Let n > 1 be a fixed integer and a, b, c, d be arbitrary integers. If a b ±odn and c d ±odn , then which of the following is true? - Mathematics MCQ Mathematics Mock Test - 1 2026 | ExamTest.live

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Let n > 1 be a fixed integer and a, b, c, d be arbitrary integers. If a \equiv b (mod n) and c \equiv d (mod n), then which of the following is true?

Solution & Step-by-step Explanation

Given a \equiv b (mod n) and c \equiv d (mod n) .This means n ∣ (a - b) and n ∣ (c - d) .So, a - b = k_{1} n and c - d = k_{2} n for some integers k_{1}, k_{2} .For addition: (a + c) - (b + d) = (a - b) + (c - d) = k_{1} n + k_{2} n = (k_{1} + k_{2}) n Since the difference is a multiple of n, a + c \equiv b + d (mod n) .For multiplication: a c = (b + k_{1} n) (d + k_{2} n) = b d + b k_{2} n + d k_{1} n + k_{1} k_{2} n^{2} a c - b d = n (b k_{2} + d k_{1} + k_{1} k_{2} n) Since the difference is a multiple of n, a c \equiv b d (mod n) .

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Let n > 1 be a fixed integer and a, b, c, d be arbitrary integers. If a \equiv b (mod n) and c \equiv d (mod n), then which of the following is true?

a + c \neq \equiv b + d (mod n) and a c \neq \equiv b d (mod n)

a + c \neq \equiv b + d (mod n) but a c \equiv b d (mod n)

a + c \equiv b + d (mod n) and a c \equiv b d (mod n)

a + c \equiv b + d (mod n) but a c \neq \equiv b d (mod n)

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