(1 point) This question concerns block cipher padding. Suppose the block cipher has a block size of 18 bytes. A certain message ends with a partial block which is 3 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Append a single byte with value 128, and then as many zero bytes as required to make the overall length a multiple of 18. Suppose the final (possibly partial) block of the message is "Ox000102" in hexadecimal. Write out the complete final block in hexadecimal. Please do not put a leading Ox as this has been written already. Ox
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00010280000000000000000000000000 This answer is not correct. Please give correct answer
![(1 point) This question concerns block cipher padding. Suppose the block cipher has a block size of 18 bytes.
A certain message ends with a partial block which is 3 bytes. We must reversibly pad out the message to use
some block cipher modes, such as CBC (even if the last block is full).
Consider the following padding strategy.
Append a single byte with value 128, and then as many zero bytes as required to make the overall length a
multiple of 18.
Suppose the final (possibly partial) block of the message is "Ox000102" in hexadecimal. Write out the
complete final block in hexadecimal. Please do not put a leading Ox as this has been written already.
Ox](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F38650a9c-63f5-423f-8b89-b03c0dad949f%2F2eee32bd-9a41-4a98-91dc-75c031667228%2Fc170b7c_processed.jpeg&w=3840&q=75)
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- This question concerns block cipher padding. Suppose the block cipher has a block size of 16 bytes. A certain message ends with a block which is 16 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full, which it is in this case). Consider the following padding strategy. Append a single byte with value 128, and then as many zero bytes as required to make the overall length a multiple of 16. Suppose the final (possibly partial) block of the message is "Ox000102030405060708090a0b0c hехad mal. Write out the complete hal block in hexadecimal. Please do not put a leading Ox as this has been written already. Because the last block is actually full, the block you add will need to be a completely new block. ОхThis question concerns block cipher padding. Suppose the block cipher has a block size of 18 bytes. A certain message ends with a partial block which is 2 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Append a single byte with value 128, and then as many zero bytes as required to make the overall length a multiple of 18. Suppose the final (possibly partial) block of the message is "0x0001" in hexadecimal. Write out the complete final block in hexadecimal. 0x = (put answer here)This question concerns block cipher padding. Suppose the block cipher has a block size of 18 bytes. A certain message ends with a partial block which is 5 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Append a single byte with value 128, and then as many zero bytes as required to make the overall length a multiple of 18. Suppose the final (possibly partial) block of the message is "Ox0001020304" in hexadecimal. Write out the complete final block in hexadecimal. Please do not put a leading Ox as this has been written already. Ox
- This question concerns block cipher padding. Suppose the block cipher has a block size of 16 bytes. A certain message ends with a partial block which is 13 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Determine the number of padding bytes required. This is a number n which satisfies 1padding Question 11 Suppose we have a block cipher with block size 8-bytes (64-bits) and we will use the padding scheme described in class and used in the homework (PKCS #7, RFC 5652). How many bytes (octets) are added to a message consisting of 11 bytes when applying the padding scheme in PKCS7? Note: The message is written in hexadecimal. Each pair of digits (number or letter) is ONE byte (octet). You answer must be a number (for example, 3, 5, 9, etc.).This question concerns block cipher padding. Suppose the block cipher has a block size of 20 bytes. A certain message ends with a partial block which is 1 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Append a single byte with value 128, and then as many zero bytes as required to make the overall length a mulitiple of 20. Suppose the final (possibly partial) block of the message is "Ox00" in hexadecimal. Write out the complete final block in hexadecimal. Please do not put a leading Ox as this has been written already. OxThis question concerns block cipher padding. Suppose the block cipher has a block size of 15 bytes. A certain message ends with a partial block which is 4 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Determine the number of padding bytes required. This is a number n which satisfies 1 ≤ n ≤ 15 and n + 1(P) is a multiple of 15, where 1(P) is the length of the unpadded plaintext. Pad the plaintext by appending n bytes, each with value n. Suppose the final (possibly partial) block of the message is "0x00010203" in hexadecimal. Write out the complete final block in hexadecimal. Please do not put a leading Ox as this has been written already. OxThis question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipher(unsigned char block, char key) { return (key+11"block)%256; } Note that the block size is 8 bits, which is one byte (and one ASCII character). We will work with the fixed key 0x08. We now encrypt various plaintexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be OXAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of OXAA and the counter is a 3 bit counter that begins at 0. In all of the problems given below, one character is one block. Each character of the plaintext should be regarded as its corresponding ASCII code. a) Encrypt the plaintext "blotto" using CTR mode. Please enter your answer in hex. (Please do **not** enter an 0x, as this has been done.) Ox b) Encrypt the plaintext "blind" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipher(unsigned char block, char key){ return (key+11*block)%256;} Note that the block size is 8 bits, which is one byte (and one ASCII character). We will work with the fixed key 0x08. We now encrypt various plaintexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be 0xAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of 0xAA and the counter is a 3 bit counter that begins at 0. In all of the problems given below, one character is one block. Each character of the plaintext should be regarded as its corresponding ASCII code. a) Encrypt the plaintext "?????" using ??? mode. Please enter your answer in hex. (Please do **not** enter an 0x, as this has been done.)0x= * Can you please answer this question // input answer next to the "=" sign.This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipher(unsigned char block, char key) { return (key+11*block)%256; } Note that the block size is 8 bits, which is one byte (and one ASCII character). We will work with the fixed key 0x08. We now encrypt various plaintexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be OxAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of OxAA and the counter is a 3 bit counter that begins at O. In all of the problems given below, one character is one block. Each character of the plaintext should be regarded as its corresponding ASCII code. a) Encrypt the plaintext "blotto" using CTR mode. Please enter your answer in hex. (Please do **not** enter an Ox, as this has been done.) Ox b) Encrypt the plaintext "blind" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipher(unsigned char block, char key) { return (key+11*block)%256; } Note that the block size is 8 bits, which is one byte (and one ASCII character). We will work with the fixed key 0x08. We now encrypt various plaintexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be OxAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of OxAA and the counter is a 3 bit counter that begins at 0. In all of the problems given below, one character is one block. Each character of the plaintext should be regarded as its corresponding ASCII code. a) Encrypt the plaintext "cheery" using CTR mode. Please enter your answer in hex. (Please do **not** enter an Ox, as this has been done.) Ox b) Encrypt the plaintext "chirper" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipher(unsigned char block, char key) { return (key+11*block)%256; } Note that the block size is 8 bits, which is one byte (and one ASCII character). We will work with the fixed key 0x08. We now encrypt various plaintexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be OXAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of OXAA and the counter is a 3 bit counter that begins at 0. In all of the problems given below, one character is one block. Each character of the plaintext should be regarded as its corresponding ASCII code. a) Encrypt the plaintext "spider" using CTR mode. Please enter your answer in hex. (Please do **not** enter an 0x, as this has been done.) Ox b) Encrypt the plaintext "spelling" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…SEE MORE QUESTIONS
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