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Solution.

Q1.

a)

i)

$π = \{π₯: π₯ β π, 1 β€ π₯ β€ 12\}=\{1,2,3,4,5,6,7,8,9,10,11,12\},\newline π΄ = \{2π₯: π₯ β π \text{πππ π₯ ππ π ππ’ππ‘ππππ ππ } 4\}=\{8,16,24\},\newline π΅ = \{π₯: π₯ β π, π₯ \text{πππ£ππππ } 2\}=\{2,4,6,8,10,12\},\newline πΆ = \{π₯: π₯ β π, π₯^2 β€ 16\}=\{1,2,3,4\}.$

ii)

$\overline{A}=\{1,2,3,4,5,6,7,9,10,11,12\},\newline \overline{A}\cap B=\{2,4,6,10,12\},\newline C-(\overline{A}\cap B)=\{1,3\}.$

iii)

$\overline{C}=\{5,6,7,8,9,10,11,12\}, \newline B\oplus \overline{C}=\{2,4,5,7,9,11\}.$

b)

$1 + 5 + 9 + β¦ + (4n β 3) = n(2n β 1) \text{for all } n β₯ 1.$

Let

$P(n):1+5+9+ . . .+(4n-3)=n(2n-1), \text{for all } n β₯ 1.\newline P(1):1=1(2\cdot 1β1), 1=1, \text{which is true.} P(1) \text{is true.}\newline \text{Assume that } P(n) \text{ is true for } n=k.\newline P(k):1+5+9+ . . .+(4k-3) = k(2k-1) \text{ is true.}\newline \text{Prove } P(k+1) \text{ is true.}\newline P(k+1):1+5+9+. . . +(4k-3)+4(k+1)-3=\newline =k(2k-1)+4(k+1)-3=2k^2βk+4k+4β3=\newline =2k^2+3k+1=2k^2+2k+k+1=(k+1)(2k+1)=\newline =(k+1)(2k+1+1-1)=(k+1)(2(k+1)-1).\newline \text{So, } P(k+1) \text{ is true, whenever } P(k) \text{ is true, hence } P(n) \text{is true.}$

c)

$π₯ = 866 \text{ and } π¦ = 732.$

i)

$866 = 2 Β· 433,\newline 732 = 2 Β· 2 Β· 3 Β· 61 = 2^2 Β· 3 Β· 61,\newline$

from here the greatest common divisor ofΒ $x$Β andΒ $y$Β isΒ $2.$

GCD$(866,732)=2.$

$ax+by=-71\cdot866+84\cdot732=2.$

ii)

$866 = 2 Β· 433,\newline 732 = 2 Β· 2 Β· 3 Β· 61 = 2^2 Β· 3 Β· 61,\newline$

from here LCM$(866,732)=2\cdot433\cdot2\cdot3\cdot61=361956.$