<Question>
What is the smallest positive multiple of 450 whose digits are all zeroes and ones?
</Question>

<think>
<Plan_1>
Okay, so I need to find the smallest positive multiple of 450 where all the digits are either 0s or 1s. Hmm, let's break this down step by step.

<prompt_1.1> Understand the problem and set up the initial conditions for finding the smallest multiple of 450 with digits 0s and 1s. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let's recall that a multiple of 450 must be divisible by 450. Since 450 factors into 2 * 3^2 * 5^2, any multiple of 450 must be divisible by 2, 5^2, and 3^2. That gives me some constraints to work with.

Divisibility by 2 is straightforward: the number must end with an even digit. Since we can only use 0s and 1s, the last digit has to be 0. That makes sense because 1 is odd, so ending with 0 ensures divisibility by 2.

Divisibility by 5^2, which is 25, requires that the number ends with at least two 0s. Wait, no, actually, the rule for 25 is that the last two digits must form a number that's divisible by 25. Since we can only use 0s and 1s, the possible endings are 00, 01, 10, or 11. Out of these, only 00 is divisible by 25. Therefore, the number must end with two 0s. So the last two digits are 00.

Additionally, the number must have digits summing to a multiple of 9 because of the divisibility rule for 9 (since 3^2 is 9). The sum of the digits (excluding the trailing zeros) must be a multiple of 9. But since the only digits are 0s and 1s, that means the number of 1s in the number must be a multiple of 9.

Putting this together, the number we're looking for is a sequence of 1s and 0s, ending with two 0s, and the total number of 1s must be a multiple of 9. Also, since we want the smallest such number, we need the least number of digits possible.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So the minimal case would be having 9 ones (since 9 is the smallest multiple of 9), followed by enough zeros to make the entire number end with two zeros. But wait, actually, the ones don't all have to be consecutive.

<prompt_2.1> Determine the minimal number of digits required for the number, considering the constraints of divisibility by 450 and digit composition. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
The number can have 1s separated by zeros, but the key is that the total number of 1s is 9, and the number ends with two zeros. However, to make the number as small as possible, we need to arrange the digits such that the number is minimized. That would mean having the 1s as far to the right as possible.

Wait, but if we have 9 ones, the minimal number would be 11111111100, right? But that's 11 digits long. But maybe there's a smaller number with more digits but arranged in a way that it's smaller.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Hmm, actually, no. Numbers are compared from left to right, so the leftmost digit has the highest place value. So, if we have fewer digits, the number is smaller. Therefore, 11111111100 is an 11-digit number. But perhaps there's a smaller number with more digits but starting with a 1 followed by zeros and ones in a way that the total number of digits is less than 11? Wait, that seems impossible because 11 digits is the minimal length required for 9 ones and 2 zeros.

<prompt_3.1> Confirm whether the number 11111111100 is the smallest possible multiple of 450 with digits 0s and 1s. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Wait, 9 ones and 2 zeros make 11 digits. So the minimal number of digits is 11. Therefore, the smallest such number would be 11111111100. But is that divisible by 450? Let's check.

Wait, but 11111111100 ends with two zeros, so it's divisible by 100, which covers the 2 and 25 divisibility. And the sum of the digits is 9, which is divisible by 9. Therefore, it should be divisible by 450.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But wait, maybe there's a smaller number with more digits but arranged differently? For example, 10111111100.

<prompt_4.1> Explore if there are smaller numbers with more digits that still meet the criteria of being a multiple of 450 with digits 0s and 1s. </prompt_4.1>

But wait, let's check if 11111111100 is indeed the smallest. Let's see if there's a smaller number with the same number of digits but a smaller value.

<prompt_4.2> Confirm if 11111111100 is the smallest number that meets all criteria. </prompt_4.2>

But hold on, maybe there's a number with fewer digits? Wait, the sum of digits must be 9 (a multiple of 9), and we need at least two zeros at the end. So total digits would be 9 + 2 = 11. Therefore, 11 digits is the minimum.

<prompt_4.3> Verify if 11111111100 is the smallest possible number by considering the minimum number of digits required. </prompt_4.3>

But let's check with fewer digits. Suppose someone claims a 9-digit number.

<prompt_4.4> Double-check the possibility of a smaller number by considering different configurations of digits. </prompt_4.4>

But let's confirm by trying to find a smaller number.

