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[CA auth] diffeehellmanPROGRAM DL : 0: Example 1. To find 1537 x 4248 modulo 10, you could multiply out and take the last digit, but a better way would be to replace 1537 by 7 and 4248 by 8 to start, find 7 x 8 = 56, and then take 56 mod 10 to get 6 as the answer. A handy standard notation is to write a tex2html_wrap_inline39 b (mod m) if a and b have the same remainder modulo m. This is read a is congruent to b modulo m . In this notation the example just mentioned looks like this: 1537 x 4248 tex2html_wrap_inline39 7 x 8 = 56 tex2html_wrap_inline39 6 (mod 10).-Example 1. To find 1537 x 4248 modulo 10, you could multiply out and take the last digit, but a better way would be to replace 1537 by 7 and 4248 by 8 to start, find 7 x 8 = 56, and then take 56 mod 10 to get 6 as the answer. A handy standard notation is to write a tex2html_wrap_inline39 b (mod m) if a and b have the same remainder modulo m. This is read a is congruent to b modulo m . In this notation the example just mentioned looks like this: 1537 x 4248 tex2html_wrap_inline39 7 x 8 = 56 tex2html_wrap_inline39 6 (mod 10).
Date : 2025-12-28 Size : 28kb User : pradeep
[CA auth] RC4-Prog DL : 0: Example 1. To find 1537 x 4248 modulo 10, you could multiply out and take the last digit, but a better way would be to replace 1537 by 7 and 4248 by 8 to start, find 7 x 8 = 56, and then take 56 mod 10 to get 6 as the answer. A handy standard notation is to write a tex2html_wrap_inline39 b (mod m) if a and b have the same remainder modulo m. This is read a is congruent to b modulo m . In this notation the example just mentioned looks like this: 1537 x 4248 tex2html_wrap_inline39 7 x 8 = 56 tex2html_wrap_inline39 6 (mod 10).-Example 1. To find 1537 x 4248 modulo 10, you could multiply out and take the last digit, but a better way would be to replace 1537 by 7 and 4248 by 8 to start, find 7 x 8 = 56, and then take 56 mod 10 to get 6 as the answer. A handy standard notation is to write a tex2html_wrap_inline39 b (mod m) if a and b have the same remainder modulo m. This is read a is congruent to b modulo m . In this notation the example just mentioned looks like this: 1537 x 4248 tex2html_wrap_inline39 7 x 8 = 56 tex2html_wrap_inline39 6 (mod 10).
Date : 2025-12-28 Size : 36kb User : pradeep
[CA auth] Yush.java.tar DL : 0: This a new algorithm in which we encrypt on the basis of position of the character. Algorithm 1 for key generation and Algorithm 2 is for encryption Algorithm A1 Step1: Let the key be ‘k’. Step2: Multiply the index number +1 with its ASCII code of the character at that index-65. Step3: Sum all the values and mod 26. Step4: Return the key. Algorithm A2 Step1: Take the string which you want to encrypt. Step2: Find the number of spaces. Step3: We add 65 to number of space. Step4: Take a string s. Step5: Add character representing number of spaces to that string. Step6: Calculate the word length of all the words present in the string and 65 to it. Step7: Now append the character that represents the word length. Step8: Subtract the number 1 and Add the number 1 with the ASCII value of the characters at alternate position. Step9: Add the key to each character in the string. Step10: Subtract 65 and the value so obtained modulo 26. Step11: Cipher text obtained.-This is a new algorithm in which we encrypt on the basis of position of the character. Algorithm 1 for key generation and Algorithm 2 is for encryption Algorithm A1 Step1: Let the key be ‘k’. Step2: Multiply the index number +1 with its ASCII code of the character at that index-65. Step3: Sum all the values and mod 26. Step4: Return the key. Algorithm A2 Step1: Take the string which you want to encrypt. Step2: Find the number of spaces. Step3: We add 65 to number of space. Step4: Take a string s. Step5: Add character representing number of spaces to that string. Step6: Calculate the word length of all the words present in the string and 65 to it. Step7: Now append the character that represents the word length. Step8: Subtract the number 1 and Add the number 1 with the ASCII value of the characters at alternate position. Step9: Add the key to each character in the string. Step10: Subtract 65 and the value so obtained modulo 26. Step11: Cipher text obtained.
Date : 2025-12-28 Size : 1kb User : Aman Virmani