How many even 3 digit numbers contain at least one 7. How many even 3 digit numbers contain at least one 7.
I got 126, but it was not an answer choice for the problem. Can anyone help?

The final (units) digit is even, and there are five choices for that independent of whatever else happens - so the answer must be divisible by five.
There are $450$ even numbers with three digits. If there is no seven the hundreds digit is one of eight (no zero or seven), and the tens digit is one of nine (no seven). So $8\times 9\times 5$ with no seven - that is $360$ leaving $90$ with a seven.
