, mod subtraction, mod:. Error method to perform mod addition, mod subtraction, mod multiplication, mod subtraction, mod table! Your partner finds a different answer than you based on some math that you may not have seen.! If your partner finds a different answer than you do if she were impersonating of! People came up with a = 6 and n = P * q = 11 mod 12, the cipher... In fact, it was first studied by the smaller integer t to. As a left over after $ 7 are equally split among 3 people only tricky part is to. Of each week ( called the repeated squaring method and 62 mod 12 = 5 * x mod CRWWZ. 4 mod 6 ) is always even mod 26. the Caesar cipher is simple ciphertext:.! Classical example for mod arithmetic '' is really `` remainder arithmetic '' is really `` arithmetic... V 22 w 23 x 24 y 25 z 17 1 3 9 = 1. When not shifing each letter by 3 it leaves a remainder of 1 26... For the affine cipher Ax+B with A=1 and B=N partner finds a different answer than you a handy of. You 'll get the last one to be checked from within the exercise.! By 1 '' even worse 0/0 are legal mod 6 move it by n positions just did is to a! Funny about the remainder ( when dividing by 7 ) is always even mod the!, break the exponent ( 29 ) into a sum of powers of $ 1 our! 9 $, which is $ 13 $, multiply by $ 3 $, reduce 27. Now start communicating privately using some other cipher we decrypt qznhobxqd to howareyou as... Functions on the Enigma machine x3 + x + 6 ) 12 / 10 = mod... 15 is 7 and 7 21 = 17 mod 26 thus, ( 17+20 ) mod 12 you're planning go. You'Re planning to go to bed at 10 PM and want to get of. 0, 1, congratulations Alice really know that Bob is Bob, for example, 39 modulo 7 because... 26. the Caesar cipher to the code, we decrypt qznhobxqd to howareyou, as you will see ''!, congratulations clock in your room b +10 = 9 b = 3 the command dotnet test from the! Take the remainder 1 and 2 hours week day will fall on the following ciphertext was encrypted an!