Explanation:
Okay, so I need to find the smallest positive multiple of 450 where all the digits are either 0 or 1. Hmm, let me think. First, let me recall what 450 factors into. 450 is 45 times 10, which is 9 times 5 times 10. Wait, breaking it down further: 450 = 2 × 3² × 5². So, any multiple of 450 must have at least these prime factors.

Now, the number we're looking for has only 0s and 1s. So, it's a number like 10, 100, 101, 110, 1000, etc., but not numbers with 2, 3, 4, 5, 6, 7, 8, or 9.

But since the number has to be divisible by 450, which includes factors of 2, 3², and 5². Let me think about what each of these factors implies for the number's structure.

First, divisibility by 5² (25). A number is divisible by 25 if its last two digits are 00, 25, 50, or 75. However, since our number can only have 0s and 1s, the only possible ending is 00. Therefore, the number must end with at least two zeros.

Next, divisibility by 2. Well, since the number ends with two zeros, it's already divisible by 2, so that's covered.

Then divisibility by 9. Wait, 450 is 9*50, so the number must also be divisible by 9. Divisibility by 9 requires that the sum of the digits is a multiple of 9. Since all the digits are 0s and 1s, the sum of the digits is just the number of 1s. Therefore, the number must have a number of 1s that is a multiple of 9.

So, summarizing:

1. The number must end with at least two zeros.
2. The number must contain exactly 9, 18, 27, etc., ones. But since we want the smallest number, we should aim for the smallest number of ones, which is 9. However, the number must also be as small as possible in terms of length. So, maybe a smaller number with more digits but fewer ones? Wait, no. Wait, the number of ones must be 9, 18, etc., so to minimize the number, we want the minimal number of digits. Since 9 is the smallest multiple, we need the number to have 9 ones and end with two zeros.

But also, the number must be divisible by 2 (which it is because of the two zeros), and by 25 (again, because of the two zeros). So, the main constraints are that it ends with two zeros and has 9 ones.

Additionally, the number must be divisible by 9 (already covered by the sum of digits being 9). Wait, but 450 is 9*50, and we already have divisibility by 25 and 2. So, since 9 and 25 and 2 are co-prime? Wait, 2 and 25 are co-prime, and 9 and 25 are co-prime, so the number has to be divisible by lcm(9, 25, 2)= 9*25*2= 450. So that's correct.

Therefore, our problem reduces to finding the smallest number consisting of 0s and 1s that ends with two zeros and has exactly nine 1s. However, wait, actually, the number of 1s must be a multiple of 9, not necessarily exactly 9. But since we want the smallest number, it's better to have exactly nine 1s, because more ones would require more digits. For example, if we have 18 ones, the number would be longer, so the smallest number is likely to have nine ones.

However, the number can be of the form 1...100, where there are 9 ones followed by two zeros. But that would be 11111111100. But maybe there is a smaller number with more zeros and fewer ones? Wait, but the number of ones has to be a multiple of 9. So, for example, if we have 9 ones, but arranged in such a way that the number is smaller than 11111111100.

But how do we arrange the digits? For example, numbers with fewer digits would be better. Let's think about the number of digits. If we need nine 1s and two 0s at the end, then total digits would be 9 + 2 = 11 digits. But maybe we can have more zeros in between? Wait, but more zeros would mean more digits. Wait, if we have more zeros, then the number would be longer. For example, if instead of 11111111100 (which is 11 digits), maybe a number with more zeros but the same number of ones? Wait, but the number of ones has to be 9, so adding more zeros would make the number longer. So, the minimal number of digits is 9 + 2 = 11, but perhaps arranging the ones in a way that the number is smaller? For example, putting the 1s as early as possible? Wait, but putting more 1s earlier would make the number larger. Wait, actually, numbers with digits 1 and 0 are compared lexicographically, so to get the minimal number, we need the smallest number of digits, and then the digits should be as small as possible. So, for example, the minimal number with nine 1s and two 0s at the end would be 10011111111? No, wait. Wait, the two zeros have to be at the end. Wait, no, the number must end with two zeros. So, the last two digits are zeros, and the rest of the digits can be 1s and 0s. But since we need the number to have nine 1s, and the rest of the digits (excluding the last two zeros) can have some 1s and 0s. Wait, but we need total of nine 1s, so the number of digits before the last two zeros is nine plus whatever zeros are in between. Wait, let me think again.

Suppose the entire number is composed of some number of digits, ending with two zeros, and the total number of ones in the entire number is nine. Then, the rest of the digits (before the last two zeros) can include some number of ones and zeros, but the total number of ones in the entire number must be nine. Therefore, the number of ones in the non-last-two-zero part is nine minus any ones in the last two zeros, but the last two zeros don't have any ones. So, all nine ones are in the part before the last two zeros. Therefore, the number is of the form [some digits with nine 1s and some 0s] followed by 00.

To make the entire number as small as possible, we need the [some digits] part to be as small as possible. How do we arrange nine 1s and some zeros in the beginning part so that the entire number is minimized?

The minimal number would be the one where the 1s are as far to the right as possible. Wait, no, to minimize the number, you want the smaller digits (0s) to be as far left as possible. Wait, for example, between the first digit and the last two zeros, you want to have as many zeros as possible on the left.

So, for example, if we have nine 1s and some zeros, to make the number as small as possible, we need to have the minimal number of digits. Wait, but we need the [some digits] part to have nine 1s. So, the minimal number of digits in that part is nine digits, all 1s. Then the total number is 11111111100.

But maybe there is a smaller number if we have more zeros in the beginning part. Wait, but adding zeros would make the number longer, but maybe the digits are smaller. Wait, but for example, if we have a number like 1010101010100. Let me count the number of ones. Let me see:

Each "10" is two digits, and if we have five of them, that's 10 digits, plus the two zeros at the end. Wait, but that would be 12 digits. But if we have 11111111100 (11 digits) versus 1010101010100 (13 digits). The first one is shorter, so it's smaller.

Wait, but maybe there's a shorter number? Let me think. For example, if we have a number with nine 1s and some zeros, but not necessarily all in the beginning. Wait, but the number has to end with two zeros. So, for example, maybe there is a number with fewer digits than 11 digits?

How many digits would be required? Well, the number must have at least two zeros at the end, and nine 1s. So, the minimal number of digits is nine 1s + two 0s = 11 digits. Therefore, the minimal possible length is 11 digits. Therefore, the minimal number is going to be an 11-digit number with nine 1s and two 0s, ending with two zeros.

Now, to find the smallest such number. How do we arrange the digits? To make the number as small as possible, we want the leftmost digits to be as small as possible. So, we need to place as many zeros as possible on the left. However, we have two zeros to place (the ones at the end are fixed). Wait, the two zeros at the end are fixed. But we can have other zeros in the digits before the last two zeros. Wait, the total number of zeros in the entire number is two (the ones at the end) plus any zeros in the earlier digits. But since we can have more zeros in the earlier digits, but we need to have nine 1s. So, the number before the last two zeros can have some number of 1s and zeros. Wait, but the total number of 1s in the entire number is nine. So, the number before the last two zeros must contain nine 1s and any number of zeros.

But to minimize the number, we want the digits before the last two zeros to be as small as possible. So, that would mean putting as many zeros as possible on the left. For example, if we have the number starting with as many zeros as possible, then the first digit cannot be zero. Wait, the number can't start with a zero. Therefore, the first digit has to be 1.

So, the first digit is 1. Then, the rest of the digits (excluding the last two zeros) can have zeros and ones, but we need a total of nine 1s. Since the first digit is already 1, we need eight more 1s in the remaining digits before the last two zeros.

So, the structure is: 1 followed by some digits (with eight 1s and some zeros), then ending with 00. So, how can we arrange those digits to make the number as small as possible?

To minimize the number, after the first 1, we want as many zeros as possible. So, let's consider the number as 1 followed by a sequence of digits (let's call this part A) then 00. Part A has eight 1s and some number of zeros. To minimize the entire number, part A should be the smallest possible number with eight 1s and the rest zeros.

So, the question now is: What is the smallest number with eight 1s and some number of zeros? Since we want the smallest number, we need to arrange the digits with the 1s as far to the right as possible. For example, the smallest number with eight 1s is 11111111, which is eight 1s. But if we add zeros, we can have more digits but smaller numbers. Wait, but adding zeros to the left would make the number larger. Wait, no, if we add zeros to the right, it would make the number longer but not necessarily smaller. Wait, but we need to have part A, which is before the final two zeros.

Wait, let me think again. The entire number is: [1][part A][00]. Part A is some digits with eight 1s and some zeros. To minimize the entire number, part A should be as small as possible. How do we make part A as small as possible? For example, part A is a number with eight 1s and some zeros. The minimal number with eight 1s and, say, k zeros is the number where the zeros are as far to the right as possible. Wait, no. For example, the minimal number with eight 1s and zeros is the one where the digits are arranged with as many zeros as possible on the left. Wait, but since part A comes after the first 1, and before the last two zeros.

Let me think of part A as a number. Let's say part A has length m. Then the entire number will have length 1 + m + 2 = m + 3. Wait, but m is the length of part A. However, the number of digits in part A is variable. However, since we need eight 1s in part A, the minimal number of digits in part A is eight (all 1s). If we add zeros to part A, the number of digits increases, but the number itself might be smaller. For example, if part A is 10000000, which is 1 followed by seven zeros, then the total number would be 1 10000000 00, which is 11000000000. But if part A is 11111111, then the total number is 1111111100, which is 11 digits. Wait, comparing 11000000000 (11 digits) and 1111111100 (11 digits). Which is smaller?

Let me write them out:

11000000000 vs 1111111100.

