2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
HW#9 (SP 22)
截止时间 3月31日 23:59 得分 113 问题 11
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此测验锁定于 3月31日 23:59。
最新 尝试 1 5,686 分钟 65,满分 113 分
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此测验的分数: 65,满分 113 分 提交时间 3月31日 23:54 此尝试进行了 5,686 分钟。
Show that if x is an odd integer, then has the form 8k+1, for some .
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2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
From Nikita Arbuzov: Use the properties of modulo
From Nikita Arbuzov: Incorrect
Show that if an integer n is not divisible by 3, then is always divisible by 3.
Similarly, show that if an integer n is not divisible by 3 then
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2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
Find GCD of 2947 and 3997 using Euclidean Theorem.
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2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
Express gcd(128469, 12818) as a linear combination of 128469 and 12818 using extended Euclid algorithm.
Prove that if with , then .
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2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
Prove that using Bezout’s identity.
didn’t prove the statement
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2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
For Z11, Find out:
Determine whether every element a of Zn has an inverse for n= 5, 6, and 7, 11
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2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
this is the answer for 9
Write the following decimal string to senary (base 6) showing work:
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2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
Find the GCD of 846 and 265.
Let’s solve this by Euclidean algorithm.
1) 846 = (q1 = 3 2) 265 = (q2 = 5 3) r1 = (q3 = 5
4) r2 = (q4 = 10
r3 is our gcd. GCD(846, 265) = 10
)*265 + (r1 =
)*r1 + (r2 = )*r2 + (r3 = 1 )*r3 + 0
We have to find the Bezout coefficients a,b such that a*846 + b*265 = GCD(846, 265). For doing this we start backward substitution. So we can write the above steps as :
1) r1 = 846 – q1*265
2) r2 = 265 – q2*r1
3) r3 = r1 – q3*r2
We start from 3 as r3 is our gcd and keep on substituting the values of r1,r2 and r3 as above.
r3 = r1 – q3*r2
= 1 *846 + 3 *265 +
*r1 (By substituting r1 and r2 in above equation)
=(a= 26 )*846 + (b= 83 )*265 (By substituting r1 in above equation)
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2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
答案 10: 3 答案 11: 5 答案 12:
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2022/4/28 21:25 HW#9 (SP 22): CMPSC 360 SP 22, Section 01: Discrete Math/Cs
测验分数: 65,满分 113 分
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