In 36,37,39 and 40, use the RSA cipher with public key and In 36 and 37, encode the messages into their numeric equivalents and encrypt them. In 39-40, decrypt the given cipher text and find the original messages.

675 089 089 048

To decrypt the given cipher text and calculate the original message:

675 089​ 089 048.

Given information:

The given cipher text 675 089​ 089 048.

Concept used:

Inverse of a ​modulo n: For all integers a and n, if gcd(a,n)=1, then there exists an integer s

such that as≡1(modn). The integer s is called the inverse of a ​modulo n.

Calculation:

Consider the cipher text,

675 089​ 089 048

The objective is to decrypt the given cipher text and find the original message.

The decryption key is d=307, since 307 is the inverse of 43.

Public key is n=713=23⋅31 and e=43.

Thus, to decrypt the message, compute the following:

675307mod713, 89307mod713,89307mod713,48307mod713

Convert 307 to a sum of powers of 2.

307=256+32+16+2+1=28+25+24+21+1

Perform 6752k for k=1,2,3,4.

675≡675(mod713)6752≡455625(mod713) since 6752=455625=18mod713 since 455625mod713=18

6754≡( 675 2)2mod713≡182mod713 since 6752mod713=18=324mod713 since​ 182=324

6758≡( 675 4)2mod713≡(324)2mod713  since 6754mod713=324≡104976mod713​  since   3242=104976≡165mod713  since  ​104976mod713=165

67516≡( 675 8)2mod713≡(165)2mod713  since​ 6758mod713=165≡27225mod713 since​  1652=27225≡131mod713​  since​  27225mod713=131

Perform 6752k for k=5,6,7,8

67532≡( 675 16)2mod713≡(131)2mod713  since​  67516mod713=131≡17161mod713 since​  1312=17161≡49mod713​  since​  17161mod713=49

67564≡( 675 32)2mod713≡(49)2mod713  since​ 67532mod713=49≡2401mod713 since​  492=2401≡262mod713​  since​  2401mod713=262

675128≡( 675 64)2mod713≡(262)2mod713  since​  67564mod713=262≡68644mod713 since​  2622=68644≡196mod713​  since​  68644mod713=196

675256≡( 675 128)2mod713≡(196)2mod713  since​  675128=196mod173≡38416mod713 since​  1962=38416≡627mod713​  since​  38416mod713=627

Use the fact that

675307mod173=( 675 256+32+16+8+2+1)mod173=( 675 256⋅ 675 32⋅ 675 16⋅ 6752⋅ 6751)mod173=(627⋅49⋅131⋅18⋅675)mod173=48900262950mod173=3mod173 using calculator=3

Thus, 675 is decrypted as 3, which corresponds to letter C.

Calculate 892k for k=1,2,3,4.

89≡89mod713 ​since by a≡a(modn)892≡892mod713=7921mod713 since 892=7921=78mod713 since 7921mod713=78

894≡( 89 2)2mod713≡(78)2mod713  since 892mod713=78≡6084mod713  since  782=6084=380mod713  since 380=6084mod713

898≡( 89 4)2mod713≡(380)2mod713  since 894mod713=380≡144400mod713  since  3802=144400=374mod713

8916≡( 89 8)2mod713≡(374)2mod713  since 898mod713=374≡139876mod713  since  3742=139876=128mod713  since 139876mod713=126

Perform 892k for k=5,6,7,8