Comparing digit by digit:

First digit: 1 vs 1

Second digit: 1 vs 1

Third digit: 0 vs 1 → Here, 0 is smaller, so 11000000000 is smaller. So, adding zeros in part A can result in a smaller number.

So, in this case, part A is 10000000 (which is 1 followed by seven zeros) making the entire number 11000000000. Wait, but part A has eight 1s? Wait no. Wait part A is supposed to have eight 1s. If I put 1 followed by seven zeros, part A has only one 1. So that's not enough.

Wait, I confused part A. So, part A is the part between the first digit (which is 1) and the last two zeros. So, the total number of 1s in the entire number is nine. Since the first digit is already a 1, part A must contain eight more 1s.

Therefore, part A must have eight 1s and some number of zeros. Therefore, part A's digits are eight 1s and zeros. Therefore, to make part A as small as possible, we need to have the leftmost digits as small as possible. Since part A starts after the first 1, let me think of part A as a number. For example, if part A has eight 1s and some zeros, the minimal number would be the one with the zeros as far to the left as possible.

Wait, but since part A comes after the first digit, and before the last two zeros, we need to arrange part A such that when combined with the first 1 and the last two zeros, the entire number is minimal.

So, for example, if part A is 10000000, which has eight 1s? No, that only has one 1. Wait, part A must have eight 1s. So, part A needs to contain eight 1s and any number of zeros. Therefore, the minimal part A would be the smallest number with eight 1s and some zeros.

For example, the minimal number with eight 1s is 11111111 (eight 1s). If I add a zero to the end, it becomes 111111110, which is nine digits. But that's still a number with eight 1s. But is there a smaller number with eight 1s? Like 101010101... but with eight 1s.

Wait, the minimal number with eight 1s is the one with the smallest number of digits, which is eight. But if we can have more digits with zeros, but the number is smaller? For example, 100000001 is 1 followed by six zeros and a 1, which is nine digits. But comparing to 11111111 (eight digits), 11111111 is smaller.

But if we want to have part A as small as possible, perhaps with more digits? For example, maybe inserting zeros in between the 1s to make the number smaller. For example, 10101010101... Wait, but how many digits?

Wait, let's think in terms of how to create the smallest number with eight 1s. It's actually similar to the problem of finding the minimal number with a certain number of 1s and 0s. The minimal number is the one where the 1s are as far to the right as possible. For example, 1000...000 with eight 1s? No. Wait, no, that would have only one 1. Wait, to have eight 1s, the minimal number is 11111111. However, if you can have leading zeros? No, because part A comes after the first 1.

Wait, maybe part A can have leading zeros? Wait, no. Because the entire number cannot have leading zeros. The first digit is already 1, so part A is between the first 1 and the last two zeros. Therefore, part A can have leading zeros? Wait, no. Because if part A starts with a zero, then the entire number would have a 1 followed by a 0, then more digits. But since part A is between the first 1 and the last two zeros, the first digit is 1, then part A can have zeros. For example, the number could be 1 0 1 0 1 0 ... 00.

Therefore, part A can have zeros in front. Therefore, to make part A as small as possible, we need the minimal number with eight 1s, allowing any number of zeros.

But the minimal number with eight 1s and some zeros is obtained by placing the 1s as far to the right as possible. For example, for eight 1s, the minimal number is 10000000...000 with eight 1s? No. Wait, if you have eight 1s, the smallest number is 11111111. If you put zeros in between, the number becomes longer but might be smaller? Wait, for example, 1010101010... Wait, let me compare two numbers:

- 11111111 (eight 1s, eight digits)
- 100000001 (nine digits, eight 1s)
- 101010101 (nine digits, eight 1s, but actually that has five 1s, not eight)
Wait, no. For eight 1s, the number 1010101010101010... Wait, maybe 100000000000000... with eight 1s. No, it's confusing.

Let me think of it as the problem of finding the minimal number (in value) with exactly eight 1s. The minimal such number is the one with the least number of digits, which is eight digits: 11111111. However, if you can have a longer number with zeros that is smaller in value, like 10000000... But 100000000... would have only one 1. So, to have eight 1s, the minimal number is 11111111. If you have zeros in between, the number becomes longer but maybe has smaller value? For example, let's compare 11111111 (eight digits) vs 101111111 (nine digits). Which is smaller?

101111111 is 1 followed by 0 and seven 1s. Let's compare digit by digit:

First digit: 1 vs 1

Second digit: 1 vs 0 → 0 is smaller, so 101111111 is smaller. Therefore, even though it's longer, it's smaller.

So, in this case, inserting a zero earlier in the number can result in a smaller number. Therefore, the minimal number isn't necessarily the one with the least digits.

Therefore, to find the minimal number with eight 1s, we can think recursively or use some strategy. It's similar to finding the minimal binary number with eight 1s, but in decimal digits. Wait, actually, it's similar to the problem of finding the minimal number made up of 0s and 1s with exactly eight 1s.

This is equivalent to finding the minimal number, which is the one where the 1s are as far to the right as possible. For example, to get the minimal number, we need the leftmost digits to be 0s as much as possible. However, the first digit cannot be 0, so the first digit is 1. Then, the remaining digits should have as many 0s as possible before the next 1.

Wait, but since we need eight 1s, the minimal number would be 10000000... but with eight 1s. Wait, perhaps the minimal number is 100000001... but how?

Alternatively, think of the problem as starting with the first 1, then inserting zeros until we have enough 1s. Wait, perhaps the minimal number is 100000001... but with eight 1s.

Actually, let me think of this as the number being a string of 0s and 1s, starting with 1, and containing eight 1s. To make it as small as possible, we want as many 0s as possible, but we need to have eight 1s. Therefore, the minimal such number is obtained by placing the 1s as far to the right as possible.

For example, if we have eight 1s, the minimal number is 1000...000 with eight 1s. Wait, but how? If you have eight 1s, you need to place them in positions. For example, the minimal number is 100...0001... but with eight 1s. Wait, actually, the minimal number with eight 1s is 11111111 (eight 1s) but that's not helpful. Wait, but inserting zeros can make the number smaller.

Wait, here's a better approach. Think of the problem as finding the minimal number with eight 1s, which is equivalent to finding the minimal number in lexicographical order. Since numbers are compared left to right, the minimal number is the one that has the earliest possible zeros. So, after the first digit (which is 1), we want to have zeros as much as possible.

So, for example, the number would be 1 followed by as many zeros as possible, then the remaining 1s. However, we have to have eight 1s. Since the first digit is already 1, we need seven more 1s.

So, to make the number as small as possible, after the first digit, we want to have zeros until we have to place the remaining 1s.

So, the number would be 1 followed by seven 0s and then seven 1s. So, total digits would be 1 + 7 + 7 = 15 digits? Wait, no. Wait, part A is between the first 1 and the last two zeros. Wait, maybe I need to think again.

Let me clarify:

Entire number structure: [1][part A][00], where part A contains eight 1s and some zeros. So, the entire number will have 1 + length(part A) + 2 digits.

To minimize the entire number, part A needs to be as small as possible. To make part A as small as possible, given that it has eight 1s, we need to place the 1s as far to the right as possible.

Therefore, part A should be a number that starts with as many zeros as possible, then the eight 1s. However, since part A is between the first 1 and the last two zeros, inserting zeros in part A would make the number smaller.

Therefore, the minimal part A is a number with eight 1s, followed by zeros. Wait, but if part A is something like 00000000... but part A cannot start with zeros because the entire number already starts with 1. Wait, part A is after the first 1, so part A can have leading zeros. For example, part A could be 00000000... but then the entire number would be 1 followed by part A (which starts with zeros) then 00.

Wait, but if part A has eight 1s, how can it start with zeros? For example, part A could be 00000000 followed by eight 1s? No, that would have eight 0s and eight 1s, but we need eight 1s. Wait, part A can have any number of 0s and eight 1s.

So, to make part A as small as possible, it should have as many leading zeros as possible. For example, part A is a sequence of zeros followed by eight 1s. However, since we need eight 1s, if we have k leading zeros, then part A is k zeros followed by eight 1s. The total number of digits in part A is k + 8.

However, the entire number is 1 followed by part A followed by 00. So, the entire number would be:

1 (first digit)

Then, k zeros

Then, eight 1s

Then, 00.

So, the entire number is 1 followed by k zeros, then eight 1s, then two zeros.

To make this as small as possible, we need to minimize the value of this number. How does varying k affect the number?

For example, if k=0: the number is 1 11111111 00 → 1111111100 (11 digits)

If k=1: the number is 1 0 11111111 00 → 101111111100 (12 digits)

Wait, but comparing 1111111100 (11 digits) vs 101111111100 (12 digits). Since 11 digits is shorter, it's smaller. Wait, but 12-digit number is larger than an 11-digit number. So, even though the 12-digit number may have a smaller digit in the second position, the total number of digits is longer, making it larger. Therefore, the shorter number is smaller. Therefore, to minimize the entire number, we want the minimal number of digits.

Therefore, k=0 is better. Wait, but let's check with k=0.

Wait, part A with k=0 is just eight 1s. So the entire number is 1 followed by eight 1s and then two zeros: 11111111100. That's 11 digits.

Alternatively, if we have k=1, making part A have one zero and eight 1s, the entire number is 1 0 11111111 00, which is 101111111100, which is 12 digits. But 12 digits is longer, so even if the first digits are smaller, it's still larger.

Therefore, the minimal number is 11111111100. However, is there a way to have a number with fewer digits? Wait, the minimal number of digits is 11, since we have nine 1s and two 0s. Therefore, 11 digits is the minimum.

