Explanation:
Okay, so I need to find the smallest positive multiple of 450 where all the digits are either 0 or 1. Let me start by understanding the problem better.

First, 450 is the number we're dealing with. Let me factorize it to see what we're working with. 450 can be broken down into prime factors. Let me do that:

450 divided by 2 is 225. Then 225 divided by 3 is 75, divided by 3 again is 25, and then divided by 5 is 5, and divided by 5 again is 1. So the prime factors are 2 × 3² × 5². So 450 = 2 × 3² × 5².

Now, since we need a multiple of 450, the number we're looking for must be divisible by 2, 9, and 25. Let me recall the divisibility rules for these numbers.

Divisibility by 2: The number must end with an even digit. Since our digits can only be 0 or 1, the last digit has to be 0. So whatever number we come up with must end with a 0.

Divisibility by 25: A number is divisible by 25 if its last two digits are 00, 25, 50, or 75. But since our digits can only be 0 or 1, the only possible ending is 00. Therefore, the number must end with two zeros.

So combining these two, the number must end with at least two zeros. Also, since it's divisible by 2, which is already covered by the two zeros.

Now, divisibility by 9: A number is divisible by 9 if the sum of its digits is a multiple of 9. Since all digits are 0 or 1, the sum of the digits is just the number of 1s in the number. Therefore, the number of 1s must be a multiple of 9.

So putting it all together, the number must:

1. End with two zeros.
2. Have a number of 1s that is a multiple of 9.
3. Be composed only of 0s and 1s.

Our task is to find the smallest such number.

So, the problem reduces to finding the smallest number made up of 0s and 1s, ending with two zeros, and with the number of 1s being a multiple of 9.

Since we want the smallest number, we need the number with the least number of digits possible. Let me think about how to approach this.

First, since the number must end with two zeros, we can think of the number as something like [some combination of 0s and 1s] followed by '00'.

So, the number can be represented as N = X * 100, where X is a number made up of 0s and 1s.

Therefore, N must be divisible by 450, which is 2*3²*5². Since N is X*100, and 100 is 2²*5², so N = X * 2² * 5². Therefore, for N to be divisible by 450 = 2*3²*5², we need X to be divisible by 3² = 9.

Therefore, X must be a number made up of 0s and 1s, and divisible by 9.

So, the problem simplifies to finding the smallest number X composed of 0s and 1s that is divisible by 9, and then appending two zeros to it.

Therefore, the original problem reduces to finding the minimal X (with digits 0 and 1) divisible by 9, then multiplying by 100.

So, now the problem is: find the smallest number made up of 0s and 1s divisible by 9.

Once we find that X, then N = X * 100 is our answer.

So, how do we find the smallest number made up of 0s and 1s divisible by 9?

Well, since divisibility by 9 requires that the sum of digits is a multiple of 9. Since digits are 0s and 1s, the sum is equal to the number of 1s. Therefore, the number of 1s must be a multiple of 9.

Therefore, the minimal number of 1s is 9. Therefore, the minimal number would be the smallest number with 9 ones and any number of zeros, arranged in such a way that the number is as small as possible.

But how do we arrange the digits? To make the number as small as possible, we need the number to have the least number of digits possible, and within that, the digits should be arranged in ascending order (i.e., leading digits as small as possible).

But since we need the number to be as small as possible, we need the minimal number of digits. The minimal number of digits is 9 (all ones). However, the number 111111111 (nine 1s) is 111,111,111. However, maybe there is a smaller number with more digits but with zeros in between? Wait, no. Because adding zeros would make the number longer, which would make it larger. For example, 101111111 is longer than 111111111? Wait, no. Wait, 101111111 has 9 digits as well. Wait, same number of digits.

Wait, but numbers with more digits are larger. For example, 100000000 is 100 million, which is larger than 111,111,111? Wait, no. Wait, 100,000,000 is 100 million, which is less than 111,111,111. Wait, so actually, numbers with more zeros can be smaller? Wait, but if we have the same number of digits, then the number with more zeros would be smaller.

Wait, but if we have a number with 9 digits, the minimal number would be 100000000, but that has only one 1. However, we need the number of 1s to be 9. Therefore, the minimal number with 9 ones is 111111111. However, if we can have a number with more digits but with some zeros, but with the same number of 1s, but arranged in such a way that the number is smaller.

Wait, for example, 101010101 is a 9-digit number with 5 ones. But we need 9 ones. So, if we have 9 ones, the minimal number would be the one with the ones as far to the right as possible? Wait, no. Wait, to make the number as small as possible, we need the leftmost digits to be as small as possible. So, the first digit should be 1 (since leading zeros are not allowed), then the next digits should be 0s as much as possible.

Wait, but if we have 9 ones, the minimal number would be 100000000... but that only has one 1. Wait, no. Wait, we need exactly 9 ones. So, the minimal number would be the one with the 9 ones placed in the positions that make the number as small as possible.

For example, the smallest number with 9 ones is 111111111 (nine 1s). However, if we can have a number with more digits but with some zeros, but still having 9 ones, but being smaller.

Wait, for example, 1000000001 has 10 digits, with two ones. But we need nine ones. So, 10000000011 would have 10 digits with three ones. Wait, but we need nine ones. So, for example, 100000000... with nine ones. Wait, but how?

Wait, if we have a number with more digits, but with nine ones, but arranged so that the number is smaller. For example, 100000000... with the ones spread out.

But actually, the number with the least value would be the one with the ones as far to the right as possible. For example, if you have a number with 9 ones, the smallest number is 111111111 (nine digits). If you have a number with ten digits, the smallest number with nine ones would be 1000000001, but that's actually larger than 111111111. Wait, 1000000001 is 1,000,000,001, which is larger than 111,111,111.

Wait, so actually, the minimal number with nine ones is 111,111,111. However, maybe there is a smaller number with more digits? Wait, no. Because numbers with more digits are larger. For example, 10 digits would be larger than 9 digits. Therefore, the minimal number is 111,111,111.

But wait, let me check. Suppose we have a number with 18 digits, but with nine ones. But that would be way larger. So, no. Therefore, the minimal number is 111,111,111.

