Shown. Exactly how many planes contain points J, K, and N?
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The smallest such number is 1055.
We want to find such that
The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.
Taking everything together, we end up with the system
Now the moduli are coprime and we can apply the CRT.
We start with
Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.
Taken modulo 2, we end up with
which means the first term is fine and doesn't require adjustment.
Taken modulo 3, we have
We want a remainder of 2, so we just need to multiply the second term by 2.
Taken modulo 5, we have
We want a remainder of 0, so we can just multiply this term by 0.
Taken modulo 7, we have
We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since , the inverse of 2 is 4.
So, we have to adjust to
and from the CRT we find
so that the general solution for all integers
.
We want a 4 digit solution, so we want
which gives .