Question

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

b) Find a number $0\leq x<5$ that solves the congruence $3x \equiv 4 \pmod{5}$.
c)Find a number $0\leq x<1000$ that solves the congruence $999x \equiv 998 \pmod{1000}$.
d)Find a number $0\leq x<21$ that solves the congruence $4x \equiv 17 \pmod{21}$.

Answer 1

A) Find a number
`0\leq x<14` 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<5` 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<1000` 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<21` 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