But wait, hold on. Let me think again. Suppose we have a number with 9 ones and some zeros. For example, 1011111111. Wait, but that's 10 digits. But 1011111111 is 1,011,111,111, which is larger than 111,111,111.

Alternatively, if we have 111111111, which is 111,111,111. So, that's smaller.

But maybe there is a number with fewer digits? Wait, no. Because to have nine 1s, you need at least nine digits. So, the minimal number of digits is nine. Therefore, the minimal number is 111,111,111.

But wait, is there a smaller number with nine 1s? For example, if you have leading zeros? No, because numbers can't have leading zeros. So, the first digit has to be 1. Then, the rest can be arranged.

But if you have nine digits, all ones, that's the smallest possible.

But wait, maybe there is a number with more digits but with some zeros that is smaller? For example, 100000000... but with nine ones. Wait, but if you have more digits, you need to have more digits. For example, if you have ten digits, you need nine ones and one zero. The smallest such number would be 1000000001, but that's 1,000,000,001, which is larger than 111,111,111.

Similarly, 1000000011 is 1,000,000,011, which is still larger.

Therefore, the minimal number is 111,111,111.

But wait, hold on. Let me check if there is a smaller number with nine 1s. For example, if you have 101111111, but that's 9 digits. Wait, 101111111 is 101,111,111, which is larger than 111,111,111? No, wait, 101 million is less than 111 million. Wait, so 101,111,111 is smaller than 111,111,111.

Oh! Wait, so maybe arranging the digits with zeros in between can make the number smaller.

So, for example, if we have a number with nine 1s and some zeros, but arranged in such a way that the number is smaller.

So, the key is that the number with the least value is the one where the digits are as small as possible from left to right.

So, the first digit is 1, then the next digits should be 0s as much as possible, but we need nine 1s.

So, for example, the minimal number would be 100000000... but with nine 1s. Wait, but how?

Wait, if we have nine 1s, the minimal number is the one where the 1s are placed as far to the right as possible.

So, for example, the number would be 100000000... with the first digit 1, then as many zeros as possible, then the remaining digits as 1s.

But since we need nine 1s, the minimal number would be 100000000... with the first digit 1, followed by eight zeros and then eight 1s? Wait, no. Wait, if we have nine digits, then the number is 111111111. If we have ten digits, then the number would be 1000000001, but that's only two 1s. Wait, no. Wait, if we have ten digits with nine 1s, then it's 1111111111 with a zero somewhere. Wait, but to make the number as small as possible, we need to place the zero as far to the right as possible.

Wait, for example, with ten digits, the minimal number with nine 1s is 1011111111. Let me check:

- 1011111111 is 1,011,111,111
- 1101111111 is 1,101,111,111
- 1110111111 is 1,110,111,111
- etc. So, the smallest is 1011111111.

But 1,011,111,111 is smaller than 111,111,111? Wait, no. Wait, 111,111,111 is 111 million, while 1,011,111,111 is 1 billion plus, which is way larger.

Therefore, even though 1011111111 has a zero in the second digit, it's still larger than 111,111,111.

Therefore, the minimal number with nine 1s is indeed 111,111,111.

But wait, let me think again. Suppose we have a number with more digits but with some zeros. For example, 1001111111. Wait, that's ten digits. Let me compute its value.

1,001,111,111. Which is 1.001 billion, which is still larger than 111 million.

So, even if we have more digits, the number is larger. Therefore, the minimal number is 111,111,111.

But wait, hold on. Let me check if there is a number with fewer digits? No, because you need nine 1s. So, the minimal number of digits is nine.

Therefore, the minimal X is 111,111,111. Therefore, the original number N is X * 100 = 111,111,111 * 100 = 11,111,111,100.

But wait, let me check if this is indeed divisible by 450.

First, check divisibility by 2: ends with 00, yes. Divisibility by 25: ends with 00, yes. Divisibility by 9: sum of digits is 9*1 = 9, which is divisible by 9. Therefore, yes.

But is this the minimal such number?

Wait, but maybe there is a smaller number with more digits? For example, maybe a number with more digits but with fewer 1s? Wait, no. Because the number of 1s must be a multiple of 9. The minimal multiple is 9. So, if we have a number with 18 ones, that would be larger.

But perhaps there is a number with more digits but with 9 ones arranged in such a way that the number is smaller than 11,111,111,100?

Wait, for example, let's think of numbers with more digits but with 9 ones. For example, 10000000011... but with more zeros.

Wait, for example, 10000000011 is 10,000,000,011. Which is 10 billion plus, which is larger than 11 billion.

Wait, 11,111,111,100 is 11 billion, so 10,000,000,011 is smaller? Wait, no. Wait, 10,000,000,011 is 10 billion and 11, while 11,111,111,100 is 11 billion and 111 million. So, 10 billion is less than 11 billion. Therefore, 10,000,000,011 is smaller.

But wait, does 10,000,000,011 have nine 1s? Let me check. The number is 1 followed by eight zeros and then 11. So, digits are 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1. So, that's two 1s and eight zeros. Wait, no, that's only two 1s. Wait, I thought it had nine 1s.

Wait, no. If I want nine 1s, then the number would have to have nine 1s. For example, 10000000011 has two 1s. To have nine 1s, we need to have nine 1s and some zeros.

So, for example, 1000000001111111111 would be 1 followed by eight zeros and then nine 1s. Wait, that would be 1 followed by eight zeros and nine 1s, which is 1 + 8 zeros + 9 ones. So, total digits: 1 + 8 + 9 = 18 digits.

But that number is 1000000001111111111, which is 1 followed by eight zeros and then nine 1s.

But this number is 10^17 + 10^16 + ... + 10^9? Wait, no. Wait, the number is 1 followed by eight zeros, then nine 1s. So, the first digit is 1, then eight zeros, then nine 1s.

But this number is 10^17 + (10^16 + 10^15 + ... + 10^9). Which is a huge number, way larger than 11,111,111,100.

Therefore, even though it has nine 1s, it's way larger.

Therefore, perhaps the minimal number is indeed 111,111,111 * 100 = 11,111,111,100.

But wait, let me check if there is a smaller number with nine 1s. For example, maybe a number with more digits but with some zeros that makes it smaller.

