Solve Problem I, 4, from the Shushu jiuzhang, which is equivalent to N = 0 (mod 11), N = 0 (mod 5), N = 4 (mod 9), N = 6 (mod 8), N = 0 (mod 7).

1. Solve each of the following. 14x = 1 (mod 45) 24x = 35 (mod 59) b. 62x = 1 (mod 99) а. С. d.

In the following question: " Let a, b ∈ Z. Show that if a ≡ 5 (mod 6) and b ≡ 3 (mod 4), then 4a + 6b ≡ 6 (mod 8)." why do we ignore the -6 when substituting the variables of a and b, if we expand "4a + 6b ≡ 6 (mod 8)" we get (4a+6b)-6=8y, when we substitute we have to end up with the 8y right? Im just confused why the -6 is ignored?

(3) Solve 66x = 111(mod 237).

Suppose b is any integer. If b mod 12 = 7, what is 8b mod 12? In other words, if division of b by 12 gives a remainder of 7, what is the remainder when 8b is divided by 12? Fill in the blanks to show that the same answer will be obtained no matter what integer is used for b at the start. . Multiply both sides of this equation by 8 and then simplify the right-hand side to find values of q Because b mod 12 = 7, there is an integer m such that b = 12m + 7 and r such that 8b = 12q + r with 0 ≤r< 12. The result is q= and r = Now 0 ≤r < 12, and q is an integer because products and differences of integers are integersex remainder obtained when 8b is divided by 12 is 62 x. So the uniqueness part of the quotient remainder theorem guarantees that the