But is there a number with nine 1s and two 0s, ending with two zeros, which is smaller than 11111111100? For example, by arranging the 1s and 0s in part A in a smarter way, not just putting all 1s first?

For example, suppose part A is not all 1s, but some combination of 1s and 0s. For instance, if part A is something like 101010... with some 1s and zeros. Let's try constructing such a number.

Let me think. If part A has eight 1s and some zeros, maybe we can have more zeros in part A, but the total number of digits is still 11 digits. Wait, if part A is 8 digits (all 1s), then adding the first 1 and the last two zeros makes 11 digits. If part A has more digits, meaning more zeros, then the total number of digits would be more than 11. Therefore, to keep the total digits at 11, part A must be 8 digits (the first digit is 1, then part A is 8 digits, then two zeros). Therefore, part A must be 8 digits with eight 1s, so all 1s. Therefore, the number is 11111111100.

But maybe if we allow part A to have more digits, then the entire number has more than 11 digits, but maybe the number is smaller? For example, let's say part A has nine digits with eight 1s and one 0. Then, the entire number is 1 followed by nine digits (eight 1s and one 0) followed by two zeros. The total digits would be 1 + 9 + 2 = 12.

But is 12-digit number smaller than 11-digit number? No, because an 11-digit number is always smaller than any 12-digit number. Therefore, we can't get a smaller number by increasing the number of digits.

Therefore, the minimal number is indeed 11111111100.

But let me check if there is a number with nine 1s and two 0s (total 11 digits) ending with 00, but arranged differently, which is smaller than 11111111100.

For example, 101111111100 is 12 digits, which is longer. But if I have an 11-digit number with some zeros in the middle and the last two zeros. For example, 11111110100. Let me count the number of 1s:

1,1,1,1,1,1,1,0,1,0,0. That's seven 1s in the first seven digits, then 0, 1, 0, 0. So total of 8 ones? Wait, first seven digits are 1s: that's seven 1s, then another 1 in the ninth digit: total eight 1s. Then where is the ninth 1?

Wait, we need nine 1s. Therefore, if I have 11111110100, that's seven 1s, then a 0, then a 1, then two zeros. So total of 8 ones. Not enough. Therefore, need to have nine 1s. Let me try another arrangement.

Suppose the number is 111111110100. Wait, that's 12 digits. No, let's stick to 11 digits.

Let me try inserting a zero somewhere in the middle. For example: 111111101100. Wait, that's 12 digits. Again, too long.

Wait, for an 11-digit number ending with two zeros, with nine ones. So, the structure is [digits][00], where [digits] has nine 1s and some zeros. Wait, no. Wait, [digits] has nine 1s and some zeros. But since the entire number ends with two zeros, and the first digit is 1. Wait, [digits] is the first 9 digits? No. Wait, total length is 11 digits, last two are zeros. So the first nine digits have nine 1s? No, because the first digit is 1, and then the remaining eight digits can have eight 1s and some zeros.

Wait, actually, the entire number is 11 digits. Last two are zeros, so first nine digits have nine 1s? No, because the first digit is 1, and the remaining eight digits can have eight 1s. Wait, no. The total number of 1s in the entire number is nine. Since the last two digits are zeros, all nine 1s are in the first nine digits. Therefore, the first nine digits are all 1s. Therefore, the only possibility is 11111111100.

Wait, that seems to be the only 11-digit number with nine 1s and two zeros at the end. Therefore, there are no other numbers with nine 1s and two zeros at the end in 11 digits.

But maybe there is a number with more than nine 1s and more than two zeros, but with the number of 1s being a multiple of 9? For example, 18 ones. However, that would require a longer number, which would be larger. Therefore, the minimal number should be 11111111100.

But hold on, the problem says "smallest positive multiple of 450". Maybe there is a smaller number than 11111111100 which is a multiple of 450 and has only 0s and 1s.

Because 11111111100 is a multiple of 450? Let me check.

First, check divisibility by 450. Since it ends with two zeros, it's divisible by 100. Also, since the sum of digits is 9, it's divisible by 9. Therefore, since 450 = 9 * 50, and 50 is 2 * 25. Wait, 450 = 2 * 3² * 5². So, since the number ends with two zeros, it's divisible by 2 and 25. And sum of digits is 9, so divisible by 9. Therefore, it is divisible by 450. Therefore, 11111111100 is a multiple of 450.

But is there a smaller number? For example, numbers with fewer digits? But we concluded that the minimal number of digits is 11. Wait, can there be a number with fewer digits? Let me check.

Let me think. The number must end with two zeros, so it must have at least two zeros. The rest of the digits can be 1s and zeros. Also, the sum of digits must be a multiple of 9.

Suppose there is a number with, say, 10 digits. Let's check if that's possible. If it has 10 digits, ending with two zeros, so the first 8 digits must contain some number of 1s. The total sum of digits is the number of 1s in the first 8 digits. For this to be a multiple of 9, the number of 1s must be 9. But there are only 8 digits in the first part. Therefore, impossible.

What about 11 digits? As discussed, possible.

What about 12 digits? Then, the number could have more 1s. But since we want the smallest number, we need the minimal number of digits. Therefore, 11 digits is the minimum.

But maybe there is an 11-digit number smaller than 11111111100 which also meets the conditions? For example, inserting zeros in the first nine digits. But since the first nine digits must contain nine 1s, you can't insert zeros there. Because inserting a zero would require removing a 1, making the total number of 1s less than nine.

Wait, for example, suppose in the first nine digits, instead of nine 1s, we have eight 1s and one 0, and then have a 1 in the last two digits? But no, the last two digits are zeros. Therefore, all nine 1s must be in the first nine digits. Therefore, the first nine digits must be all 1s. Therefore, the number must be 11111111100.

Therefore, that is the only 11-digit number with nine 1s and two zeros at the end. Hence, it must be the minimal number.

But wait, maybe there's a number with more than nine 1s (like 18) but shorter? Wait, no. Because with more 1s, you need more digits. For example, if you have 18 ones and two zeros, the number is longer than 11 digits.

Alternatively, maybe a number that ends with two zeros but has fewer digits? But as shown earlier, with fewer than 11 digits, it's impossible.

Wait, let me test with 10 digits. Suppose a number with 10 digits, ending with 00. Then the first 8 digits have to have sum of digits = multiple of 9. The maximum sum of digits in first 8 digits is 8. Which is less than 9. Hence, impossible.

For 11 digits, first 9 digits have sum of digits = 9 (since last 2 digits are 0). Hence, possible.

For 12 digits, first 10 digits have sum of digits multiple of 9. For example, sum could be 9 or 18. If sum is 9, then the number would be 12 digits: first 10 digits have 9 ones and one zero, then two zeros. For example, 101111111100. Which is 12 digits. But 12-digit number is larger than 11-digit number. So, 11111111100 is smaller.

But maybe there's a smaller number with 11 digits? Like, if instead of nine 1s followed by two zeros, there's some zeros in the first nine digits, but with the total number of 1s being nine. Wait, but the first nine digits must have nine 1s, because the last two are zeros. Hence, they have to be all 1s.

Wait, another thought. What if the number has more than two zeros, but not at the end? Wait, but the number must be divisible by 25, which requires last two digits to be 00. Therefore, the number must end with 00. Therefore, the other zeros can be anywhere else. However, the total number of 1s must be a multiple of 9.

But if we have more zeros in the first part, but keeping the last two digits as 00.

Wait, for example, if the number is 1010101010100. Let me count the number of digits: 1 0 1 0 1 0 1 0 1 0 1 0 0. That's 13 digits. The sum of digits is 6 ones. Not multiple of 9. Not good.

If we have more ones. For example, 1110000000000. That's 13 digits. Sum of digits is 3. Not multiple of 9.

What if we have 11111111100 (11 digits) vs something else. Let me think. Suppose we have a number like 1001111111100. Wait, that's 13 digits. Again, longer.

Alternatively, is there a way to have an 11-digit number with nine 1s and two zeros, not necessarily all the 1s first? For example, 11111110100. Wait, let's check:

Number: 1 1 1 1 1 1 1 0 1 0 0.

Digits: positions 1-11:

1,1,1,1,1,1,1,0,1,0,0.

Number of 1s: Let's count: first seven 1s, then a 0, then a 1, then two 0s. So total 7 + 1 = 8 ones. Not enough. Need nine.

If I change one of the zeros to a 1: 111111110100. Wait, that's 12 digits.

Alternatively, 111111110100 (12 digits). But that's longer.

Alternatively, what about 11111111100? That's 11 digits.

Wait, perhaps inserting a zero and adjusting somewhere else? Let's try:

Suppose the number is 111111110100. But that's 12 digits.

Wait, maybe there's a way to have a number with nine 1s and two zeros in 11 digits but not all 1s first.

For example: 111111101100. But that's 12 digits.

Wait, no. Wait, 11 digits. Let me try:

Positions 1 to 11:

If I have eight 1s and one 0 in the first nine digits, then the tenth and eleventh digits are 0. So total digits: 11. Number of 1s: 8. Need one more 1. Therefore, if in the first nine digits, we have nine 1s, which is all. Hence, no other way.

Therefore, it's impossible to have an 11-digit number with nine 1s and two zeros at the end unless the first nine digits are all 1s. Therefore, the only such number is 11111111100.

But let me check if 11111111100 is indeed the minimal. For example, is there a smaller number with more 1s? Wait, but more 1s would require longer numbers.

Alternatively, maybe there is a number with nine 1s and two zeros but arranged such that it is smaller. For example, if the number is 101111111100 (12 digits), which is larger than 11111111100.

Alternatively, if there's a number like 110111111100 (12 digits). Still larger.

Alternatively, if there is a number with a different arrangement. Wait, but how?