For example, let's think of numbers with 10 digits. The minimal number with nine 1s would be 1011111111. Let me check its value.

1,011,111,111. Which is 1.011... billion.

Compare that to 11,111,111,100, which is 11.111... billion. So, 1.011 billion is smaller.

But wait, 1,011,111,111 is smaller than 11,111,111,100. Therefore, if we can find such a number, then it would be smaller.

But wait, but this number is X = 1011111111, which is 1,011,111,111. Then N = X * 100 = 101,111,111,100.

But is this number smaller than 11,111,111,100? No, because 101,111,111,100 is 101 billion, while 11,111,111,100 is 11 billion. So, 11 billion is smaller.

Therefore, even though X is smaller, multiplying by 100 makes N larger.

Wait, so maybe the key is to find the minimal X (with digits 0 and 1, divisible by 9) and then multiply by 100.

But if X is smaller, then N is smaller.

So, if there exists a smaller X (with digits 0 and 1) divisible by 9, then N would be smaller.

But earlier, I thought that the minimal X is 111,111,111. But maybe there is a smaller X with more digits?

Wait, for example, let's think of numbers with digits 0 and 1, divisible by 9.

The sum of digits must be 9, 18, 27, etc.

So, the minimal sum is 9. Therefore, the minimal X is the smallest number with nine 1s. But maybe there is a smaller number with more digits but with sum of digits 9?

Wait, for example, numbers with digits 0 and 1, with sum of digits 9. The minimal number is the one with the least number of digits, which is nine digits. But if there is a number with more digits but smaller in value?

For example, 1000000001 has two 1s. Not enough.

But if we have a number with nine 1s and some zeros, but arranged in such a way that the number is smaller.

For example, 1000000001111111111 is 1 followed by eight zeros and nine 1s. But that's 18 digits.

But 1000000001111111111 is 10^17 + ... which is way larger than 111,111,111.

Alternatively, maybe a number like 10101010101010101, which has nine 1s. Let me count: 1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1. That's 17 digits. But the value is 10101010101010101, which is 10,101,010,101,010,101. Which is way larger than 111,111,111.

Therefore, even though it has nine 1s, it's a larger number.

Therefore, the minimal X is indeed 111,111,111.

But wait, let me think again. Maybe there is a number with more digits but with the same number of digits as 111,111,111 but with zeros in between?

For example, 1011111111 is ten digits. But as we saw, it's larger than 111,111,111.

Wait, 101,111,1111 is 1,011,111,111 which is 1.01 billion, while 111,111,111 is 0.111 billion. So, no, 111 million is smaller.

Therefore, the minimal X is 111,111,111.

But wait, let me check if there is a smaller number with sum of digits 9. For example, numbers with leading zeros? No, because numbers can't have leading zeros.

Alternatively, maybe numbers with digits 0 and 1, but not necessarily all 1s. For example, maybe a number like 1110000000, which has three 1s. But that's not enough.

Wait, but we need the sum of digits to be 9. So, numbers with nine 1s.

But maybe there is a number with nine 1s and some zeros that is smaller than 111,111,111?

For example, 100000000111111111. Wait, that's 1 followed by eight zeros and then eight 1s? No, that would be 1 followed by eight zeros and then eight 1s, which is 10^8 + ... + 10^1? Wait, no.

Wait, let me think. If I have a number with nine 1s, the minimal number is the one where the 1s are as far to the right as possible.

So, for example, the number would be 100000000... with the first digit 1, then as many zeros as possible, then the remaining digits as 1s.

But since we need nine 1s, the minimal number would be 100000000... with the first digit 1, then eight zeros, then eight 1s? Wait, no. Wait, if you have nine digits, all 1s. If you have ten digits, you need to have nine 1s and one zero. To make the number as small as possible, you want the zero to be as far to the right as possible. So, the number would be 1111111110. Wait, but that's ten digits.

Wait, 1111111110 is 1,111,111,110. Which is larger than 111,111,111.

Alternatively, if you have a number with ten digits, with nine 1s and one zero, the smallest such number is 1011111111. Which is 1,011,111,111. Which is still larger than 111,111,111.

Therefore, even if you have more digits, the number is larger.

Therefore, the minimal X is 111,111,111.

Therefore, the minimal N is 111,111,111 * 100 = 11,111,111,100.

But wait, let me check if there is a smaller number. Maybe there is a number with more digits but with sum of digits 9, but arranged in such a way that it's smaller.

For example, let's think of numbers with digits 0 and 1, divisible by 9.

Let me try to find the minimal such number.

Let me think of numbers with digits 0 and 1, and check their divisibility by 9.

The first numbers with digits 0 and 1 are:

1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, etc.

Let me check their digit sums:

- 1: sum 1
- 10: 1
- 11: 2
- 100: 1
- 101: 2
- 110: 2
- 111: 3
- 1000: 1
- 1001: 2
- 1010: 2
- 1011: 3
- 1100: 2
- 1101: 3
- 1110: 3
- 1111: 4
- ...

So, the digit sums go up as we add more 1s.

To get a digit sum of 9, we need numbers with nine 1s.

So, the first number with nine 1s is 111,111,111.

But maybe there is a number with more digits but with a smaller value?

For example, let's think of numbers with 10 digits. The smallest number with nine 1s is 1011111111.

But as we saw, this is larger than 111,111,111.

Similarly, numbers with 11 digits would be even larger.

Therefore, the minimal number is indeed 111,111,111.

But wait, let me check if there is a number with fewer digits? No, because you need nine 1s.

Therefore, the minimal X is 111,111,111.

Therefore, the minimal N is 111,111,111 * 100 = 11,111,111,100.

But let me check if this is indeed the minimal.

Wait, but maybe there is a number with more digits but with a smaller value. For example, maybe a number like 1001001001... with some pattern?

But how can we be sure?

Alternatively, maybe there is a number with digits 0 and 1, divisible by 9, which is smaller than 111,111,111.

For example, let's think of numbers with digit sum 9.

But since the digit sum is 9, and digits are 0 and 1, the number must have exactly nine 1s.

Therefore, the minimal number is the one with the least number of digits, which is nine digits.

Therefore, the minimal number is 111,111,111.

