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In a RSA cryptosystem a participant uses two prime numbers p and q is 17 and 11 respectively to generate his/her public and private keys...If the public keybof participant is 7 and cipher text C is 11 then the original message M is_________
in Computer Networks by Active (1.2k points) | 61 views
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@HeadShot 2 @Mk Utkarsh .how to solve the last step? Pls mention the source or tell the method...even i have applied Fermat's theorem..still i am mot able to solve it

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https://gateoverflow.in/39588/gate2016-2-29

see laxman ans comments same one discussed.

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11^2mod187 = 121

11^3mod187 = 22

then brk then in powers

$11^{23}mod187 = ( (11^3mod187)^7 * 11^2mod187 )m od187$

$=> (22^7 * 121)mod187$

$=>(22^7mod187 * 121)mod187$

$=>(((22^2mod187)^3 * 22 )mod187 * 121 )mod187 $

$=> ((110^3*22)mod187 *121)mod187$

$=> (44*121)mod187$

$=> 5324mod187$

$=> 88$

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