An application using the Chinese Remainder Theorem

Converting $b_{1} = (100)_{2}, \: b_{2} = (110)_{2}, \: b_{3} = (1010)_{2}, \: b_{4} = (1101)_{2}$ to base 10 we obtain: $$

$b_{1} = 4, \: b_{2} = 6, \: b_{3} = 10, \: b_{4} = 13.$ $$

Using Chinese Remainder theorem:

$X \equiv 4 \mod{5}, \: X \equiv 6 \mod{7}, \: X \equiv 10 \mod{11}, \: X \equiv 13 \mod{16} \implies$

$m = 5 * 7 * 11 * 16 = 6160, \: M_{1} = 7 * 11 * 16 = 1232, \: M_{2} = 5 * 11 * 16 = 880,$ $$

$M_{3} = 5 * 7 * 16 = 560, \: M_{4} = 5 * 7 * 11 = 385$

Let $1232 y_{1} \equiv 1 \mod{5}, \: 880 y_{2} \equiv 1 \mod{7}, \: 560 y_{3} \equiv 1 \mod{11}, 385 y_{4} \equiv 1 \mod{16}$ :

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To find the inverses $y_{1}, y_{2},$ , $y_{3}, y_{4}$ : $$

The inverses in this problem can be easily found by inspection, but for clarity I will show all the work. $$

$1232 y_{1} \equiv 1\mod{5} \implies 1232 y_{1} - 5 j_{1} = 1$ $$

Using a repeated application of the Euclidean Algorithm we obtain: $$

$1232 = 5 * 246 + 2, \: 5 = 2 * 2 + 1 \implies$ $$

$1 = 5 - 2 * 2 = 5 - (1232 - 5 * (246)) * 2 = 5 * (493) - 1232 * (2) = 1232 * (-2) - 5 * (-493)$ $$

$\implies y_{1} \equiv -2 \mod{5} \equiv 3 \mod 5$ $$

$\boxed{y_{1} \equiv 3 \mod{5}}$ $$

$880 y_{2} \equiv 1 \mod{7} \implies 880 y_{2} - 7 j_{2} = 1$ $$

Using a repeated application of the Euclidean Algorithm we obtain: $$

$880 = 125 * 7 + 5, \: 7 = 5 * 1 + 2, \: 5 = 2 * 2 + 1 \implies$ $$

$1 = 5 - 2 * 2 = 5 - (7 - 5) * 2 = 5 * 3 - 7 * 2 = (880 - 125 * 7) * 3 - 7 * 2 = 880 * (3) - 7 * (377)$ $$

$\implies \boxed{y_{2} \equiv 3 \mod{7}}$ $$

$560 y_{3} \equiv 1 \mod{11} \implies 560 y_{3} - 11 j_{3} = 1$ $$

Using a repeated application of the Euclidean Algorithm we obtain: $$

$560 = 11 * 50 + 10, \: 11 = 10 * 1 + 1 \implies$ $$

$1 = 11 - 10 = 11 - (560 - 11 * 50) = 11 * (51) - 560 * (1) = 560 * (-1) - 11 * (-51)$ $$

$\implies y_{3} \equiv -1 \mod{11} \equiv 10 \mod{11}$ $$

$\boxed{y_{3} \equiv 10 \mod{11}}$ $$

$385 y_{4} \equiv 1 \mod{16} \implies 385 y_{4} - 16 j_{4} = 1$ $$

Using the Euclidean Algorithm we obtain: $$

$385 = 16 * 24 + 1 \implies 1 = 385 * (1) - 16 * (24) \implies$

$\boxed{ y_{4} \equiv 1 \mod 16}$

$\therefore X \equiv \sum_{j = 1}^{4} b_{j} * M_{j} * y_{j} \mod 6160 \equiv$ $91629 \mod 6160 \equiv 5389 \mod 6160$ . $$

$\therefore$ Cipher text $\boxed{X = 5389}$ . $$

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$\textbf{Note:}$ To decipher the data base $B$ we use the Cipher text $X = 5389$ and the divisors $m_{1} = 5, \: m_{2} = 7, \: m_{3} = 11, \: m_{4} = 16$ to obtain: $$

$b_{1} \equiv 5389 \mod{5} \equiv 4 \mod 5$ $$

$b_{2} \equiv 5389 \mod{7} \equiv 6 \mod 7$ $$

$b_{3} \equiv 5389 \mod{11} \equiv 10 \mod 11$ $$

$\ b_{4} \equiv 5389 \mod{16} \equiv 13 \mod{16}$ $$

Converting to base 2 we obtain: $$

$b_{1} = (100)_{2}, \: b_{2} = (110)_{2}, \: b_{3} = (1010)_{2}, \: b_{4} = (1101)_{2}$ . $$

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$\textbf{Note:}$ Knowledge of Cipher text $X$ and $m_{j}$ permits access only to file $j$ , for access to the other files, it is necessary to know moduli other then $m_{j}$ .