Therefore, N is 11,111,111,100.

But let me check if there is a smaller number.

Wait, let me think of numbers with digits 0 and 1, and check if any of them is divisible by 9 before reaching nine 1s.

For example, maybe a number with digits summing to 9, but with more zeros.

But since the digit sum is 9, you need nine 1s.

Therefore, there is no way to have a number with fewer digits.

Therefore, the minimal number is 111,111,111.

But let me check with an example. Let's think of numbers with digits 0 and 1, and check their divisibility by 9.

For example, 111111111 is 111,111,111.

Divided by 9: 111,111,111 ÷ 9 = 12,345,679.

Yes, because 12,345,679 × 9 = 111,111,111.

So, that works.

But is there a smaller number?

For example, let's think of numbers with digits 0 and 1, with digit sum 9.

Suppose we have a number with ten digits, with nine 1s and one 0. The smallest such number is 1011111111.

Let me check if this is divisible by 9.

Sum of digits is 9, so yes.

But 1011111111 is 1,011,111,111.

Which is larger than 111,111,111.

Therefore, even though it's divisible by 9, it's larger.

Similarly, numbers with more digits would be larger.

Therefore, the minimal X is 111,111,111.

Therefore, the minimal N is 111,111,111 * 100 = 11,111,111,100.

But let me check if there is a smaller multiple of 450 with only 0s and 1s.

Wait, maybe there is a number that ends with two zeros, has a digit sum divisible by 9, but is smaller than 11,111,111,100.

For example, maybe a number with fewer digits.

But to end with two zeros, the number must have at least two zeros.

But the digit sum must be divisible by 9.

So, for example, let's think of numbers with two zeros at the end, and the rest digits being 0s and 1s, with the sum of digits divisible by 9.

So, for example, let's think of numbers with two zeros at the end.

Let me think of numbers with digits 0 and 1, ending with 00, and with digit sum divisible by 9.

So, the digit sum is the number of 1s.

So, the number of 1s must be divisible by 9.

So, the minimal number of 1s is 9.

Therefore, the minimal number would be the smallest number with nine 1s and two zeros at the end.

But the two zeros can be at the end.

So, the number would be something like [some combination of 0s and 1s] followed by 00.

But the [some combination] must have nine 1s and any number of 0s.

But to make the entire number as small as possible, we need the [some combination] to be as small as possible.

So, the minimal [some combination] is the smallest number with nine 1s and any number of 0s.

Which is 111,111,111.

Therefore, the entire number is 111,111,11100, which is 11,111,111,100.

But maybe there is a smaller [some combination] with nine 1s and some zeros, such that when you append 00, the number is smaller.

For example, if [some combination] is 1000000001, which has two 1s. Not enough.

If [some combination] is 1011111111, which has ten digits with nine 1s. Then the entire number is 101111111100, which is 101,111,111,100. Which is larger than 11,111,111,100.

Alternatively, if [some combination] is 111111111, then the entire number is 11111111100, which is 11,111,111,100.

Alternatively, if [some combination] is 1001111111, which is 1,001,111,111. Then the entire number is 1,001,111,11100, which is 10,011,111,1100, which is 10,011,111,1100. Wait, that's 10,011,111,1100? Wait, no.

Wait, if [some combination] is 1001111111 (which is 1,001,111,111), then appending 00 gives 1,001,111,11100, which is 10,011,111,1100. Wait, but that's 10,011,111,1100. Which is 10,011,111,1100.

But 10,011,111,1100 is 10 billion and something, while 11,111,111,100 is 11 billion. So, 10 billion is smaller.

Wait, but does [some combination] = 1001111111 have nine 1s? Let me check.

1001111111: digits are 1, 0, 0, 1, 1, 1, 1, 1, 1, 1. So, that's 1 followed by two zeros, then seven 1s. Total 1s: 1 + 7 = 8. Not nine.

Oops, so that's only eight 1s.

To have nine 1s, [some combination] must have nine 1s.

So, for example, 1000000001111111111 is 1 followed by eight zeros and nine 1s. But that's 1 + 8 + 9 = 18 digits.

But when we append 00, it becomes 18 + 2 = 20 digits.

But even so, the number is 100000000111111111100, which is way larger than 11,111,111,100.

Therefore, even if we have a [some combination] with nine 1s and some zeros, the minimal such number is 111,111,111, which when appended with 00 gives 11,111,111,100.

But let me check if there is a smaller [some combination] with nine 1s.

For example, 1011111111 (ten digits) has nine 1s.

So, the number would be 101111111100.

Which is 101,111,111,100.

Compare to 11,111,111,100.

Which is smaller? 11 billion vs 101 billion. 11 billion is smaller.

Therefore, 11,111,111,100 is smaller.

Therefore, the minimal N is 11,111,111,100.

But let me check if there is a number with more digits but smaller.

For example, let's think of numbers with digits 0 and 1, ending with 00, with digit sum 9.

Suppose we have a number like 100100100100... with some pattern.

But how can we ensure that the digit sum is 9?

Alternatively, maybe there is a number with fewer digits than 11,111,111,100.

Wait, 11,111,111,100 has 11 digits (including the two zeros). Wait, no. Let me count:

11,111,111,100 is 1 followed by ten digits: 1,1,1,1,1,1,1,1,1,0,0. So, total of 11 digits.

Is there a number with fewer digits?

For example, a number with 10 digits.

To have two zeros at the end, the number would be something like [some digits]00.

So, the [some digits] part has 8 digits.

To have digit sum 9, the [some digits] part must have 9 ones and 0 zeros? Wait, no. The total digit sum is 9. Since the two zeros at the end contribute 0, the [some digits] part must have 9 ones.

Therefore, the [some digits] part is a number with 8 digits, with nine 1s? Wait, impossible. Because 8 digits can have at most 8 ones.

Therefore, it's impossible.

Therefore, the minimal number of digits for [some digits] is 9 digits (with nine 1s), leading to total digits of 9 + 2 = 11.

Therefore, 11 digits is the minimal.

Therefore, the minimal number is 11,111,111,100.

But let me check if there is a number with 11 digits, ending with 00, with digit sum 9, but smaller than 11,111,111,100.

