If a number is divisible by 450, then it must be divisible by all divisors of 450, including 9 and 50.

For a number to be divisible by 9, the sum of its digits must be divisible by 9. Since a positive number must have at least one digit which is not 0, the number we're looking for is forced to have at least 9 ones among its digits.

The number we're looking for must also be divisible by 50, which means that it ends in 50 or 00. Since the digit 5 is not allowed, our number must end in 00, which means the smallest candidate is  $\boxed{11,\! 111,\! 111,\! 100}$. In fact, because 9 and 50 $\emph{do}$ divide this number, and because 450 is the least common multiple of 9 and 50, we know that 450 does divide 11,111,111,100; so that number is the correct answer.
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
What is the smallest positive multiple of 450 whose digits are all zeroes and ones?
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
11,\! 111,\! 111,\! 100
What is the smallest positive multiple of 450 whose digits are all zeroes and ones?
test