Okay, so I need to find the smallest positive multiple of 450 where all the digits are either 0 or 1. Hmm, let's start by understanding the problem. 

First, 450 is the number we're looking at. Its prime factorization is 2 * 3^2 * 5^2. So any multiple of 450 must be divisible by 2, 9 (since 3^2 is 9), and 25 (since 5^2 is 25). That's a key point. 

A multiple of 450 can be written as 450 * k, where k is a positive integer. We need 450 * k to be a number composed only of 0s and 1s. The smallest such number would be the one with the least number of digits, starting from the smallest possible and moving upwards.

Let me break down the requirements. Since the number must be divisible by 25, the last two digits must be 00, 25, 50, or 75. But since all digits have to be 0 or 1, the only possible ending is 00. So the number must end with two 0s. That simplifies things a bit.

Also, the number must be divisible by 9. The rule for divisibility by 9 is that the sum of the digits must be a multiple of 9. Since the digits can only be 0 or 1, the sum of the digits is equal to the number of 1s. Therefore, the number of 1s in the number must be a multiple of 9. So the number of 1s in the number must be 9, 18, 27, etc.

Additionally, the number must be divisible by 2. Since the last digit is already 0 (from the divisibility by 25 requirement), that's already satisfied.

So putting it all together, the number we're looking for must end with two 0s, have a number of 1s that is a multiple of 9, and be the smallest such number possible.

Let me start by considering the smallest possible numbers that end with 00 and have a number of 1s equal to 9. The smallest such number would be 11111111100. But wait, is that the smallest? Let me check.

Wait, the number must be as small as possible. So the number of digits should be minimized. Let's see. The number must end with two 0s, so the last two digits are 0. The rest of the digits should be 1s, but the total number of 1s must be a multiple of 9. So the minimal number of 1s is 9, so the total number of digits would be 9 (1s) + 2 (0s) = 11 digits. So the number would be 11111111100. But is that the smallest possible?

Wait, but maybe the number can have 1s and 0s in the middle as well. For example, maybe a number like 10101010100 is smaller. But wait, how do we compare numbers with different lengths? A shorter number is always smaller than a longer one. So the smallest possible number would be the one with the fewest digits. So 11111111100 has 11 digits. Is there a smaller number with fewer digits?

Wait, 9 digits would be impossible because we need at least two zeros at the end, so 9 digits would end with two zeros, but then the first seven digits would have to be 1s, but 7 is not a multiple of 9. So the next possible is 9 + 2 = 11 digits. Therefore, 11111111100 is the smallest possible. But wait, maybe there's a number with more 0s and 1s in between that's smaller? For example, 10011111100. Wait, but that's 13 digits, which is longer than 11, so it's bigger. So no, 11111111100 is smaller.

But let me verify. Let's check if 11111111100 is divisible by 450. Since it ends with 00, it's divisible by 100, which is 4*25, so divisible by 25. The sum of the digits is 9, which is divisible by 9. And it's even, so divisible by 2. Therefore, 11111111100 is divisible by 450. So that's a candidate.

But wait, is there a smaller number? Let's think. Maybe a number with fewer digits but arranged differently. For example, 1000000000000... but that's 1 followed by zeros, but the sum of digits is 1, which is not a multiple of 9. So that wouldn't work. Alternatively, maybe 11111111111100, but that's longer. Hmm.

Wait, perhaps there's a number with fewer digits. Let's see. The minimal number of digits is 11, as we concluded before. So 11111111100 is the smallest possible. But wait, let's check if there's a number with 9 1s and 2 0s arranged in a way that's smaller than 11 digits. Wait, no. Because 11 digits is the minimal. So 11111111100 is the answer. But let me check with the divisibility rules again.

Wait, 11111111100 divided by 450. Let me compute that. 11111111100 / 450. Let me see. 11111111100 divided by 100 is 111111111. Then 111111111 divided by 4.5. Wait, 450 is 45*10. So 11111111100 / 450 = (11111111100 / 10) / 45 = 1111111110 / 45. Let me compute 1111111110 divided by 45. 45 * 24691358 = 1111111110. So yes, 11111111100 is divisible by 450. Therefore, it is a valid multiple.

But wait, is there a smaller number? For example, maybe a number with 9 1s and 2 0s, but arranged in such a way that the number is smaller. For example, 10111111100. Let me check the sum of digits. 1+0+1+1+1+1+1+1+1+0+0 = 8. That's not a multiple of 9. So that's invalid. How about 11011111100? Sum is 1+1+0+1+1+1+1+1+1+0+0 = 8. Still 8. Not good. Hmm. Maybe 11101111100. Sum is 1+1+1+0+1+1+1+1+1+0+0 = 8. Still 8. Hmm. What if we try to have 18 1s? Then the number would be 18 digits long, which is longer, so it's worse.

Alternatively, is there a way to have the 1s spread out more? For example, 100100100100100, but that's 16 digits, which is more than 11. So 11111111100 is smaller.

Wait, but maybe there's a number with 9 1s and 2 0s, but arranged in a way that the number is smaller. For example, putting the 0s towards the end. But since the number has to end with 00, the last two digits are fixed as 00. So the 1s are all in the first nine digits. So the smallest possible number would be 11111111100, which is nine 1s followed by two 0s. Any other arrangement would require moving the 1s, but since all digits are 1s and 0s, the smallest number is achieved by having all 1s first, then 0s. So 11111111100 is indeed the smallest.

But wait, let me confirm with a smaller example. Suppose we have a number that's a multiple of 100 (ends with 00) and sum of digits is 9. The smallest such number is 11111111100. Let me check numbers with fewer digits. For example, 10011111100 is 11 digits as well, but it's larger than 11111111100 because the first digit is 1, then 0, then 1, etc. So 11111111100 is indeed the smallest.

Therefore, the answer should be 11111111100. Let me check once more. The digits are all 1s and 0s. The number ends with 00, so divisible by 25 and 2. The sum of digits is 9, which is divisible by 9. Therefore, it's divisible by 450. And it's the smallest such number because any other arrangement would either have a higher number of digits or a higher value with the same number of digits. 

So, I think that's the answer. But just to be thorough, let me check if there's a number with 9 digits that ends with two 0s and has 9 1s. But that would be 11111111100, which is 11 digits, as we already established. So no, 9 digits can't have two 0s and 9 1s because that would require 11 digits. Therefore, the minimal number is 11 digits long. 

Therefore, the smallest positive multiple of 450 with all digits 0 or 1 is 11111111100.
However, upon further reflection, I realize that maybe there's a number with fewer digits that still satisfies the conditions. Let me think again.

Wait, the number must be a multiple of 450, which requires it to be divisible by 2, 9, and 25. We know it must end with 00. The sum of the digits must be a multiple of 9. The smallest number of digits is 11, but maybe there's a number with 10 digits? Let's check.

If the number has 10 digits, it would need to end with 00, so the last two digits are 0. Then there are 8 digits left. If all those 8 digits are 1s, the sum of digits would be 8, which is not a multiple of 9. So that's invalid. Alternatively, if some of the digits are 0s, the sum would be less than 8. But 8 is not a multiple of 9. Therefore, 10 digits is impossible.

Similarly, 9 digits: 9 digits, ending with 00. Then the first 7 digits can be 1s, sum is 7, which is not a multiple of 9. So 9 digits is also impossible. Therefore, 11 digits is indeed the minimal.

Therefore, 11111111100 is the answer.
</think>