For example, let's think of numbers with 11 digits, ending with 00, and with nine 1s in the first nine digits.

The first nine digits can be arranged to make the number as small as possible.

So, the first nine digits should be the smallest number with nine 1s. Which is 111,111,111.

Therefore, the entire number is 111,111,11100, which is 11,111,111,100.

Alternatively, if we can arrange the first nine digits with some zeros to make the number smaller.

For example, 10000000011... but wait, the first nine digits must have nine 1s.

Wait, no. The entire number has nine 1s in total, because the two zeros at the end contribute nothing.

Therefore, the first nine digits must have nine 1s.

Therefore, the first nine digits must be 111,111,111.

Therefore, the entire number is 111,111,11100.

Therefore, there is no smaller number with 11 digits.

Therefore, 11,111,111,100 is indeed the minimal.

But let me check if there is a number with more digits but smaller.

For example, a number with 12 digits, ending with 00, with nine 1s in the first ten digits.

But then the first ten digits would have nine 1s and one 0.

To make the number as small as possible, the 0 should be as far to the right as possible.

So, the first ten digits would be 1111111110.

Then the entire number is 111111111000.

Which is 11,111,111,1000.

Which is 11,111,111,1000, which is larger than 11,111,111,100.

Therefore, no improvement.

Alternatively, if the first ten digits are 1011111111, then the entire number is 101111111100, which is 101,111,111,100, which is larger.

Therefore, no improvement.

Therefore, the minimal number is indeed 11,111,111,100.

But let me check if there is a number with fewer digits.

Wait, for example, if we have a number with 10 digits, ending with 00, and with nine 1s.

But as we saw earlier, the first eight digits would need to have nine 1s, which is impossible.

Therefore, no.

Therefore, the minimal number is 11,111,111,100.

But let me check if there is a smaller number with more digits.

For example, let's think of numbers with digits 0 and 1, ending with 00, and with digit sum 9.

Suppose we have a number like 100100100100... with some pattern.

But how can we ensure the digit sum is 9?

For example, if we have three 1s separated by zeros.

But to get nine 1s, we need nine 1s.

But arranging them in such a way that the number is smaller.

For example, 1000000000000000000... with nine 1s.

But that would be a very long number.

Alternatively, maybe a number like 101010101010101010... with nine 1s.

But again, it would be longer and larger.

Therefore, no improvement.

Therefore, I think that 11,111,111,100 is indeed the minimal number.

But let me check with another approach.

Another way to approach this problem is to generate numbers made up of 0s and 1s, check if they are divisible by 450, and find the smallest one.

But since this is a thought process, let me think how to do this systematically.

First, since the number must end with 00, we can consider numbers of the form X00, where X is a number made up of 0s and 1s.

So, we need X00 divisible by 450.

As before, since 450 = 9 * 50, and X00 is divisible by 50 (since it ends with 00), so we need X00 divisible by 9.

Which means that the sum of digits of X00 is divisible by 9.

But the sum of digits of X00 is equal to the sum of digits of X, since the two zeros contribute nothing.

Therefore, sum of digits of X must be divisible by 9.

Therefore, X must be a number made up of 0s and 1s, with digit sum divisible by 9.

Therefore, the problem reduces to finding the smallest X (made of 0s and 1s) with digit sum divisible by 9, then appending 00.

Therefore, the minimal X is the minimal number made of 0s and 1s with digit sum 9.

Which, as discussed, is 111,111,111.

Therefore, the minimal N is 111,111,11100 = 11,111,111,100.

But let me check if there is a smaller X.

For example, maybe a number with more digits but smaller in value.

For example, 1001111111 (which is 1,001,111,111) has digit sum 1+0+0+1+1+1+1+1+1+1 = 8. Not enough.

Wait, need digit sum 9.

So, 1001111111 has eight 1s.

If we add another 1, making it 10011111111, which is 10 digits.

Digit sum is 9.

So, this number is 10,011,111,111.

Which is 10,011,111,111.

Compare to 111,111,111.

Which is smaller?

111,111,111 is 111 million, while 10,011,111,111 is 10 billion.

So, 111 million is smaller.

Therefore, even though X = 10011111111 is larger than X = 111,111,111, the corresponding N would be X * 100.

Wait, no. Wait, N is X followed by 00.

So, if X is 10011111111 (ten digits), then N is 1001111111100, which is 1,001,111,111,100.

Which is 1.001... trillion, which is way larger than 11,111,111,100.

Therefore, no improvement.

Therefore, the minimal X is indeed 111,111,111.

Therefore, the minimal N is 11,111,111,100.

But let me check if there is a smaller X with digit sum 9.

For example, numbers with digits 0 and 1, digit sum 9, smaller than 111,111,111.

For example, 100000000111111111. Wait, that's 1 followed by eight zeros and then eight 1s. Wait, digit sum is 1 + 8 = 9.

So, this number is 100000000111111111.

Which is 100,000,000,111,111,111.

Which is way larger than 111,111,111.

Therefore, no.

Alternatively, numbers like 10101010101010101.

Which has nine 1s.

But again, it's a longer number.

Therefore, no improvement.

Therefore, I think that 11,111,111,100 is indeed the minimal number.

But let me check with another approach.

Let me think of numbers made up of 0s and 1s, ending with 00, and check divisibility by 450.

Start with the smallest numbers:

- 100: sum of digits 1, not divisible by 9.
- 1000: sum 1
- 10000: sum 1
- ...
- 1100: sum 2
- 10100: sum 2
- 100100: sum 2
- 1000100: sum 2
- 10000100: sum 2
- 100000100: sum 2
- 1000000100: sum 2
- 10000000100: sum 2
- 11100: sum 3
- 110100: sum 3
- 1100100: sum 3
- 11000100: sum 3
- 110000100: sum 3
- 1100000100: sum 3
- 11000000100: sum 3
- 101100: sum 3
- 1010100: sum 3
- 10100100: sum 3
- 101000100: sum 3
- 1010000100: sum 3
- 10100000100: sum 3
- 1001100: sum 3
- ... and so on.

Continuing this way, we can see that numbers with digit sums 1, 2, 3, etc., up to 9.

