After the mechanic it cost $1.88 cheaper to run the car
If
N = a (mod 10)
N = b (mod 13)
gcd(10,13) = 1
then
N = 10 bx + 13 ay (mod 130)
Where
10x + 13y = 1
<span>-> </span>(10x + 13)
(mod 2) = 1 (mod 2)
<span>-> </span>y (mod 2) = 1
y = -3, x = 4
-> N = 40b – 39a
(mod 130)
It is given that ra + sb
should be non-negative:
N = 40b – 39a (mod 130)
N = 40b + (130 – 39)a (mod 130)
N = 40b + 91a (mod 130)
Therefore, N modulo 130, in terms of a and b is: <span>N = 40b + 91a
(mod 130).</span>
Yup and why would it be hard do it backward as multiplication
1 2x^2y +x^2 -3x +4
2. 3x +y +z4
3. x +2xyz3.
4. 9x^2yz2.
