Question

What is the original message encrypted using the RSA system with $\mathrm{n}=53\cdot 61\mathrm{and}\mathrm{e}=17$if the encrypted message is $3185203824602550$? (To decrypt, first find the decryption exponent d, which is the inverse of $\mathrm{e}=17\mathrm{modulo}52\cdot 60$).

Short answer

SQUIRREL.

Step-by-step solution

Step 1

Given encryption: $3185203824602550$

$$\begin{gathered} \mathrm{n}=53\cdot 61=3233 \\ \mathrm{e}=17 \end{gathered}$$

Since $n=52\cdot 60=3120,d$ is then the inverse of $17\mathrm{modulo}3120$(which means $17\mathrm{modulo}3120=1$and $d$integer with $0\le d<3120$).

$d=2753$

Decrypt each block using the mapping $M=C^{d}mod3233$

$$\begin{gathered} M_1={3185}^{2753}mod3233=1816 \\ {M}_2={2038}^{2753}mod3233=2008 \\ {M}_3={2460}^{2753}mod3233=1717 \\ {M}_4={2550}^{2753}mod3233=0411 \end{gathered}$$

Thus, the message is then:

$1816200817170411$

Let,

$$\begin{gathered} \mathrm{A}=00,\mathrm{B}=01,\mathrm{C}=02,\mathrm{D}=03,\mathrm{E}=04,\mathrm{F}=05,\mathrm{G}=06,\mathrm{H}=07,\mathrm{I}=08,\mathrm{J}=09,\mathrm{K}=10,\mathrm{L}=11,\mathrm{M}=12 \\ \mathrm{N}=13,\mathrm{O}=14,\mathrm{P}=15,\mathrm{Q}=16,\mathrm{R}=17,\mathrm{S}=18,\mathrm{T}=19,\mathrm{U}=20,\mathrm{V}=21,\mathrm{W}=22,\mathrm{X}=23,\mathrm{Y}=24,\mathrm{Z}=25. \end{gathered}$$

SQ UI RR EL

Rewriting the blocks such that English words form:

SQUIRREL