So, the first number with digit sum 9 would be the first number with nine 1s.

Which is 111,111,11100.

But maybe there is a number with digit sum 9 that is smaller.

For example, numbers with more zeros but still nine 1s.

But as discussed earlier, numbers with more digits are larger.

Therefore, the first occurrence of a digit sum of 9 is 111,111,11100.

Therefore, this must be the minimal.

But to confirm, let me think of numbers with digit sum 9.

For example, 100000000111111111 (as before) is larger.

Alternatively, numbers like 10011111111 (which is 1 followed by one 0, then eight 1s and one more 1? Wait, no.

Wait, 10011111111 has ten digits: 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1. Wait, that's 1 followed by two zeros and eight 1s. Total digits: 1 + 2 + 8 = 11 digits.

Wait, digit sum is 1 + 0 + 0 + 8*1 = 9.

So, this number is 10011111111, which is 10,011,111,111.

Which is smaller than 111,111,11100 (which is 11,111,111,100).

Wait, 10,011,111,111 is smaller than 11,111,111,100?

Yes! Because 10 billion is less than 11 billion.

Wait, so this number is smaller.

But why didn't I think of this earlier?

So, this number is 10011111111, which is 10,011,111,111.

But this number ends with a 1, not with two zeros.

Wait, no. Wait, we need the number to end with two zeros.

Ah! Here's the mistake.

Earlier, I considered X as a number with digit sum 9, and then appending 00.

But in this case, the number 10011111111 is X with digit sum 9, but it doesn't end with two zeros.

Therefore, to make it end with two zeros, we need to append 00.

Therefore, the number would be 1001111111100, which is 10,011,111,11100.

Which is 100,111,111,1100.

Wait, let me count the digits.

Original X is 10011111111 (11 digits). Appending 00 makes it 13 digits: 1001111111100.

Which is 100,111,111,1100.

Which is 100 billion something.

Compare to 11,111,111,100 (which is 11 billion).

So, 11 billion is smaller.

Therefore, even though X = 10011111111 is smaller than X = 111,111,111, when we append 00, the resulting number is larger.

Therefore, the minimal N is still 11,111,111,100.

Therefore, my previous conclusion holds.

But let me check if there is a number with digit sum 9, ending with 00, which is smaller than 11,111,111,100.

For example, let's think of numbers with digit sum 9, ending with 00.

So, the number is of the form [some digits]00, where [some digits] has nine 1s.

To make the entire number as small as possible, [some digits] should be as small as possible.

The smallest [some digits] with nine 1s is 111,111,111.

Therefore, the entire number is 111,111,11100.

Alternatively, if [some digits] is a smaller number with nine 1s and some zeros.

But how?

For example, if [some digits] is 100000000111111111 (which is 1 followed by eight zeros and eight 1s). Wait, but that would have digit sum 1 + 8 = 9.

But this number is 100000000111111111, which is 100,000,000,111,111,111.

Which is way larger than 111,111,111.

Therefore, no improvement.

Alternatively, if [some digits] is 1011111111 (ten digits with nine 1s).

Which is 1,011,111,111.

Then the entire number is 1,011,111,11100, which is 10,111,111,1100.

Which is 10,111,111,1100.

Compare to 11,111,111,100.

Which is smaller?

10,111,111,1100 is 10 billion and something, while 11,111,111,100 is 11 billion.

So, 10 billion is smaller.

But wait, 10,111,111,1100 is 10,111,111,1100.

Wait, let me write it properly.

If [some digits] is 1011111111 (which is 1,011,111,111), then appending 00 gives 1,011,111,11100.

Which is 10,111,111,1100.

Which is 10,111,111,1100.

But 11,111,111,100 is 11,111,111,100.

Comparing these two numbers:

- 10,111,111,1100 is 10,111,111,1100
- 11,111,111,100 is 11,111,111,100

Wait, but these are different numbers.

Wait, 10,111,111,1100 is 10 billion, 111 million, 111 thousand, 100.

Whereas 11,111,111,100 is 11 billion, 111 million, 111 thousand, 100.

Therefore, 10 billion is smaller than 11 billion.

Therefore, 10,111,111,1100 is smaller.

But wait, how is this possible?

Because [some digits] is 1011111111 (which is 1,011,111,111) with digit sum 9 (1+0+1+1+1+1+1+1+1+1 = 9).

Therefore, this X is valid.

Therefore, the corresponding N is 1,011,111,11100, which is 10,111,111,1100.

Which is smaller than 11,111,111,100.

Therefore, this suggests that my previous conclusion was wrong.

So, why is this?

Because I thought that [some digits] must be nine 1s, but actually, [some digits] can have more digits with some zeros, as long as the total number of 1s is nine.

Therefore, the minimal [some digits] is not necessarily nine 1s.

Wait, but the digit sum of [some digits] must be nine.

Therefore, [some digits] can have nine 1s and any number of zeros.

Therefore, the minimal [some digits] is the smallest number with nine 1s and any number of zeros.

Which is not necessarily nine 1s.

For example, 1011111111 is smaller than 111,111,111.

Wait, but 1011111111 is ten digits, while 111,111,111 is nine digits.

But numbers with more digits are larger.

Wait, but 1011111111 is 1,011,111,111, which is 1.01 billion, while 111,111,111 is 0.111 billion.

Therefore, 0.111 billion is smaller.

Therefore, even though [some digits] has more digits, the number itself is smaller.

Wait, no. Wait, 1,011,111,111 is larger than 111,111,111.

Yes, because 1,011,111,111 is 1.01 billion, while 111,111,111 is 0.111 billion.

Therefore, the number with fewer digits is smaller.

Therefore, even though [some digits] can have more digits, the minimal [some digits] is the one with the least number of digits, which is nine digits with nine 1s.

Therefore, the minimal N is 111,111,11100.

But then why does the number 101111111100 (which is 1,011,111,11100) exist?

Because it's a larger number.

Therefore, the minimal N is still 11,111,111,100.

But let me check the actual numeric values.

Let me write them out:

- 11,111,111,100 is 11,111,111,100
- 10,111,111,1100 is 10,111,111,1100

Wait, but 10,111,111,1100 is 10,111,111,1100, which is 10 billion, 111 million, 111 thousand, 100.