Wait, maybe numbers with leading 1s, then zeros, then more 1s... but since all the 1s must be in the first nine digits, and the last two digits are zeros, there is no other way.

Therefore, seems like 11111111100 is the answer.

But let me check with another approach. Maybe using BFS or something.

Let me think. Since we need numbers made up of 0s and 1s, ending with 00, and divisible by 450. Let me think about building numbers incrementally and checking divisibility.

But that might take a long time, but maybe for small numbers.

The number must end with 00, so let's consider numbers ending with 00. Let's think about numbers composed of 0s and 1s, ending with 00, and check if they're divisible by 450.

The smallest such numbers would be:

- 100: Divisible by 100, but sum of digits is 1, not divisible by 9. Not divisible by 450.

- 1000: Same, sum of digits 1. No.

- 1100: sum digits 2. No.

- 10000: sum 1.

- 10100: sum 2.

- 11000: sum 2.

- 11100: sum 3.

- 100000: sum 1.

- 100100: sum 2.

- 101000: sum 2.

- 101100: sum 3.

- 110000: sum 2.

- 110100: sum 3.

- 111000: sum 3.

- 111100: sum 4.

- 1000000: sum 1.

Continuing this way, it's going to take too long. However, the sum of digits needs to be 9. So, we need numbers where the sum of digits is 9, and ending with 00.

So, the number is made of 0s and 1s, ends with 00, and sum of digits is 9. Therefore, the number has nine 1s and any number of zeros (at least two for the ending).

The minimal number is the one with the fewest digits, which is nine 1s + two zeros = 11 digits. As we discussed earlier.

Therefore, the minimal number is 11111111100.

But let me verify with another example. Suppose the number is 11111111100. Divided by 450, let me compute 11111111100 / 450.

First, divide by 100: 11111111100 / 100 = 111111111. Then divide by 4.5? Wait, 450 is 4.5 * 100. Wait, 450 = 9 * 50. Let me check divisibility.

Sum of digits: 9, so divisible by 9. Ends with 00, so divisible by 100. Therefore, divisible by 450.

Therefore, yes, 11111111100 is divisible by 450.

But maybe there is a smaller number? Let me think. What if the number has more zeros in the middle? For example, 100111111100. Wait, this is 12 digits. But 11 digits number is smaller.

Alternatively, maybe 101111111100. Still 12 digits.

But what about numbers with more than nine 1s but with more zeros? For example, 11111111111100 (14 digits). But this is bigger.

Alternatively, maybe a number like 10000000000000 (but sum is 1), nope.

Alternatively, is there a number with nine 1s and two zeros, arranged such that it is smaller than 11111111100? For example, 101111111100 (12 digits) is larger.

Alternatively, what if the number is 111111110100 (12 digits). Still larger.

Therefore, it seems that 11111111100 is indeed the smallest.

But just to be thorough, let me check if there's an 11-digit number with nine 1s and two 0s, not all 1s first, but with the zeros placed in such a way that the number is smaller.

For example, suppose the number is 101111111100. Wait, that's 12 digits. No.

If I try to make an 11-digit number with nine 1s and two 0s, where one of the zeros is in the middle. For example:

First digit: 1, then a zero, then eight 1s, then a zero. But the last two digits must be zero. Therefore:

Structure: 1 0 [eight 1s] 0 0.

So, the number would be 10 followed by eight 1s and two zeros: 101111111100. Wait, that's 12 digits again.

Alternatively, if we have two zeros somewhere in the first nine digits. For example:

Number: 1 followed by eight digits with eight 1s and one zero, then two zeros. Wait, total digits would be 1 + 9 + 2 = 12 digits. Again longer.

Alternatively, if we have two zeros in the first nine digits. For example: 1 followed by seven 1s, one 0, one 1, then two zeros. That would be: 1 1 1 1 1 1 1 0 1 0 0. Which is 11 digits. Let's check:

Digits: 1,1,1,1,1,1,1,0,1,0,0.

Number of 1s: 7 + 1 = 8. Not enough. Need nine. Therefore, need to replace one of the zeros with a 1. Then we have 1 followed by seven 1s, one 1, one 0, then two zeros. Wait, that would be eight 1s and one 0 in the first nine digits. So total number of 1s would be eight. Still not nine.

Therefore, need to have nine 1s in the first nine digits. Therefore, the only way is to have all 1s in the first nine digits.

Therefore, the number must be 11111111100.

So, I think that's the answer.

But just to be 100% sure, let me check if there's a multiple of 450 with digits 0 and 1 smaller than 11111111100.

Let me think about numbers with fewer digits.

As we saw earlier, with 10 digits, it's impossible because the sum of digits can't be 9.

What about numbers with more than nine 1s? For example, with 18 ones? But that would require more digits.

Alternatively, maybe a number with nine 1s and more zeros, but shorter? No.

Alternatively, numbers that end with 00, have sum of digits multiple of 9, but with more zeros in between.

Wait, for example, let's think about the number 11111111100. Is there a smaller number? Let me think in terms of decimal numbers.

What is 11111111100? It's 11,111,111,100.

Is there a smaller number composed of 0s and 1s, ending with 00, with sum of digits 9?

For example, 10000000000 (sum 1). No.

100000000100: sum 2.

...

What about 100100100100? Let's check:

Digits: 1,0,0,1,0,0,1,0,0,1,0,0. Sum is 4. Not enough.

How about 111000000000 (sum 3). No.

How about 111100000000 (sum 4).

How about 111110000000 (sum 5).

Continuing, 111111000000 (sum 6), 111111100000 (sum 7), 111111110000 (sum 8), 111111111000 (sum 9). Wait, this is 111111111000, which is 12 digits.

Compare with 11111111100 (11 digits). Which is smaller? The 11-digit number is smaller.

So, 11111111100 is smaller than 111111111000.

Therefore, 11111111100 is smaller.

Alternatively, is there a number like 1010101010100? Let me check the sum.

Digits: 1,0,1,0,1,0,1,0,1,0,1,0,0. Sum is 6. Not enough.

If I add more 1s. For example, 1110000000000 (sum 3). No.

Alternatively, 11111111100.

Therefore, after checking all these possibilities, I can't find a smaller number than 11111111100.

Therefore, I think the answer is 11111111100.

But to confirm, let me check if this number is indeed the minimal.

Another way: think of 450 in terms of multiples. The question is essentially asking for the smallest number consisting of 0s and 1s divisible by 450.

Since 450 = 9 * 50, and the number must have at least two zeros at the end (for divisibility by 25), and the sum of digits divisible by 9.

Therefore, the minimal such number would be the minimal number with sum of digits 9, ending with 00.

Which is 11111111100.

Therefore, I think that's the answer.

**Final Answer**
The smallest positive multiple of 450 whose digits are all zeroes and ones is \boxed{11111111100}.

To solve the problem of finding the **smallest positive multiple of 450** whose **digits are all zeroes and ones**, we proceed by systematically analyzing the conditions and constraints.

---

### Step 1: Prime Factorization of 450

We start with the prime factorization of 450:

$$
450 = 2 \times 3^2 \times 5^2
$$

Thus, the number we are looking for must be divisible by:

- **2** and **5² = 25** → Divisibility by 2 and 25.
- **3² = 9** → Divisibility by 9.

---

### Step 2: Constraints from the Digit Set

The number must consist **only of digits 0 and 1**, so it's made up of these two values exclusively.

---

### Step 3: Divisibility by 2 and 25

- **Divisibility by 2** is automatically satisfied if the number ends in **0** or **1** — but **actually**, more specifically, it must be even. Since the only even digits are **0**, it suffices if the number ends in **0**.
- **Divisibility by 25** requires the number to end in **00** (i.e., the last two digits must both be 0).

So, the number **must end in two zeros**.

---

### Step 4: Divisibility by 9

A number is divisible by 9 **if the sum of its digits is a multiple of 9**. Since the only digits are **0** and **1**, the sum of the digits is **equal to the number of 1s** in the number.

Therefore, the number must have **exactly 9 ones** and **some number of zeroes**, with at least **two zeroes at the end** to satisfy the divisibility by 25.

---

### Step 5: Minimize the Number

To find the **smallest** such number:

- We need at least **nine 1s**.
- We need **two zeroes at the end**.
- The remaining digits (if any) can include **additional zeroes** or more ones.

The **smallest such number** is the one with:

- **9 ones**
- **Exactly two zeroes at the end**
- The rest of the digits (if any) are zeroes
- The number is as small as possible in numeric value

Since the number must have nine 1s, and the minimal way to do this is to place all 9 ones first and then the two zeroes at the end, the number becomes:

$$
11111111100
$$

This number has:

- Nine 1s
- Two trailing zeroes (so divisible by 100 and 25)
- The sum of its digits is **9**, which is divisible by 9
- Thus, it is divisible by **450**

---

### Step 6: Verify Minimality

We check if a smaller number is possible:

- **Fewer digits than 11** would not be possible because the number must contain **at least nine 1s** (for divisibility by 9) plus **two trailing zeroes** (for divisibility by 25), totaling at least **11 digits**.
- Any arrangement that places the ones and zeroes in different orders (e.g., interspersing zeros among the ones) results in a **longer** number or a **larger** one in value.
- There is no way to reduce the number of digits or digits' value without violating the required divisibility and digit constraints.

---

### Final Verification

Let’s confirm:

- $11111111100$ is divisible by 2 (ends in 0) ✅
- $11111111100$ is divisible by 25 (ends in 00) ✅
- $11111111100$ is divisible by 9 (sum of digits = 9) ✅
- It contains only 0s and 1s ✅
- It has the smallest number of digits and the smallest value that satisfies all conditions

---

### Final Answer

$$
\boxed{11111111100}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest positive multiple of 450 where all the digits are either 0 or 1. Hmm, let me think. First, let me recall what 450 factors into. 450 is 45 times 10, which is 9 times 5 times 10. Wait, breaking it down further: 450 = 2 × 3² × 5². So, any multiple of 450 must have at least these prime factors.