<prompt_4.5> Attempt to find a smaller number by considering different configurations of digits. </prompt_4.5>

But let's check if there's a smaller number with the same number of digits but arranged differently. For example, 10000000000 is 10 digits, but sum of digits is 1, which is not divisible by 9.

<prompt_4.6> Confirm if 11111111100 is the smallest number by checking other configurations. </prompt_4.6>

But let's check another angle. Suppose we model this as a number composed of 1s and 0s, ending with 00, with the sum of digits (the number of 1s) divisible by 9. The smallest such number would be 11111111100.

<prompt_4.7> Re-evaluate the approach and confirm if 11111111100 is the correct answer. </prompt_4.7>
</Plan_4>

<Execution_4>
<execution_4.1>
But 11111111100 is 11 digits. If we place a 0 between some of the 1s, like 10111111100, but that would be larger than 11111111100 because the second digit is 0 vs. 1. So numbers starting with 1 followed by 0s and 1s would be larger if the 0s are towards the left.
</execution_4.1>

<execution_4.2>
Wait, no. If we have 9 ones, the smallest number would be the one with the 1s as far to the right as possible. So starting with a 1, then as many zeros as possible, then the remaining ones. Wait, but if you have 9 ones and 2 zeros, the smallest number would be 1000000000000... but no, that's not right. Wait, actually, to minimize the number, you want the highest place values to be as small as possible. So the leftmost digit should be 1, followed by as many zeros as possible, and then the remaining ones. Wait, but we have 9 ones in total. So if we have the first digit as 1, then the next digits should be zeros until we place the remaining 8 ones. But since the number must end with two zeros, the last two digits are fixed as 00. Therefore, the minimal number would have the 1s grouped together at the end. So, for example, 11111111100. If we spread the ones out, the number would be larger. For instance, 10111111100 is larger than 11111111100 because the second digit is 0 vs. 1. Similarly, 11011111100 would be larger. So yes, 11111111100 is the smallest arrangement.
</execution_4.2>

<execution_4.3>
So 11111111100 (which is 11 digits) is the minimal length. But numbers with 11 digits can't be smaller than that. For example, 10000000000 is 10 digits but sum of digits is 1, which isn't divisible by 9. Similarly, 100000000000 is 12 digits, but maybe that's smaller? Wait, no. Wait, 11111111100 is 11 digits, which is smaller than any 12-digit number starting with 1. So 11111111100 is the minimal length.
</execution_4.3>

<execution_4.4>
If we have 9 digits, we need 9 digits where the sum is 9 (so all digits are 1s), but ending with two zeros. But 9 digits can't end with two zeros because that would require at least two digits. So 9 digits is impossible. Similarly, 10 digits: if we have 10 digits, we need 9 ones and 1 zero, but the last two digits must be zeros. So that would require at least two zeros, so 10 digits would need 8 ones and 2 zeros. But 8 isn't a multiple of 9, so that's not allowed. Therefore, 11 digits is indeed the minimum.
</execution_4.4>

<execution_4.5>
Wait, no. Because to have 9 ones, you need at least 9 digits. But since we need two zeros at the end, that's two more digits. So 11 digits. Therefore, 11111111100 is the smallest possible.
</execution_4.5>

<execution_4.6>
Not divisible by 9. The next would be 11111111100, which is 11 digits. But maybe there's a number with more digits but smaller in value? For example, 10000000000 is 10 digits, but sum of digits is 1. 100000000000 is 12 digits, but sum is 1. Not helpful. Alternatively, numbers like 101000000000, but sum is 2. Not divisible by 9. 110000000000, sum is 2. Still not. So perhaps the minimal number is indeed 11111111100.
</execution_4.6>

<execution_4.7>
But let's test with smaller numbers.

For example, 11111111100 is 11 digits. Let's see if there's a smaller number with more digits but starting with 1 followed by zeros and ones in a way that it's smaller. Wait, but numbers with more digits are actually larger. For example, a 10-digit number starting with 1 is smaller than an 11-digit number starting with 1. But we can't have a 10-digit number because to have the sum of digits as 9, you need at least 9 digits (all ones), plus two zeros at the end, making 11 digits. Therefore, 11 digits is the minimum.

Therefore, the number should be 11111111100. But let's verify if this is indeed divisible by 450.

