Question :- A four-digit number when divided by 7,8 and 9 leave remainders of 4,1,2 respectively. What could be a possible remainder when the number is divided by 216.
I need help in understanding the solution given by the professor in class
Number has to be of form $7a+4, 8b+1, 9c+2$
On combining these three kind of numbers into one family we get $504k+137$
Now from the next step I am getting confused :-
He said since $gcd(504,216)=72$
and $504k+137$ is $72a+65$ kind of number
therefore with 216 we will have $65, 137, 209$ as remainders
Why will these three remainders exist for sure and none of them is getting eliminated ?
This is similar to a problem I had asked earlier If p is a prime number greater than 3, then what all remainders are possible when $(p^3+15)$ is divided by 18? , where in by extending the remainders from mod 6 to mod 18 we got a list of "possible" remainders finally from which only 2 had survived upon checking , however in this current question none of the three remainders are getting eliminated. Why is this happening ?