Now, the number we're looking for has only 0s and 1s. So, it's a number like 10, 100, 101, 110, 1000, etc., but not numbers with 2, 3, 4, 5, 6, 7, 8, or 9.

But since the number has to be divisible by 450, which includes factors of 2, 3², and 5². Let me think about what each of these factors implies for the number's structure.

First, divisibility by 5² (25). A number is divisible by 25 if its last two digits are 00, 25, 50, or 75. However, since our number can only have 0s and 1s, the only possible ending is 00. Therefore, the number must end with at least two zeros.

Next, divisibility by 2. Well, since the number ends with two zeros, it's already divisible by 2, so that's covered.

Then divisibility by 9. Wait, 450 is 9*50, so the number must also be divisible by 9. Divisibility by 9 requires that the sum of the digits is a multiple of 9. Since all the digits are 0s and 1s, the sum of the digits is just the number of 1s. Therefore, the number must have a number of 1s that is a multiple of 9.

So, summarizing:

1. The number must end with at least two zeros.
2. The number must contain exactly 9, 18, 27, etc., ones. But since we want the smallest number, we should aim for the smallest number of ones, which is 9. However, the number must also be as small as possible in terms of length. So, maybe a smaller number with more digits but fewer ones? Wait, no. Wait, the number of ones must be 9, 18, etc., so to minimize the number, we want the minimal number of digits. Since 9 is the smallest multiple, we need the number to have 9 ones and end with two zeros.

But also, the number must be divisible by 2 (which it is because of the two zeros), and by 25 (again, because of the two zeros). So, the main constraints are that it ends with two zeros and has 9 ones.

Additionally, the number must be divisible by 9 (already covered by the sum of digits being 9). Wait, but 450 is 9*50, and we already have divisibility by 25 and 2. So, since 9 and 25 and 2 are co-prime? Wait, 2 and 25 are co-prime, and 9 and 25 are co-prime, so the number has to be divisible by lcm(9, 25, 2)= 9*25*2= 450. So that's correct.

Therefore, our problem reduces to finding the smallest number consisting of 0s and 1s that ends with two zeros and has exactly nine 1s. However, wait, actually, the number of 1s must be a multiple of 9, not necessarily exactly 9. But since we want the smallest number, it's better to have exactly nine 1s, because more ones would require more digits. For example, if we have 18 ones, the number would be longer, so the smallest number is likely to have nine ones.

However, the number can be of the form 1...100, where there are 9 ones followed by two zeros. But that would be 11111111100. But maybe there is a smaller number with more zeros and fewer ones? Wait, but the number of ones has to be a multiple of 9. So, for example, if we have 9 ones, but arranged in such a way that the number is smaller than 11111111100.

But how do we arrange the digits? For example, numbers with fewer digits would be better. Let's think about the number of digits. If we need nine 1s and two 0s at the end, then total digits would be 9 + 2 = 11 digits. But maybe we can have more zeros in between? Wait, but more zeros would mean more digits. Wait, if we have more zeros, then the number would be longer. For example, if instead of 11111111100 (which is 11 digits), maybe a number with more zeros but the same number of ones? Wait, but the number of ones has to be 9, so adding more zeros would make the number longer. So, the minimal number of digits is 9 + 2 = 11, but perhaps arranging the ones in a way that the number is smaller? For example, putting the 1s as early as possible? Wait, but putting more 1s earlier would make the number larger. Wait, actually, numbers with digits 1 and 0 are compared lexicographically, so to get the minimal number, we need the smallest number of digits, and then the digits should be as small as possible. So, for example, the minimal number with nine 1s and two 0s at the end would be 10011111111? No, wait. Wait, the two zeros have to be at the end. Wait, no, the number must end with two zeros. So, the last two digits are zeros, and the rest of the digits can be 1s and 0s. But since we need the number to have nine 1s, and the rest of the digits (excluding the last two zeros) can have some 1s and 0s. Wait, but we need total of nine 1s, so the number of digits before the last two zeros is nine plus whatever zeros are in between. Wait, let me think again.

Suppose the entire number is composed of some number of digits, ending with two zeros, and the total number of ones in the entire number is nine. Then, the rest of the digits (before the last two zeros) can include some number of ones and zeros, but the total number of ones in the entire number must be nine. Therefore, the number of ones in the non-last-two-zero part is nine minus any ones in the last two zeros, but the last two zeros don't have any ones. So, all nine ones are in the part before the last two zeros. Therefore, the number is of the form [some digits with nine 1s and some 0s] followed by 00.

To make the entire number as small as possible, we need the [some digits] part to be as small as possible. How do we arrange nine 1s and some zeros in the beginning part so that the entire number is minimized?

The minimal number would be the one where the 1s are as far to the right as possible. Wait, no, to minimize the number, you want the smaller digits (0s) to be as far left as possible. Wait, for example, between the first digit and the last two zeros, you want to have as many zeros as possible on the left.

So, for example, if we have nine 1s and some zeros, to make the number as small as possible, we need to have the minimal number of digits. Wait, but we need the [some digits] part to have nine 1s. So, the minimal number of digits in that part is nine digits, all 1s. Then the total number is 11111111100.

But maybe there is a smaller number if we have more zeros in the beginning part. Wait, but adding zeros would make the number longer, but maybe the digits are smaller. Wait, but for example, if we have a number like 1010101010100. Let me count the number of ones. Let me see:

Each "10" is two digits, and if we have five of them, that's 10 digits, plus the two zeros at the end. Wait, but that would be 12 digits. But if we have 11111111100 (11 digits) versus 1010101010100 (13 digits). The first one is shorter, so it's smaller.

Wait, but maybe there's a shorter number? Let me think. For example, if we have a number with nine 1s and some zeros, but not necessarily all in the beginning. Wait, but the number has to end with two zeros. So, for example, maybe there is a number with fewer digits than 11 digits?

How many digits would be required? Well, the number must have at least two zeros at the end, and nine 1s. So, the minimal number of digits is nine 1s + two 0s = 11 digits. Therefore, the minimal possible length is 11 digits. Therefore, the minimal number is going to be an 11-digit number with nine 1s and two 0s, ending with two zeros.

Now, to find the smallest such number. How do we arrange the digits? To make the number as small as possible, we want the leftmost digits to be as small as possible. So, we need to place as many zeros as possible on the left. However, we have two zeros to place (the ones at the end are fixed). Wait, the two zeros at the end are fixed. But we can have other zeros in the digits before the last two zeros. Wait, the total number of zeros in the entire number is two (the ones at the end) plus any zeros in the earlier digits. But since we can have more zeros in the earlier digits, but we need to have nine 1s. So, the number before the last two zeros can have some number of 1s and zeros. Wait, but the total number of 1s in the entire number is nine. So, the number before the last two zeros must contain nine 1s and any number of zeros.

But to minimize the number, we want the digits before the last two zeros to be as small as possible. So, that would mean putting as many zeros as possible on the left. For example, if we have the number starting with as many zeros as possible, then the first digit cannot be zero. Wait, the number can't start with a zero. Therefore, the first digit has to be 1.

So, the first digit is 1. Then, the rest of the digits (excluding the last two zeros) can have zeros and ones, but we need a total of nine 1s. Since the first digit is already 1, we need eight more 1s in the remaining digits before the last two zeros.

So, the structure is: 1 followed by some digits (with eight 1s and some zeros), then ending with 00. So, how can we arrange those digits to make the number as small as possible?

To minimize the number, after the first 1, we want as many zeros as possible. So, let's consider the number as 1 followed by a sequence of digits (let's call this part A) then 00. Part A has eight 1s and some number of zeros. To minimize the entire number, part A should be the smallest possible number with eight 1s and the rest zeros.

So, the question now is: What is the smallest number with eight 1s and some number of zeros? Since we want the smallest number, we need to arrange the digits with the 1s as far to the right as possible. For example, the smallest number with eight 1s is 11111111, which is eight 1s. But if we add zeros, we can have more digits but smaller numbers. Wait, but adding zeros to the left would make the number larger. Wait, no, if we add zeros to the right, it would make the number longer but not necessarily smaller. Wait, but we need to have part A, which is before the final two zeros.

Wait, let me think again. The entire number is: [1][part A][00]. Part A is some digits with eight 1s and some zeros. To minimize the entire number, part A should be as small as possible. How do we make part A as small as possible? For example, part A is a number with eight 1s and some zeros. The minimal number with eight 1s and, say, k zeros is the number where the zeros are as far to the right as possible. Wait, no. For example, the minimal number with eight 1s and zeros is the one where the digits are arranged with as many zeros as possible on the left. Wait, but since part A comes after the first 1, and before the last two zeros.

Let me think of part A as a number. Let's say part A has length m. Then the entire number will have length 1 + m + 2 = m + 3. Wait, but m is the length of part A. However, the number of digits in part A is variable. However, since we need eight 1s in part A, the minimal number of digits in part A is eight (all 1s). If we add zeros to part A, the number of digits increases, but the number itself might be smaller. For example, if part A is 10000000, which is 1 followed by seven zeros, then the total number would be 1 10000000 00, which is 11000000000. But if part A is 11111111, then the total number is 1111111100, which is 11 digits. Wait, comparing 11000000000 (11 digits) and 1111111100 (11 digits). Which is smaller?

Let me write them out:

11000000000 vs 1111111100.

