Starz

thanks!

Thanks! I wasn't sure if Modulo 7^2019 worked on this kind of problem, and yeah, I think it does.

So my way (that I didn't know that worked) was to see the ones digit for powers of the remainder of 7 mod 5.

2, 4, 8, 6

It would be 8, and then divide that by 5 and get the remainder of 3.

Thanks!

Thanks! 

a²+b²=25

ab = 5

a/b + b/a = ?

So, getting rid of the denominator, (a/b + b/a) * ab = a²+b²

so a²+b² / ab = ?

25/5 = \(5\)

-\(S\)

Greatest Odd Factor would be 5^5 * 3^3. 3125 * 27 = 84375

S

Just found the correct answer. Thank you for the help! 

There are 7! ways to put the beads on the grid, not considering rotations and reflections. Arrangements can be reflected or not reflected and can be rotated by 0, 60, 120, 180, 240, or 300 degrees, so they come in groups of twelve equivalent arrangements. Correcting for the symmetries, we find that there are 7!/12 or 420 distinct arrangements.

Thank you!