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?
 A: Count the number of $3$ digit even numbers that do not contain $7$ and subtract from the total number of $3$ digit even numbers.

Let $xyz$ be the $3$ digit even number. Total number of $3$ digit even numbers is $450$, since $x$ has $9$ options, $y$ has $10$ options and $z$ has $5$ options. To obtain the number of $3$ digit even numbers that do not contain $7$, $x$ has $8$ options, while $y$ has $9$ options and $z$ continues to have $5$ options, i.e., $8 \times 9 \times 5 = 360$. Hence, the number of $3$ digit even numbers that contain at-least one $7$ is $450 - 360 = 90$.

A: 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.
A: The $7$ can't be in the ones place if the number is even.  That leaves the tens and hundreds place.
For all hundreds except the $700$s, there are five evens with at least one $7$: $n70, n72, n74, n76, n78$.
That accounts for $40$.
For the $700$s, there are $50$:  $700, 702, 704, ... 796, 798$.
So there are $90$ total.
A: There are $10$ digits, $0$ to $9$.
The probability of a $3$ digit even number without a $7$ is $$8/9 \times 9/10 \times 5/10 = 360/900$$
Which is $360$ even numbers without a $7$.
So the number with at least one $7$ is $450 - 360 = 90$.
A: A more numerical approach using python
In [2]: import itertools

In [3]: I=itertools.ifilter(lambda x:"7" in str(x), range(100,1000,2))

In [4]: len(list(I))
Out[4]: 90

