For integers a, b and k0 which is NOT correct.

Question

For integers a, b and k>0 which is NOT correct.

BlackTom AI · Accepted Answer

For any integer k > 0, k mod k = 0 because dividing k by itself gives quotient 1 and remainder 0. Modulo is defined so a mod k is the remainder r when a = kq + r with 0 ≤ r < k.

1) Relevant concepts
- Definition of a mod k: a = kq + r with 0 ≤ r < k, so a mod k = r.
- Congruence properties: a ≡ a mod k (mod k). From this follow the reduction rules for addition and multiplication:
  (a + b) mod k = [(a mod k) + (b mod k)] mod k
  (ab) mod k = [(a mod k)(b mod k)] mod k
  Also ab ≡ (a mod k)b (mod k).

2) Step-by-step reasoning for each option
A: "k mod k = 1"
- Write k = k·1 + 0 so remainder r = 0. Therefore k mod k = 0, not 1. So this statement is false.

B: "ab mod k = [(a mod k) · (b mod k)] mod k"
- Let a = kq1 + r1 and b = kq2 + r2 with 0 ≤ r1,r2 < k.
- ab = (kq1 + r1)(kq2 + r2) = k·(other terms) + r1·r2.
- Thus ab mod k = (r1·r2) mod k = [(a mod k)·(b mod k)] mod k. This is true.

C: "ab mod k = [(a mod k)b] mod k."
- Since a ≡ (a mod k) (mod k), multiply both sides by b: ab ≡ (a mod k)·b (mod k).
- Hence ab mod k = [(a mod k)b] mod k. This is true.

D: "a + b mod k = [(a mod k) + (b mod k)] mod k"
- Similar to B: a + b = k·(q1+q2) + (r1 + r2). Reducing modulo k gives the stated equality. This is true.

3) Why the correct choice is correct
- The question asks which statement is NOT correct. A is the only false statement because the remainder of dividing k by k is 0, not 1.

Conclusion: The statement "k mod k = 1" is NOT correct, so the correct choice is "k mod k = 1", which matches the provided answer.

For integers a, b and k>0 which is NOT correct.单项选择题

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