Comparing digit by digit:

First digit: 1 vs 1

Second digit: 1 vs 1

Third digit: 0 vs 1 → Here, 0 is smaller, so 11000000000 is smaller. So, adding zeros in part A can result in a smaller number.

So, in this case, part A is 10000000 (which is 1 followed by seven zeros) making the entire number 11000000000. Wait, but part A has eight 1s? Wait no. Wait part A is supposed to have eight 1s. If I put 1 followed by seven zeros, part A has only one 1. So that's not enough.

Wait, I confused part A. So, part A is the part between the first digit (which is 1) and the last two zeros. So, the total number of 1s in the entire number is nine. Since the first digit is already a 1, part A must contain eight more 1s.

Therefore, part A must have eight 1s and some number of zeros. Therefore, part A's digits are eight 1s and zeros. Therefore, to make part A as small as possible, we need to have the leftmost digits as small as possible. Since part A starts after the first 1, let me think of part A as a number. For example, if part A has eight 1s and some zeros, the minimal number would be the one with the zeros as far to the left as possible.

Wait, but since part A comes after the first digit, and before the last two zeros, we need to arrange part A such that when combined with the first 1 and the last two zeros, the entire number is minimal.

So, for example, if part A is 10000000, which has eight 1s? No, that only has one 1. Wait, part A must have eight 1s. So, part A needs to contain eight 1s and any number of zeros. Therefore, the minimal part A would be the smallest number with eight 1s and some zeros.

For example, the minimal number with eight 1s is 11111111 (eight 1s). If I add a zero to the end, it becomes 111111110, which is nine digits. But that's still a number with eight 1s. But is there a smaller number with eight 1s? Like 101010101... but with eight 1s.

Wait, the minimal number with eight 1s is the one with the smallest number of digits, which is eight. But if we can have more digits with zeros, but the number is smaller? For example, 100000001 is 1 followed by six zeros and a 1, which is nine digits. But comparing to 11111111 (eight digits), 11111111 is smaller.

But if we want to have part A as small as possible, perhaps with more digits? For example, maybe inserting zeros in between the 1s to make the number smaller. For example, 10101010101... Wait, but how many digits?

Wait, let's think in terms of how to create the smallest number with eight 1s. It's actually similar to the problem of finding the minimal number with a certain number of 1s and 0s. The minimal number is the one where the 1s are as far to the right as possible. For example, 1000...000 with eight 1s? No. Wait, no, that would have only one 1. Wait, to have eight 1s, the minimal number is 11111111. However, if you can have leading zeros? No, because part A comes after the first 1.

Wait, maybe part A can have leading zeros? Wait, no. Because the entire number cannot have leading zeros. The first digit is already 1, so part A is between the first 1 and the last two zeros. Therefore, part A can have leading zeros? Wait, no. Because if part A starts with a zero, then the entire number would have a 1 followed by a 0, then more digits. But since part A is between the first 1 and the last two zeros, the first digit is 1, then part A can have zeros. For example, the number could be 1 0 1 0 1 0 ... 00.

Therefore, part A can have zeros in front. Therefore, to make part A as small as possible, we need the minimal number with eight 1s, allowing any number of zeros.

But the minimal number with eight 1s and some zeros is obtained by placing the 1s as far to the right as possible. For example, for eight 1s, the minimal number is 10000000...000 with eight 1s? No. Wait, if you have eight 1s, the smallest number is 11111111. If you put zeros in between, the number becomes longer but might be smaller? Wait, for example, 1010101010... Wait, let me compare two numbers:

- 11111111 (eight 1s, eight digits)
- 100000001 (nine digits, eight 1s)
- 101010101 (nine digits, eight 1s, but actually that has five 1s, not eight)
Wait, no. For eight 1s, the number 1010101010101010... Wait, maybe 100000000000000... with eight 1s. No, it's confusing.

Let me think of it as the problem of finding the minimal number (in value) with exactly eight 1s. The minimal such number is the one with the least number of digits, which is eight digits: 11111111. However, if you can have a longer number with zeros that is smaller in value, like 10000000... But 100000000... would have only one 1. So, to have eight 1s, the minimal number is 11111111. If you have zeros in between, the number becomes longer but maybe has smaller value? For example, let's compare 11111111 (eight digits) vs 101111111 (nine digits). Which is smaller?

101111111 is 1 followed by 0 and seven 1s. Let's compare digit by digit:

First digit: 1 vs 1

Second digit: 1 vs 0 → 0 is smaller, so 101111111 is smaller. Therefore, even though it's longer, it's smaller.

So, in this case, inserting a zero earlier in the number can result in a smaller number. Therefore, the minimal number isn't necessarily the one with the least digits.

Therefore, to find the minimal number with eight 1s, we can think recursively or use some strategy. It's similar to finding the minimal binary number with eight 1s, but in decimal digits. Wait, actually, it's similar to the problem of finding the minimal number made up of 0s and 1s with exactly eight 1s.

This is equivalent to finding the minimal number, which is the one where the 1s are as far to the right as possible. For example, to get the minimal number, we need the leftmost digits to be 0s as much as possible. However, the first digit cannot be 0, so the first digit is 1. Then, the remaining digits should have as many 0s as possible before the next 1.

Wait, but since we need eight 1s, the minimal number would be 10000000... but with eight 1s. Wait, perhaps the minimal number is 100000001... but how?

Alternatively, think of the problem as starting with the first 1, then inserting zeros until we have enough 1s. Wait, perhaps the minimal number is 100000001... but with eight 1s.

Actually, let me think of this as the number being a string of 0s and 1s, starting with 1, and containing eight 1s. To make it as small as possible, we want as many 0s as possible, but we need to have eight 1s. Therefore, the minimal such number is obtained by placing the 1s as far to the right as possible.

For example, if we have eight 1s, the minimal number is 1000...000 with eight 1s. Wait, but how? If you have eight 1s, you need to place them in positions. For example, the minimal number is 100...0001... but with eight 1s. Wait, actually, the minimal number with eight 1s is 11111111 (eight 1s) but that's not helpful. Wait, but inserting zeros can make the number smaller.

Wait, here's a better approach. Think of the problem as finding the minimal number with eight 1s, which is equivalent to finding the minimal number in lexicographical order. Since numbers are compared left to right, the minimal number is the one that has the earliest possible zeros. So, after the first digit (which is 1), we want to have zeros as much as possible.

So, for example, the number would be 1 followed by as many zeros as possible, then the remaining 1s. However, we have to have eight 1s. Since the first digit is already 1, we need seven more 1s.

So, to make the number as small as possible, after the first digit, we want to have zeros until we have to place the remaining 1s.

So, the number would be 1 followed by seven 0s and then seven 1s. So, total digits would be 1 + 7 + 7 = 15 digits? Wait, no. Wait, part A is between the first 1 and the last two zeros. Wait, maybe I need to think again.

Let me clarify:

Entire number structure: [1][part A][00], where part A contains eight 1s and some zeros. So, the entire number will have 1 + length(part A) + 2 digits.

To minimize the entire number, part A needs to be as small as possible. To make part A as small as possible, given that it has eight 1s, we need to place the 1s as far to the right as possible.

Therefore, part A should be a number that starts with as many zeros as possible, then the eight 1s. However, since part A is between the first 1 and the last two zeros, inserting zeros in part A would make the number smaller.

Therefore, the minimal part A is a number with eight 1s, followed by zeros. Wait, but if part A is something like 00000000... but part A cannot start with zeros because the entire number already starts with 1. Wait, part A is after the first 1, so part A can have leading zeros. For example, part A could be 00000000... but then the entire number would be 1 followed by part A (which starts with zeros) then 00.

Wait, but if part A has eight 1s, how can it start with zeros? For example, part A could be 00000000 followed by eight 1s? No, that would have eight 0s and eight 1s, but we need eight 1s. Wait, part A can have any number of 0s and eight 1s.

So, to make part A as small as possible, it should have as many leading zeros as possible. For example, part A is a sequence of zeros followed by eight 1s. However, since we need eight 1s, if we have k leading zeros, then part A is k zeros followed by eight 1s. The total number of digits in part A is k + 8.

However, the entire number is 1 followed by part A followed by 00. So, the entire number would be:

1 (first digit)

Then, k zeros

Then, eight 1s

Then, 00.

So, the entire number is 1 followed by k zeros, then eight 1s, then two zeros.

To make this as small as possible, we need to minimize the value of this number. How does varying k affect the number?

For example, if k=0: the number is 1 11111111 00 → 1111111100 (11 digits)

If k=1: the number is 1 0 11111111 00 → 101111111100 (12 digits)

Wait, but comparing 1111111100 (11 digits) vs 101111111100 (12 digits). Since 11 digits is shorter, it's smaller. Wait, but 12-digit number is larger than an 11-digit number. So, even though the 12-digit number may have a smaller digit in the second position, the total number of digits is longer, making it larger. Therefore, the shorter number is smaller. Therefore, to minimize the entire number, we want the minimal number of digits.

Therefore, k=0 is better. Wait, but let's check with k=0.

Wait, part A with k=0 is just eight 1s. So the entire number is 1 followed by eight 1s and then two zeros: 11111111100. That's 11 digits.

Alternatively, if we have k=1, making part A have one zero and eight 1s, the entire number is 1 0 11111111 00, which is 101111111100, which is 12 digits. But 12 digits is longer, so even if the first digits are smaller, it's still larger.

Therefore, the minimal number is 11111111100. However, is there a way to have a number with fewer digits? Wait, the minimal number of digits is 11, since we have nine 1s and two 0s. Therefore, 11 digits is the minimum.

But is there a number with nine 1s and two 0s, ending with two zeros, which is smaller than 11111111100? For example, by arranging the 1s and 0s in part A in a smarter way, not just putting all 1s first?