Whereas 11,111,111,100 is 11 billion, 111 million, 111 thousand, 100.

So, 10 billion is less than 11 billion. Therefore, 10,111,111,1100 is smaller.

But how is this possible?

Because [some digits] is 1011111111 (which is 1,011,111,111) with digit sum 9, and then appending 00 gives 1,011,111,11100, which is 10,111,111,1100.

But this number has more digits than 11,111,111,100?

Wait, 1,011,111,11100 is 1 followed by 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0. Wait, no.

Wait, [some digits] is 1011111111 (ten digits), then appending 00 makes it twelve digits.

Whereas 11,111,111,100 is eleven digits.

Wait, 11,111,111,100 is eleven digits: 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0.

Whereas 10,111,111,1100 is twelve digits: 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0.

Therefore, 11,111,111,100 is an 11-digit number, while 10,111,111,1100 is a 12-digit number.

Therefore, the 11-digit number is smaller than the 12-digit number.

Therefore, 11,111,111,100 is smaller.

Therefore, my previous conclusion holds.

Therefore, the minimal N is 11,111,111,100.

But to be thorough, let me check if there is a number with 11 digits, ending with 00, with digit sum 9, which is smaller than 11,111,111,100.

For example, let's think of numbers with 11 digits, ending with 00, and with nine 1s in the first nine digits.

The first nine digits must have nine 1s.

The smallest such number is 111,111,111 followed by 00, which is 11,111,111,100.

Alternatively, if we can arrange the first nine digits with some zeros to make the number smaller.

But since the first nine digits must have nine 1s, the smallest number is 111,111,111.

Therefore, no improvement.

Therefore, the minimal N is indeed 11,111,111,100.

But let me check if there is a number with 11 digits, ending with 00, with digit sum 9, but with the first nine digits having fewer than nine 1s, but compensated by some 1s in the remaining digits.

Wait, no. Because the total digit sum is nine. The last two digits are zeros, so the first nine digits must have nine 1s.

Therefore, the first nine digits must have nine 1s.

Therefore, the minimal number is 111,111,11100.

Therefore, I think this is the correct answer.

But to be absolutely sure, let me check if there is a smaller number with digit sum 9, ending with 00.

For example, let's think of numbers with 11 digits, ending with 00, and with nine 1s in the first nine digits.

The first nine digits can be arranged to be smaller than 111,111,111.

But since they must have nine 1s, the smallest number is 111,111,111.

Because any other arrangement would have a higher value.

For example, 10111111111 (but this is 11 digits, with nine 1s? Let me check: 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1. That's ten digits? No, 11 digits: 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1. Yes, that's 11 digits.

Wait, this number is 10111111111, which is 10,111,111,111.

But this number ends with a 1, not with 00.

To make it end with 00, we need to have the last two digits as 00.

Therefore, the number would be 101111111100, which is 10,111,111,1100.

Which is 12 digits.

As discussed earlier, this is larger than 11,111,111,100.

Therefore, no improvement.

Therefore, I think I can conclude that the smallest positive multiple of 450 with all digits 0 or 1 is 11,111,111,100.

But to confirm, let me check if this number is indeed divisible by 450.

Divisible by 2: ends with 00, yes.

Divisible by 25: ends with 00, yes.

Divisible by 9: sum of digits is 9*1 = 9, which is divisible by 9.

Yes.

Therefore, it is divisible by 450.

Is there a smaller number?

Let me think of numbers with fewer digits.

For example, 111,111,11100 is 11 digits.

Is there a number with 10 digits?

To have 10 digits ending with 00, the first 8 digits must have nine 1s. But 8 digits can have at most 8 ones. Therefore, impossible.

Therefore, no.

Therefore, the minimal number is indeed 11,111,111,100.

But let me check if there is a number with 11 digits, ending with 00, with digit sum 9, but smaller than 11,111,111,100.

For example, let's think of numbers like 1000000001100.

Wait, this is 1000000001100, which is 1 followed by eight zeros, then 11, then 00.

Wait, that would be 1000000001100, which is 1,000,000,001,100.

Which is 1.000... trillion, which is larger than 11 billion.

Therefore, no.

Alternatively, 100111111100.

Which is 100,111,111,100.

Which is 100 billion, which is larger than 11 billion.

Therefore, no.

Therefore, I think I can confidently say that the smallest positive multiple of 450 with all digits 0 or 1 is 11,111,111,100.

But just to be thorough, let me check if there is a number with digit sum 9, ending with 00, with fewer than 11 digits.

As discussed, with 10 digits, it's impossible.

With 9 digits: the number would be X00, where X is 7 digits. But the digit sum of X must be 9.

But X is 7 digits, so maximum digit sum is 7. Not enough.

Therefore, impossible.

Therefore, the minimal number is indeed 11,111,111,100.

Therefore, the answer is 11111111100, which is 11,111,111,100.

But let me write it in the required format.

The number is 11111111100.

But let me check if there is a smaller number with more digits.

For example, let's think of numbers with digit sum 9, ending with 00, and with more digits but smaller value.

For example, 1000000000000000000... with nine 1s.

But this would be a very long number, which is larger.

Therefore, no.

Therefore, I think I can conclude that the answer is 11111111100.

But to make sure, let me check if there is a number with digit sum 9, ending with 00, and smaller than 11111111100.

For example, let's think of numbers like 1000000001100 (which is 1,000,000,001,100) or 100000001100 (which is 1,000,000,011,00).

But these are all larger.

Alternatively, numbers like 100000011100 (which is 100,000,011,100).

Still larger.

Therefore, no.

Therefore, the answer is 11111111100.

But let me check if there is a number with digit sum 9, ending with 00, and with the digits arranged in a way that it's smaller.

For example, 10101010100.

Wait, let's check the digit sum: 1+0+1+0+1+0+1+0+1+0+0 = 5. Not enough.

If I add more 1s: 1010101010100. Digit sum is 6. Still not enough.

To get digit sum 9, need nine 1s.

For example, 1010101010101010100. This has nine 1s.

But this number is 1010101010101010100, which is 101,010,101,010,101,0100. Which is way larger.

Therefore, no improvement.

