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EppDiscMath5 8.4.040. Use the RSA cipher with public key $(pq, e) = (23 \cdot 31, 43) = (713, 43)$ and private key $(pq, d)$, where $d = 307$, to decrypt the given ciphertext and find the original message. (Assume the letters of the alphabet are encoded as follows: $A = 01, B = 02, C = 03, \dots, Z = 26$.) 137 089 046 Since $d = 307 = 256 + 32 + 16 + 2 + 1$, find the first letter of the decrypted message by computing $137^d \pmod{713}$. Now $137^1 \equiv a \pmod{713}$ $137^8 \equiv d \pmod{713}$ $137^{64} \equiv g \pmod{713}$ $137^2 \equiv b \pmod{713}$ $137^{16} \equiv e \pmod{713}$ $137^{128} \equiv h \pmod{713}$ $137^4 \equiv c \pmod{713}$ $137^{32} \equiv f \pmod{713}$ $137^{256} \equiv i \pmod{713}$. The result is that $a = 137$ $e = 576$ $i = 576$ , $b = 231$ , $f = 231$ , $c = 599$ , $g = 599$ , $d = 162$ , $h = 162$ , and Thus, $137^{307} \pmod{713} = (a \cdot b \cdot e \cdot f \cdot i) \pmod{713} = \boxed{\phantom{123}} \times \boxed{\phantom{123}}$, and so the first letter in the encrypted message is $\boxed{\phantom{123}}$. Repeat these computations for each encrypted letter and find the completed decrypted message. $\boxed{\phantom{123}}$

Name: eppdiscmath5 84040 use the rsa cipher with public key pq e 23 cdot 31 43 713 43 and private key pq d where d 307 to decrypt the given ciphertext and find the original message assume the lett 85152
Uploaded: 2022-11-24T07:05:23-08:00
Duration: 2 min 2 s
Channel: Sri K
Description: eppdiscmath5 84040 use the rsa cipher with public key pq e 23 cdot 31 43 713 43 and private key pq d where d 307 to decrypt the given ciphertext and find the original message assume the lett 85152

          EppDiscMath5 8.4.040.
Use the RSA cipher with public key $(pq, e) = (23 \cdot 31, 43) = (713, 43)$ and private key $(pq, d)$, where
$d = 307$, to decrypt the given ciphertext and find the original message. (Assume the letters of the alphabet
are encoded as follows: $A = 01, B = 02, C = 03, \dots, Z = 26$.)

137 089 046

Since $d = 307 = 256 + 32 + 16 + 2 + 1$, find the first letter of the decrypted message by computing
$137^d \pmod{713}$. Now

$137^1 \equiv a \pmod{713}$
$137^8 \equiv d \pmod{713}$
$137^{64} \equiv g \pmod{713}$

$137^2 \equiv b \pmod{713}$
$137^{16} \equiv e \pmod{713}$
$137^{128} \equiv h \pmod{713}$

$137^4 \equiv c \pmod{713}$
$137^{32} \equiv f \pmod{713}$
$137^{256} \equiv i \pmod{713}$.

The result is that $a = 137$
$e = 576$
$i = 576$

, $b = 231$
, $f = 231$

, $c = 599$
, $g = 599$

, $d = 162$
, $h = 162$

, and

Thus, $137^{307} \pmod{713} = (a \cdot b \cdot e \cdot f \cdot i) \pmod{713} = \boxed{\phantom{123}} \times \boxed{\phantom{123}}$, and so the first letter in the
encrypted message is $\boxed{\phantom{123}}$.

Repeat these computations for each encrypted letter and find the completed decrypted message.
$\boxed{\phantom{123}}$

eppdiscmath5 84040 use the rsa cipher with public key pq e 23 cdot 31 43 713 43 and private key pq d where d 307 to decrypt the given ciphertext and find the original message assume the lett 85152

Added by Kelly T.

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