For example, suppose part A is not all 1s, but some combination of 1s and 0s. For instance, if part A is something like 101010... with some 1s and zeros. Let's try constructing such a number.

Let me think. If part A has eight 1s and some zeros, maybe we can have more zeros in part A, but the total number of digits is still 11 digits. Wait, if part A is 8 digits (all 1s), then adding the first 1 and the last two zeros makes 11 digits. If part A has more digits, meaning more zeros, then the total number of digits would be more than 11. Therefore, to keep the total digits at 11, part A must be 8 digits (the first digit is 1, then part A is 8 digits, then two zeros). Therefore, part A must be 8 digits with eight 1s, so all 1s. Therefore, the number is 11111111100.

But maybe if we allow part A to have more digits, then the entire number has more than 11 digits, but maybe the number is smaller? For example, let's say part A has nine digits with eight 1s and one 0. Then, the entire number is 1 followed by nine digits (eight 1s and one 0) followed by two zeros. The total digits would be 1 + 9 + 2 = 12.

But is 12-digit number smaller than 11-digit number? No, because an 11-digit number is always smaller than any 12-digit number. Therefore, we can't get a smaller number by increasing the number of digits.

Therefore, the minimal number is indeed 11111111100.

But let me check if there is a number with nine 1s and two 0s (total 11 digits) ending with 00, but arranged differently, which is smaller than 11111111100.

For example, 101111111100 is 12 digits, which is longer. But if I have an 11-digit number with some zeros in the middle and the last two zeros. For example, 11111110100. Let me count the number of 1s:

1,1,1,1,1,1,1,0,1,0,0. That's seven 1s in the first seven digits, then 0, 1, 0, 0. So total of 8 ones? Wait, first seven digits are 1s: that's seven 1s, then another 1 in the ninth digit: total eight 1s. Then where is the ninth 1?

Wait, we need nine 1s. Therefore, if I have 11111110100, that's seven 1s, then a 0, then a 1, then two zeros. So total of 8 ones. Not enough. Therefore, need to have nine 1s. Let me try another arrangement.

Suppose the number is 111111110100. Wait, that's 12 digits. No, let's stick to 11 digits.

Let me try inserting a zero somewhere in the middle. For example: 111111101100. Wait, that's 12 digits. Again, too long.

Wait, for an 11-digit number ending with two zeros, with nine ones. So, the structure is [digits][00], where [digits] has nine 1s and some zeros. Wait, no. Wait, [digits] has nine 1s and some zeros. But since the entire number ends with two zeros, and the first digit is 1. Wait, [digits] is the first 9 digits? No. Wait, total length is 11 digits, last two are zeros. So the first nine digits have nine 1s? No, because the first digit is 1, and then the remaining eight digits can have eight 1s and some zeros.

Wait, actually, the entire number is 11 digits. Last two are zeros, so first nine digits have nine 1s? No, because the first digit is 1, and the remaining eight digits can have eight 1s. Wait, no. The total number of 1s in the entire number is nine. Since the last two digits are zeros, all nine 1s are in the first nine digits. Therefore, the first nine digits are all 1s. Therefore, the only possibility is 11111111100.

Wait, that seems to be the only 11-digit number with nine 1s and two zeros at the end. Therefore, there are no other numbers with nine 1s and two zeros at the end in 11 digits.

But maybe there is a number with more than nine 1s and more than two zeros, but with the number of 1s being a multiple of 9? For example, 18 ones. However, that would require a longer number, which would be larger. Therefore, the minimal number should be 11111111100.

But hold on, the problem says "smallest positive multiple of 450". Maybe there is a smaller number than 11111111100 which is a multiple of 450 and has only 0s and 1s.

Because 11111111100 is a multiple of 450? Let me check.

First, check divisibility by 450. Since it ends with two zeros, it's divisible by 100. Also, since the sum of digits is 9, it's divisible by 9. Therefore, since 450 = 9 * 50, and 50 is 2 * 25. Wait, 450 = 2 * 3² * 5². So, since the number ends with two zeros, it's divisible by 2 and 25. And sum of digits is 9, so divisible by 9. Therefore, it is divisible by 450. Therefore, 11111111100 is a multiple of 450.

But is there a smaller number? For example, numbers with fewer digits? But we concluded that the minimal number of digits is 11. Wait, can there be a number with fewer digits? Let me check.

Let me think. The number must end with two zeros, so it must have at least two zeros. The rest of the digits can be 1s and zeros. Also, the sum of digits must be a multiple of 9.

Suppose there is a number with, say, 10 digits. Let's check if that's possible. If it has 10 digits, ending with two zeros, so the first 8 digits must contain some number of 1s. The total sum of digits is the number of 1s in the first 8 digits. For this to be a multiple of 9, the number of 1s must be 9. But there are only 8 digits in the first part. Therefore, impossible.

What about 11 digits? As discussed, possible.

What about 12 digits? Then, the number could have more 1s. But since we want the smallest number, we need the minimal number of digits. Therefore, 11 digits is the minimum.

But maybe there is an 11-digit number smaller than 11111111100 which also meets the conditions? For example, inserting zeros in the first nine digits. But since the first nine digits must contain nine 1s, you can't insert zeros there. Because inserting a zero would require removing a 1, making the total number of 1s less than nine.

Wait, for example, suppose in the first nine digits, instead of nine 1s, we have eight 1s and one 0, and then have a 1 in the last two digits? But no, the last two digits are zeros. Therefore, all nine 1s must be in the first nine digits. Therefore, the first nine digits must be all 1s. Therefore, the number must be 11111111100.

Therefore, that is the only 11-digit number with nine 1s and two zeros at the end. Hence, it must be the minimal number.

But wait, maybe there's a number with more than nine 1s (like 18) but shorter? Wait, no. Because with more 1s, you need more digits. For example, if you have 18 ones and two zeros, the number is longer than 11 digits.

Alternatively, maybe a number that ends with two zeros but has fewer digits? But as shown earlier, with fewer than 11 digits, it's impossible.

Wait, let me test with 10 digits. Suppose a number with 10 digits, ending with 00. Then the first 8 digits have to have sum of digits = multiple of 9. The maximum sum of digits in first 8 digits is 8. Which is less than 9. Hence, impossible.

For 11 digits, first 9 digits have sum of digits = 9 (since last 2 digits are 0). Hence, possible.

For 12 digits, first 10 digits have sum of digits multiple of 9. For example, sum could be 9 or 18. If sum is 9, then the number would be 12 digits: first 10 digits have 9 ones and one zero, then two zeros. For example, 101111111100. Which is 12 digits. But 12-digit number is larger than 11-digit number. So, 11111111100 is smaller.

But maybe there's a smaller number with 11 digits? Like, if instead of nine 1s followed by two zeros, there's some zeros in the first nine digits, but with the total number of 1s being nine. Wait, but the first nine digits must have nine 1s, because the last two are zeros. Hence, they have to be all 1s.

Wait, another thought. What if the number has more than two zeros, but not at the end? Wait, but the number must be divisible by 25, which requires last two digits to be 00. Therefore, the number must end with 00. Therefore, the other zeros can be anywhere else. However, the total number of 1s must be a multiple of 9.

But if we have more zeros in the first part, but keeping the last two digits as 00.

Wait, for example, if the number is 1010101010100. Let me count the number of digits: 1 0 1 0 1 0 1 0 1 0 1 0 0. That's 13 digits. The sum of digits is 6 ones. Not multiple of 9. Not good.

If we have more ones. For example, 1110000000000. That's 13 digits. Sum of digits is 3. Not multiple of 9.

What if we have 11111111100 (11 digits) vs something else. Let me think. Suppose we have a number like 1001111111100. Wait, that's 13 digits. Again, longer.

Alternatively, is there a way to have an 11-digit number with nine 1s and two zeros, not necessarily all the 1s first? For example, 11111110100. Wait, let's check:

Number: 1 1 1 1 1 1 1 0 1 0 0.

Digits: positions 1-11:

1,1,1,1,1,1,1,0,1,0,0.

Number of 1s: Let's count: first seven 1s, then a 0, then a 1, then two 0s. So total 7 + 1 = 8 ones. Not enough. Need nine.

If I change one of the zeros to a 1: 111111110100. Wait, that's 12 digits.

Alternatively, 111111110100 (12 digits). But that's longer.

Alternatively, what about 11111111100? That's 11 digits.

Wait, perhaps inserting a zero and adjusting somewhere else? Let's try:

Suppose the number is 111111110100. But that's 12 digits.

Wait, maybe there's a way to have a number with nine 1s and two zeros in 11 digits but not all 1s first.

For example: 111111101100. But that's 12 digits.

Wait, no. Wait, 11 digits. Let me try:

Positions 1 to 11:

If I have eight 1s and one 0 in the first nine digits, then the tenth and eleventh digits are 0. So total digits: 11. Number of 1s: 8. Need one more 1. Therefore, if in the first nine digits, we have nine 1s, which is all. Hence, no other way.

Therefore, it's impossible to have an 11-digit number with nine 1s and two zeros at the end unless the first nine digits are all 1s. Therefore, the only such number is 11111111100.

But let me check if 11111111100 is indeed the minimal. For example, is there a smaller number with more 1s? Wait, but more 1s would require longer numbers.

Alternatively, maybe there is a number with nine 1s and two zeros but arranged such that it is smaller. For example, if the number is 101111111100 (12 digits), which is larger than 11111111100.

Alternatively, if there's a number like 110111111100 (12 digits). Still larger.

Alternatively, if there is a number with a different arrangement. Wait, but how?

