[tex]\bold{n\equiv 91a+40b\ mod \ 130}[/tex]
Given that a is the remainder when n is divided by 10. Then, [tex]n\equiv a\ mod 10[/tex]
Also b is the remainder when n is divided by 13 i.e., [tex]n\equiv b\ mod 13[/tex]
Then by Division algorithm, there are integers k and m such that n = a + 10k and n = b + 13m
These two equations are equivalent to
13n = 13a +130k and 10n = 10b + 130m
Adding these two equations, 23n = 13a + 10b + 130(k + m)
⇒ [tex]23n\equiv 13a+10b\ mod\ 130[/tex]
gcd (23,130) = 1
So 23 has an inverse modulo 130.
Therefore, [tex]n\equiv 23^{-1} (13a+10b) \ mod\ 130[/tex]
Applying Euclidean algorithm for 130 and 23,
130 = 23 x 5 + 15
23 = 1 x 15 +8
15 = 1 x 8 +7
8 = 1 x 7 + 1
Therefore, 1 = 8 - 7 = 8 - (15 -8 ) = 2 x 8 - 15 = 2( 23 - 15 ) - 15
= 2 x 23 -3 x 15 = 2(23) -3(130-5x23) =17(23) -3(130)
So inverse of 23 = 17
So, [tex]n\equiv 17 (13a+10b) \ mod\ 130\equiv 221a +170b \ mod \ 130 \equiv \bold{91a+40b \ mod\ 130}[/tex]
Learn more about Euclidean algorithm at https://brainly.com/question/24836675
#SPJ4
