Find an integer that leaves a remainder of 2 when divided by either 3 or 5, but that is divisible by 4.

We want an integer such that



Note that the moduli are all relatively prime, so we can use the Chinese remainder theorem right away. As a first step, let's suppose



Taken modulo 3, the last two terms immediately vanish, and . We want a remainder of 2, so we just multiply this term by 2.



Next, taken modulo 4, all terms vanish, so we're good here.
Then, taken modulo 5, the first two terms vanish and we're left with . We want a remainder of 2. To rectify this, we can first multiply this term by the inverse of 4 modulo 5, then multiply again by 2. This guarantees that



The inverse of 4 modulo 5 is 4, since , so we end up with



You can confirm for yourself that 272 satisfies the desired conditions. The CRT says that any integer of the form



will work, i.e. where , and in particular 32 is the smallest positive solution.

2+2-9-$29: ! 3$3$282! 2! 2

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.