Wait, maybe numbers with leading 1s, then zeros, then more 1s... but since all the 1s must be in the first nine digits, and the last two digits are zeros, there is no other way.

Therefore, seems like 11111111100 is the answer.

But let me check with another approach. Maybe using BFS or something.

Let me think. Since we need numbers made up of 0s and 1s, ending with 00, and divisible by 450. Let me think about building numbers incrementally and checking divisibility.

But that might take a long time, but maybe for small numbers.

The number must end with 00, so let's consider numbers ending with 00. Let's think about numbers composed of 0s and 1s, ending with 00, and check if they're divisible by 450.

The smallest such numbers would be:

- 100: Divisible by 100, but sum of digits is 1, not divisible by 9. Not divisible by 450.

- 1000: Same, sum of digits 1. No.

- 1100: sum digits 2. No.

- 10000: sum 1.

- 10100: sum 2.

- 11000: sum 2.

- 11100: sum 3.

- 100000: sum 1.

- 100100: sum 2.

- 101000: sum 2.

- 101100: sum 3.

- 110000: sum 2.

- 110100: sum 3.

- 111000: sum 3.

- 111100: sum 4.

- 1000000: sum 1.

Continuing this way, it's going to take too long. However, the sum of digits needs to be 9. So, we need numbers where the sum of digits is 9, and ending with 00.

So, the number is made of 0s and 1s, ends with 00, and sum of digits is 9. Therefore, the number has nine 1s and any number of zeros (at least two for the ending).

The minimal number is the one with the fewest digits, which is nine 1s + two zeros = 11 digits. As we discussed earlier.

Therefore, the minimal number is 11111111100.

But let me verify with another example. Suppose the number is 11111111100. Divided by 450, let me compute 11111111100 / 450.

First, divide by 100: 11111111100 / 100 = 111111111. Then divide by 4.5? Wait, 450 is 4.5 * 100. Wait, 450 = 9 * 50. Let me check divisibility.

Sum of digits: 9, so divisible by 9. Ends with 00, so divisible by 100. Therefore, divisible by 450.

Therefore, yes, 11111111100 is divisible by 450.

But maybe there is a smaller number? Let me think. What if the number has more zeros in the middle? For example, 100111111100. Wait, this is 12 digits. But 11 digits number is smaller.

Alternatively, maybe 101111111100. Still 12 digits.

But what about numbers with more than nine 1s but with more zeros? For example, 11111111111100 (14 digits). But this is bigger.

Alternatively, maybe a number like 10000000000000 (but sum is 1), nope.

Alternatively, is there a number with nine 1s and two zeros, arranged such that it is smaller than 11111111100? For example, 101111111100 (12 digits) is larger.

Alternatively, what if the number is 111111110100 (12 digits). Still larger.

Therefore, it seems that 11111111100 is indeed the smallest.

But just to be thorough, let me check if there's an 11-digit number with nine 1s and two 0s, not all 1s first, but with the zeros placed in such a way that the number is smaller.

For example, suppose the number is 101111111100. Wait, that's 12 digits. No.

If I try to make an 11-digit number with nine 1s and two 0s, where one of the zeros is in the middle. For example:

First digit: 1, then a zero, then eight 1s, then a zero. But the last two digits must be zero. Therefore:

Structure: 1 0 [eight 1s] 0 0.

So, the number would be 10 followed by eight 1s and two zeros: 101111111100. Wait, that's 12 digits again.

Alternatively, if we have two zeros somewhere in the first nine digits. For example:

Number: 1 followed by eight digits with eight 1s and one zero, then two zeros. Wait, total digits would be 1 + 9 + 2 = 12 digits. Again longer.

Alternatively, if we have two zeros in the first nine digits. For example: 1 followed by seven 1s, one 0, one 1, then two zeros. That would be: 1 1 1 1 1 1 1 0 1 0 0. Which is 11 digits. Let's check:

Digits: 1,1,1,1,1,1,1,0,1,0,0.

Number of 1s: 7 + 1 = 8. Not enough. Need nine. Therefore, need to replace one of the zeros with a 1. Then we have 1 followed by seven 1s, one 1, one 0, then two zeros. Wait, that would be eight 1s and one 0 in the first nine digits. So total number of 1s would be eight. Still not nine.

Therefore, need to have nine 1s in the first nine digits. Therefore, the only way is to have all 1s in the first nine digits.

Therefore, the number must be 11111111100.

So, I think that's the answer.

But just to be 100% sure, let me check if there's a multiple of 450 with digits 0 and 1 smaller than 11111111100.

Let me think about numbers with fewer digits.

As we saw earlier, with 10 digits, it's impossible because the sum of digits can't be 9.

What about numbers with more than nine 1s? For example, with 18 ones? But that would require more digits.

Alternatively, maybe a number with nine 1s and more zeros, but shorter? No.

Alternatively, numbers that end with 00, have sum of digits multiple of 9, but with more zeros in between.

Wait, for example, let's think about the number 11111111100. Is there a smaller number? Let me think in terms of decimal numbers.

What is 11111111100? It's 11,111,111,100.

Is there a smaller number composed of 0s and 1s, ending with 00, with sum of digits 9?

For example, 10000000000 (sum 1). No.

100000000100: sum 2.

...

What about 100100100100? Let's check:

Digits: 1,0,0,1,0,0,1,0,0,1,0,0. Sum is 4. Not enough.

How about 111000000000 (sum 3). No.

How about 111100000000 (sum 4).

How about 111110000000 (sum 5).

Continuing, 111111000000 (sum 6), 111111100000 (sum 7), 111111110000 (sum 8), 111111111000 (sum 9). Wait, this is 111111111000, which is 12 digits.

Compare with 11111111100 (11 digits). Which is smaller? The 11-digit number is smaller.

So, 11111111100 is smaller than 111111111000.

Therefore, 11111111100 is smaller.

Alternatively, is there a number like 1010101010100? Let me check the sum.

Digits: 1,0,1,0,1,0,1,0,1,0,1,0,0. Sum is 6. Not enough.

If I add more 1s. For example, 1110000000000 (sum 3). No.

Alternatively, 11111111100.

Therefore, after checking all these possibilities, I can't find a smaller number than 11111111100.

Therefore, I think the answer is 11111111100.

But to confirm, let me check if this number is indeed the minimal.

Another way: think of 450 in terms of multiples. The question is essentially asking for the smallest number consisting of 0s and 1s divisible by 450.

Since 450 = 9 * 50, and the number must have at least two zeros at the end (for divisibility by 25), and the sum of digits divisible by 9.

Therefore, the minimal such number would be the minimal number with sum of digits 9, ending with 00.

Which is 11111111100.

Therefore, I think that's the answer.

**Final Answer**
The smallest positive multiple of 450 whose digits are all zeroes and ones is \boxed{11111111100}.

To solve the problem of finding the **smallest positive multiple of 450** whose **digits are all zeroes and ones**, we proceed by systematically analyzing the conditions and constraints.

---

### Step 1: Prime Factorization of 450

We start with the prime factorization of 450:

$$
450 = 2 \times 3^2 \times 5^2
$$

Thus, the number we are looking for must be divisible by:

- **2** and **5² = 25** → Divisibility by 2 and 25.
- **3² = 9** → Divisibility by 9.

---

### Step 2: Constraints from the Digit Set

The number must consist **only of digits 0 and 1**, so it's made up of these two values exclusively.

---

### Step 3: Divisibility by 2 and 25

- **Divisibility by 2** is automatically satisfied if the number ends in **0** or **1** — but **actually**, more specifically, it must be even. Since the only even digits are **0**, it suffices if the number ends in **0**.
- **Divisibility by 25** requires the number to end in **00** (i.e., the last two digits must both be 0).

So, the number **must end in two zeros**.

---

### Step 4: Divisibility by 9

A number is divisible by 9 **if the sum of its digits is a multiple of 9**. Since the only digits are **0** and **1**, the sum of the digits is **equal to the number of 1s** in the number.

Therefore, the number must have **exactly 9 ones** and **some number of zeroes**, with at least **two zeroes at the end** to satisfy the divisibility by 25.

---

### Step 5: Minimize the Number

To find the **smallest** such number:

- We need at least **nine 1s**.
- We need **two zeroes at the end**.
- The remaining digits (if any) can include **additional zeroes** or more ones.

The **smallest such number** is the one with:

- **9 ones**
- **Exactly two zeroes at the end**
- The rest of the digits (if any) are zeroes
- The number is as small as possible in numeric value

Since the number must have nine 1s, and the minimal way to do this is to place all 9 ones first and then the two zeroes at the end, the number becomes:

$$
11111111100
$$

This number has:

- Nine 1s
- Two trailing zeroes (so divisible by 100 and 25)
- The sum of its digits is **9**, which is divisible by 9
- Thus, it is divisible by **450**

---

### Step 6: Verify Minimality

We check if a smaller number is possible:

- **Fewer digits than 11** would not be possible because the number must contain **at least nine 1s** (for divisibility by 9) plus **two trailing zeroes** (for divisibility by 25), totaling at least **11 digits**.
- Any arrangement that places the ones and zeroes in different orders (e.g., interspersing zeros among the ones) results in a **longer** number or a **larger** one in value.
- There is no way to reduce the number of digits or digits' value without violating the required divisibility and digit constraints.

---

### Final Verification

Let’s confirm:

- $11111111100$ is divisible by 2 (ends in 0) ✅
- $11111111100$ is divisible by 25 (ends in 00) ✅
- $11111111100$ is divisible by 9 (sum of digits = 9) ✅
- It contains only 0s and 1s ✅
- It has the smallest number of digits and the smallest value that satisfies all conditions

---

### Final Answer

$$
\boxed{11111111100}
$$

Solution:
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.