Find the smallest positive $n$ such that \begin{align*} n & \equiv 3 \pmod{4}, \\ n & \equiv 2 \pmod{5}, \\ n & \equiv 6 \pmod{7}. \end{align*}

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number such that taking it mod 4, 5, and 7 leaves the desired remainders:

so we multiply the first term by 3.

so we multiply the second term by 2.

so we multiply the last term by 7.

Now,

By the CRT, the system of congruences has a general solution

or all integers , , the least (and positive) of which is 27.

cheater

step-by-step explanation:

41*67/2=1373.5

step-by-step explanation:

i just used a calculator