A 3-digit integer is chosen randomly. Find the probability that it's possible to add a digit to its right end s.t. the result is a multiple of 45? A three-digit integer is chosen at random. What is the probability that it is possible to add a digit to its right end such that the resulting four-digit number is a multiple of $45$?
Edit: I got $\frac{1}{15}$ as my answer can anyone confirm
 A: How many 3-digit numbers are there? How many are of the form $9k$? How many are of the form $9k+4$? 
For total 3-digit numbers, you have $999-99 = 900$. 
For 3-digit numbers of the form $9k$, you have $111-11 = 100$. 
For 3-digit numbers of the form $9k+4$, you have $994 = 9k-5 \Longrightarrow k = 111$ and $94 = 9k-5 \Longrightarrow k=11$, so there are $100$ 3-digit numbers of that form. 
Total probability: $$\dfrac{200}{900} = \dfrac{2}{9}$$
A: Simply,
We have the resulting numbers: $1000,1001,1002,1003,\dots,9999$
This is the sample space (All possible outcomes).
Next, we find out the series of integers that are divisible by $45$: these integers are:
$1035,1080,1126,\dots,9990$
Now, using the a result of arithmetic progression, that the number of terms $n=\frac{\text{last term}-\text{first term}}{\text{common ratio}}+1$
So, $n=\frac{9990-1035}{45}+1=200$
Therefore, the required probability $=\frac{200}{9000}=\frac{2}{90}$
May be I am wrong, I do not believe in $\frac{2}{9}$ as InterstellarProbe answered.
EDIT (Because I was wrong):
The $3$-digit numbers that are available: $100,101,102,103,\dots,999$
So a total of $900$ numbers.
Out of these $900$ integers, $200$ integers can be used to create numbers that are divisible by $45$ by adding a digit to the right.
Hence the required probability is $200/900=2/9$
