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The binary expansion of the decimal number 105 is ___________two. Only type the digits; do not include the base.

To complete the division algorithm equation, a = mq + r, using a = – 38 and m = 7, which of the following gives appropriate values for integers q and r, with r expressed as a non-negative integer between 0 and (m-1), inclusive.

For arbitrary integers a, b, and d, with d ≠ 0, if there are integers s and t such that d = as + bt, then d = GCD(a, b).

Determine which of these set identities are supported by the entries in the membership table given below.  There may be more than one or none. Select ‘True’ if the identity is supported by this given membership table; otherwise select ‘False’. [1]  (A – B) – C ⊈ (A – B) [2]  (A – C) – B ⊂ (A – C) [3]  (A – B) ⊄ (A – C) – B [4]  (A – C) ≠ (A – B) – C [5]  (A – C) ⊆ (A – B) [6]  (A – C) – B = (A – B) – C A B C A – C A – B (A – C) – B (A – B) – C 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 1 1 1 1 0 1 0 1 0 0 1 1 0 1 0 0 0 1 1 1 0 0 0 0

Use the Euclidean algorithm to determine the GCD(286, 273).  Show your work. Then express the GCD(286, 273) value you identify as a linear combination of 286 and 273.  Show your work.

Prove the following statement using induction. “For all positive integers n, 3|(n3 + 2n).”  Use good proof technique.  Note:  To avoid the need for typing superscript exponents, you may use the expression ‘n^3’ to represent n3. Grading rubric:1 pt. State the basis step, then prove it.1 pt. State the inductive hypothesis.2 pt. Complete the proof of the inductive step.1 pt. State the final conclusion at the end of the proof.1 pt. Label each part: the basis step, inductive hypothesis, inductive step, and conclusion.  

Complete your choice of one of the proofs given below. PROOF 1: Prove the following statement using a proof by cases.   [Hint: there are 3 cases] “For all positive integers n with 2 ≤ n ≤ 4, n!/2 ≤ n2+1.” Use good proof technique.   Grading rubric:1 pt. State any givens and assumptions.3 pt. Clearly identify the cases and prove each case.1 pt. State the final conclusion at the end of the proof. Note:  Remember that n factorial, written as n!, is defined as n(n-1)…(2)1, the product of n times every positive integer less than n.   To avoid the need for typing superscript exponents, you may use the expression ‘2^n’ to represent 2n.  Also the ≤ symbol can be written as

Let the function f : ℕ → ℝ be defined recursively as follows:      Initial Condition:  f (0) = 18Recursive Part:  f (n + 1) = 1/9 * f (n), for n > 0 Consider how to prove the following statement about this given function f using induction. For all nonnegative integers n, f (n) = 2/9(n-1). Select the best response for each question below about how this proof by induction should be done.  Q1.  Which of the following would be a correct Basis step for this proof?   [Basis] A.  For n = 0, f(n) = f(0) = 18; also 2/9(n-1) = 2/9-1 = 2*9 = 18, so f(n) = 2/9(n-1) for n = 0. B.  For n = k, assume f(k) = 2/9(k-1) for some integer k ≥ 0, so f(n) = 2/9(n-1) for n = k. C.  For n = k+1, f(k+1) = 2/9(k+1-1) when f(k) = 2/9(k-1) for some integer k ≥ 0, so f(n) = 2/9(n-1) for n = k+1. D.  For n = 1, f(n) = f(1) = 1/9*f(0) = 1/9*18 = 2; also 2/9(n-1) = 2/90 = 2, so f(n) = 3n for n = 1. Q2.  Which of the following would be a correct Inductive Hypothesis for this proof?   [InductiveHypothesis] A.  Prove f(k) = 2/9(k-1) for some integer k ≥ 0. B.  Assume f(k+1) = 2/9(k) when f(k) = 2/9(k-1) for some integer k ≥ 0. C.  Prove f(k) = 2/9(k-1) for all integers k ≥ 0. D.  Assume f(k) = 2/9(k-1) for some integer k ≥ 0. Q3.  Which of the following would be a correct completion of the Inductive Step for this proof?   [InductiveStep] A.  f(k+1) = 1/9*f(k), which confirms the recursive part of the definition. B.  When the inductive hypothesis is true, f(k+1) = 1/9*f(k) = 1/9*2/9(k-1) = 2/9(k-1)+1 = 2/9(k) = 2/9(k+1)-1 C.  When the inductive hypothesis is true, f(k+1) = 2/9(k) = 2/9(k+1)-1 = 1/9*2/9(k-1) = 1/9*f(k), which confirms the recursive part of the definition. D.  When f(k+1) = 2/9(k+1)-1 = 2/9(k) = 1/9*2/9(k-1); also f(k+1) = 1/9*f(k), so f(k) = 2/9(k-1), confirming the induction hypothesis. Q4.  Which of the following would be a correct conclusion for this proof?   [Conclusion] A.  By the principle of mathematical induction, f(n+1) =1/9* f(n) for all integers n ≥ 0. B.  By the principle of mathematical induction, f(k) = f(k+1) for all integers k ≥ 0. C.  By the principle of mathematical induction, f(n) = 2/9(n-1) for all integers n ≥ 0. D.  By the principle of mathematical induction, f(k) = 2/9(k-1) implies f(k+1) = 2/9(k) for all integers k ≥ 0.  

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The binary expansion of the decimal number 205 is ___________two. Show your work or explain your answer.