The six-digit number \(\overline{579ABC}\) is divisible by 5, 7, and 9. Find all possible values of this six-digit number.
+45 for a number to be divisible by 5, it must end in either 5 or 0, which means that \(C\) is either 5 or 0. for a number to be divisible by 7... well, we'll deal with that later. and for it to be divisible by 9, the sum of the digits must be divisible by 9. since \(5+7+6=21\), \(A+B+C\) is equal to 6, or a multiple of 9 plus 6.
if \(C\) is 0, then \(A+B\) is equal to 6 or a mutliple of 6 plus 1, and if \(C\) is 5, \(A+B\) is equal to 1 or a multiple of 9 plus 1. this is the point where i begin to bash, case by case, and where there is almost definitely a more graceful solution.
if \(A+B=6\), then \(A\) and \(B\) can be \((0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1),\) or \((6, 0)\). we check if each of these make the original number divisible by 7 (using a calculator in my case), and we have \((4, 2, 5)\) as an ordered triplet.
if \(A+B=6+1(9)=15\), then \(A\) and \(B\) can be \((6, 9), (7, 8), (8, 7),\) or \((9, 6)\). once again, we check if any of these will make the original number divisible by 7, but this time, none of the values for \(A\) and \(B\) work.
if \(C = 5\), then \(A\) and \(B\) can be 1 or 10. \(A+B=1\) yields no working ordered triplets and \(A+B=10\) yields \((4, 6, 0)\).
all in all, the only possible values for the number are \(\boxed{579425}\) and \(\boxed{579460}\).
note: this solution, once again, is extremely messy, and i would love it if someone else could offer some suggestions for a more graceful answer. also, when i mention ordered pairs, i mean in the form \((A, B)\), and when i mention ordered triplets, i mean in the form \((A, B, C)\).
to the original poster: if you could post the official solution once you get it (or if it exists), that would be amazing. thanks for the awesome problem!