First, check divisibility by 2: ends with 0, yes. Divisible by 5: ends with 0, yes. Divisible by 25: ends with 00, yes. Divisible by 9: sum of digits is 9, which is divisible by 9. Therefore, yes, 11111111100 is divisible by 450.

But is this the smallest such number? Let's see if there's a smaller number with more digits but arranged in a way that it's smaller. Wait, but numbers with more digits are larger. For example, a 12-digit number would be larger than an 11-digit number. Therefore, 11111111100 is the smallest.

But wait, another thought: maybe there's a number with 9 ones and 2 zeros, but arranged such that the ones are not all at the beginning. For example, 10111111100.



But wait, let's check if there's a smaller number with more digits but starting with 1 followed by zeros in a way that the total number of digits is less. Wait, but how?



But wait, let's check 11111111100 divided by 450: 11111111100 ÷ 450. Let's compute that.
</execution_4.7>
</Execution_4>

<Plan_5>
Based on execution_4.7:
First, 450 is 45 * 10, so dividing by 10 gives 1111111110. Then divide by 45.

<prompt_5.1> Perform the division of 11111111100 by 450 to confirm it is a multiple. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
1111111110 ÷ 45. Let's compute 1111111110 ÷ 5 = 222222222. Then divide by 9: 222222222 ÷ 9 = 24691358. So 450 * 24691358 = 11111111100.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
So that's correct.

But is there a smaller number?

<prompt_6.1> Confirm that 11111111100 is the smallest multiple of 450 with digits 0s and 1s. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Suppose there is a smaller number. Let's think about numbers with fewer digits. As discussed earlier, 9 digits can't end with two zeros and have 9 ones. 10 digits: 8 ones and 2 zeros, but 8 isn't divisible by 9. 11 digits: 9 ones and 2 zeros. So 11 digits is the minimum. Therefore, 11111111100 is the smallest.

But wait, another angle: maybe the ones are not all consecutive. For example, if we have 10011111100. But this has 10 ones? Wait, no. Let's count. 1,0,0,1,1,1,1,1,1,0,0. That's 9 ones. Wait, same number of ones. But the number is 10011111100, which is 11 digits. But numerically, 10011111100 is larger than 11111111100 because the second digit is 0 vs. 1. Therefore, 11111111100 is smaller.

Alternatively, 10101111100. But again, the leading digits would make it larger. So indeed, grouping all the ones at the beginning gives the smallest number.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Therefore, after considering all possibilities, the smallest positive multiple of 450 with all digits 0s and 1s is 11111111100.

But wait, let me check another multiple. Suppose we consider 450 times some integer.

<prompt_7.1> Explore an alternative method by considering multiples of 450. </prompt_7.1>

But let me check with another approach. Let's model this as a binary number (though with decimal digits). We need the smallest binary-looking number in decimal that's divisible by 450.

<prompt_7.2> Conclude the solution by confirming 11111111100 is the smallest multiple. </prompt_7.2>
</Plan_7>

<Execution_7>
<execution_7.1>
Let's see, 450 * n = N, where N is composed of 0s and 1s. Let's see, 450 is 9*50, so N must be a multiple of 9 and 50. As before, N must end with 00, and sum of digits multiple of 9. So the minimal N would be the minimal number ending with 00, with sum of digits 9. Therefore, 11111111100. There's no smaller number because with fewer digits, either the sum of digits can't be 9 or the number can't end with 00.
</execution_7.1>

<execution_7.2>
Since the digits are 0 and 1, it's like finding the smallest binary number (in decimal terms) divisible by 450. But binary numbers in decimal are numbers like 1, 10, 11, 100, etc. So the problem is similar to finding the smallest binary multiple of 450 in decimal. There might be existing algorithms or methods for this.

Alternatively, since we need the number to be divisible by 2, 25, and 9, and we know it must end with 00 and have digits summing to 9, 18, etc. The minimal sum is 9, so 9 ones. Therefore, the minimal number is 11111111100. Since we can't have fewer digits, this is the answer.
</execution_7.2>
</Execution_7>

<Plan_8>
Based on execution_7.2:
**Final Answer**
The smallest positive multiple of 450 whose digits are all zeroes and ones is \boxed{11111111100}.
</Plan_8>

<Execution_8>
No prompt to execute.
</Execution_8>

<Final_answer>
Thus, the smallest positive multiple of 450 whose digits are all zeros and ones is $\boxed{11111111100}$.
</Final_answer>
</think>