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remainder 2 divisor 3; remainder 3 divisor 4; and remainder 1 divisor 5. Multiply
the other two divisors (4 x 5 = 20) and by the remainder (2, so =40). Subtract 1
and divide by the first divisor (3) so we get 13. Repeating using the second divisor
we get 3 * 5 * 3 = 45 - 1 = 44 / 4 = 11. For the third combination 3 x 4 = 12 - 1 = 11.
To get an even division by 5 we use -5. Now we sum three expressions of the form
remainder * ((factor * divisor) + 1): 2 * ((13 * 3) + 1) = 80 + 3 * (11 * 4) + 1) = 135
and (1 * (-5 * 5) + 1) = -24. The sum is then 191 modulo the product of the divisors
(3 * 4 * 5 = 60). We can reduce 191 by two (2) 60s, hence 71.
the other two divisors (4 x 5 = 20) and by the remainder (2, so =40). Subtract 1
and divide by the first divisor (3) so we get 13. Repeating using the second divisor
we get 3 * 5 * 3 = 45 - 1 = 44 / 4 = 11. For the third combination 3 x 4 = 12 - 1 = 11.
To get an even division by 5 we use -5. Now we sum three expressions of the form
remainder * ((factor * divisor) + 1): 2 * ((13 * 3) + 1) = 80 + 3 * (11 * 4) + 1) = 135
and (1 * (-5 * 5) + 1) = -24. The sum is then 191 modulo the product of the divisors
(3 * 4 * 5 = 60). We can reduce 191 by two (2) 60s, hence 71.
The original problem is attacked using pairs of congruences:
