Question

A) Find a number $0\leq x<14$ that solves the congruence $9x \equiv 9 \pmod{14}$.

Answer 1

 

A) Find a number $0\leq x$ that solves the congruence $9x \equiv 9 \pmod{14}$.

$9x \equiv 9 \pmod{14}$ means that 9x-9 is a multiple of 14, 

that is 9(x-1) is a multiple of 14. A clear solution in the set is x=1



b) Find a number $0\leq x$ that solves the congruence $3x \equiv 4 \pmod{5}$.


$3x \equiv 4 \pmod{5}$ means that 3x-4 is a multiple of 5, 

x∈{0, 1, 2, 3, 4} , so for these values 3x-4 becomes {-4, -1, 2, 5, 8}.

so x= 3 


c)Find a number $0\leq x$ that solves the congruence $999x \equiv 998 \pmod{1000}$.


$999x \equiv 998 \pmod{1000}$ means that 999x-998 is a multiple of 1,000

we can check that for x=2,         999*2-998=1,998-998=1,000


d)Find a number $0\leq x$ that solves the congruence $4x \equiv 17 \pmod{21}$

$4x \equiv 17 \pmod{21}$ means that 4x-17 is a multiple of 21, 

fro x=7, 4x=28 and so 4x-17=28-17=21 is a multiple of 21.



Answers:

A) 1
B) 3
C) 2
D) 4