A two-digit integer is written next to itself twice, forming a six-digit number. If the resulting number is divisible by 6, then how many possibilities are there for the original two-digit number?

I'm not sure about this, but I think it always must be divisible by 3, so all we need is for the original to be divisible by 2. So there are 9 digits for the tens place of the original and 5 places for the ones digit, so 45. Can someone check this?