An application using the Chinese Remainder Theorem

I will solve this problem twice using two methods. First I will use the Chinese Remainder Theorem then I will use an algebraic approach. If your not familiar with the Chinese Remainder Theorem you can view the algebraic approach.. $$

Using Chinese Remainder theorem:(Assuming your familiar with it.)

$X \equiv 6 \mod{13}, \: X \equiv 14 \mod{20}, \: X \equiv 4 \mod{11} \implies$

$m = 13 * 20 * 11 = 2860, \: M_{1} = 20 * 11 = 220, \: M_{2} = 13 * 11 = 143, \: M_{3} = 13 * 20 = 260$

Let $220 y_{1} \equiv 1 \mod{13}, \: 143 y_{2} \equiv 1 \mod{20}, \: 260 y_{3} \equiv 1 \mod{11}$ :

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

$220 y_{1} \equiv 1 \mod{13} \implies 220 y_{1} - 13 j_{1} = 1$ $$

Using a repeated application of the Euclidean Algorithm we obtain: $$ $220 = 13 * 16 + 12, \: 13 = 12 * (1) + 1 \implies$ $$

$1 = 13 - 12 = 13 - (220 - 13 * 6) = 13 * (17) - 220 = 220 * (-1) - 13 * (-17) \implies$

$\boxed{y_{1} \equiv 12 \mod {13}}$ $$

$143 y_{2} \equiv 1 \mod{20} \implies 143 y_{2} - 20 j_{2} = 1$ $$

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

$143 = 20 * 7 + 3, \: 12 = 3 * 6 + 12, \: 3 = 2 * 1 + 1 \implies$ $$

$1 = 3 - 2 = 3 - (20 - 3 * 6) = 3 * 7 - 20 = (143 - 20 * 7) * 7 - 20 = 143 * (7) - 20 * (50) \implies$ $$

$\boxed{y_{2} \equiv 7 \mod {20}}$ $$

$260 y_{3} \equiv 1 \mod{11} \implies 260 y_{3} - 11 j_{3} = 1$ $$

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

$260 = 11 * (23) + 7, \: 7 = 4 * (1) + 3, \: 4 = 3 * (1) + 1 \implies$ $$

$1 = 4 - 3 = 4 - (7 - 4) = 4 * 2 - 7 = (11 - 7) * 2 - 7 = 11 * (2) - 7 * 3 =$ $$

$11 * 2 - (260 - 11 * 23) * 3 = 11 * (71) - 260 * (3) = 260 * (-3) - 11 * (-71) \implies$ $$

$\boxed{y_{3} \equiv 8 \mod {11}}$ $$

For $b_{1} = 6, \: b_{2} = 14, \: b_{3} = 4$ $\implies X \equiv \sum_{j = 1}^{3} b_{j} * M{j} * y_{j} \mod {2860} =$ $$

$15840 + 14014 + 8320 = 38174 \equiv \boxed{994 \mod{2860}}.$ $$

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$\therefore$ In the 2860 day year, day 6 in the 13 day cycle, day 14 in the 20 day cycle, and day 4 in the 11 day cycle would coincide on day $\boxed{994}$ $$

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Using an algebraic Approach: $$

$X \equiv 6 \mod{13}, \: X \equiv 14 \mod{20}, \: X \equiv 4 \mod{11} \implies$ $$

$X = 13 j_{1} + 6 = 20 j_{2} + 14 = 11 j_{3} + 4 \implies$ $$

$13 j_{1} - 20 j_{2} = 8$ $$

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

$20 = 13 * 1 + 7, \: 13 = 7 * 1 + 6, \: 7 = 6 * 1 + 1 \implies$ $$

$1 = 7 - 6 * 1 = 7 - (13 - 7 * 1) = 7 * 2 - 13 = (20 - 13) * 2 - 13 - 20 * (2) - 13 * (3)$ $$ $= 13 * (-3) - 20 * (-2) \implies$ $$

$13 * (-24) - 20 * (-16) = 8 \implies$ $$

$\boxed{j_{1} = -24 + 20 * t}, \boxed{j_{2} = -16 + 13 * t}$ $$

and, $11 j_{3} - 13 j_{1} = 2$ $$

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

$13 = 11 * 1 + 2, \: 11 = 2 * 5 + 1 \implies$ $$

$1 = 11 - 2 * 5 = 11 - (13 - 11) * 5 = 11 * (6) - 13 * (5) \implies$ $$

$11 * (12) -13 * (10) = 2 \implies$ $\boxed{j_{3} = 12 + 13 * t^{*}},$ $\boxed{j_{1} = 10 + 11 * t^{*}}$

$-24 + 20 t = 10 + 11 t^{*} \implies 20 t - 11 t^{*} = 34$ $$

$20 = 11 * 1 + 9, \: 11 = 9 * 1 + 2, \: 9 = 2 * 4 + 1 \implies$ $$

$1 = 9 - 2 * 4 = 9 - (11 - 9) 8 4 = 9 * 5 - 11 * 4 = (20 - 11) * 5 - 11 * 4 =$ $$

$20 * (5) - 11 * (9) \implies 20 * (170) - 11 * (306) = 34 \implies$ $$

$\boxed{t = 170 + 11 m},$ $\boxed{t^{*} = 306 + 20 m}$ $$

$\implies$ $$

$\boxed{j_{1} = 3376 + 20 * 11 m}$ $\boxed{j_{2} = 2194 + 13 * 11 m}$ $\boxed{j_{3} = 3990 + 20 * 13 m}$ $$

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$j_{1}, j_{2},$ and $j_{3}$ satisfy all three equations $\implies X = 13 * (3376 + 20 * 11 m) + 6 = 43894 + 13 * 20 * 11 m = 43894 + 2860 m$ $$

$\implies \boxed{X \equiv 43894 \mod 2860 \equiv 994 \mod 2860.}.$ $$

$\therefore$ In the 2860 day year, day 6 in the 13 day cycle, day 14 in the 20 day cycle, and day 4 in the 11 day cycle would coincide on day $\boxed{994}$