Therefore, I think I can conclude that the smallest positive multiple of 450 with all digits 0 or 1 is 11111111100.

**Final Answer**
The smallest positive multiple of 450 with all digits 0 or 1 is \boxed{11111111100}.

To solve the problem of finding the **smallest positive multiple of 450** whose **digits are all zeroes and ones**, we begin by analyzing the properties of the number 450 and the constraints on the digits.

---

### Step 1: Prime Factorization of 450

We start by factoring 450:

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

This tells us that any multiple of 450 must be divisible by:

- **2** (i.e., the number must be even),
- **9** (i.e., the sum of its digits must be divisible by 9),
- **25** (i.e., the number must end in **00**).

---

### Step 2: Constraints on the Number

Given that the number must consist **only of digits 0 and 1**, and must be divisible by 450, we can deduce the following:

- **Divisibility by 2 and 25** implies the number must end in **00**.
- **Divisibility by 9** implies the **sum of the digits** must be a multiple of 9.
- Since the digits are only 0s and 1s, the **sum of the digits is equal to the number of 1s**.
- Therefore, the number must contain **exactly 9 ones** (the smallest multiple of 9).

---

### Step 3: Constructing the Smallest Valid Number

We now seek the **smallest number** composed of only 0s and 1s, with **exactly 9 ones**, and **ending in 00**.

- The smallest number with 9 ones is **111111111** (nine 1s).
- To satisfy the divisibility by 25, we must append **00** to the end.
- Therefore, the number becomes:

$$
11111111100
$$

This number:
- Has **9 ones** (sum = 9, divisible by 9),
- Ends in **00** (divisible by 25),
- Is **even** (divisible by 2),
- Therefore, it is divisible by **450**.

---

### Step 4: Verifying Minimality

We must ensure that **no smaller number** satisfies all the conditions.

- Any number with **fewer than 9 ones** would not be divisible by 9.
- Any number with **more than 9 ones** would be larger.
- Any number with **fewer than 11 digits** (i.e., less than 9 digits + 2 zeros) would not have enough digits to accommodate 9 ones and the required ending of 00.
- Any number with **more digits** would be larger than 11111111100.

Thus, **11111111100** is the **smallest** such number.

---

### 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. Let me start by understanding the problem better.

First, 450 is the number we're dealing with. Let me factorize it to see what we're working with. 450 can be broken down into prime factors. Let me do that:

450 divided by 2 is 225. Then 225 divided by 3 is 75, divided by 3 again is 25, and then divided by 5 is 5, and divided by 5 again is 1. So the prime factors are 2 × 3² × 5². So 450 = 2 × 3² × 5².

Now, since we need a multiple of 450, the number we're looking for must be divisible by 2, 9, and 25. Let me recall the divisibility rules for these numbers.

Divisibility by 2: The number must end with an even digit. Since our digits can only be 0 or 1, the last digit has to be 0. So whatever number we come up with must end with a 0.

Divisibility by 25: A number is divisible by 25 if its last two digits are 00, 25, 50, or 75. But since our digits can only be 0 or 1, the only possible ending is 00. Therefore, the number must end with two zeros.

So combining these two, the number must end with at least two zeros. Also, since it's divisible by 2, which is already covered by the two zeros.

Now, divisibility by 9: A number is divisible by 9 if the sum of its digits is a multiple of 9. Since all digits are 0 or 1, the sum of the digits is just the number of 1s in the number. Therefore, the number of 1s must be a multiple of 9.

So putting it all together, the number must:

1. End with two zeros.
2. Have a number of 1s that is a multiple of 9.
3. Be composed only of 0s and 1s.

Our task is to find the smallest such number.

So, the problem reduces to finding the smallest number made up of 0s and 1s, ending with two zeros, and with the number of 1s being a multiple of 9.

Since we want the smallest number, we need the number with the least number of digits possible. Let me think about how to approach this.

First, since the number must end with two zeros, we can think of the number as something like [some combination of 0s and 1s] followed by '00'.

So, the number can be represented as N = X * 100, where X is a number made up of 0s and 1s.

Therefore, N must be divisible by 450, which is 2*3²*5². Since N is X*100, and 100 is 2²*5², so N = X * 2² * 5². Therefore, for N to be divisible by 450 = 2*3²*5², we need X to be divisible by 3² = 9.

Therefore, X must be a number made up of 0s and 1s, and divisible by 9.

So, the problem simplifies to finding the smallest number X composed of 0s and 1s that is divisible by 9, and then appending two zeros to it.

Therefore, the original problem reduces to finding the minimal X (with digits 0 and 1) divisible by 9, then multiplying by 100.

So, now the problem is: find the smallest number made up of 0s and 1s divisible by 9.

Once we find that X, then N = X * 100 is our answer.

So, how do we find the smallest number made up of 0s and 1s divisible by 9?

Well, since divisibility by 9 requires that the sum of digits is a multiple of 9. Since digits are 0s and 1s, the sum is equal to the number of 1s. Therefore, the number of 1s must be a multiple of 9.

Therefore, the minimal number of 1s is 9. Therefore, the minimal number would be the smallest number with 9 ones and any number of zeros, arranged in such a way that the number is as small as possible.

But how do we arrange the digits? To make the number as small as possible, we need the number to have the least number of digits possible, and within that, the digits should be arranged in ascending order (i.e., leading digits as small as possible).

But since we need the number to be as small as possible, we need the minimal number of digits. The minimal number of digits is 9 (all ones). However, the number 111111111 (nine 1s) is 111,111,111. However, maybe there is a smaller number with more digits but with zeros in between? Wait, no. Because adding zeros would make the number longer, which would make it larger. For example, 101111111 is longer than 111111111? Wait, no. Wait, 101111111 has 9 digits as well. Wait, same number of digits.

Wait, but numbers with more digits are larger. For example, 100000000 is 100 million, which is larger than 111,111,111? Wait, no. Wait, 100,000,000 is 100 million, which is less than 111,111,111. Wait, so actually, numbers with more zeros can be smaller? Wait, but if we have the same number of digits, then the number with more zeros would be smaller.

Wait, but if we have a number with 9 digits, the minimal number would be 100000000, but that has only one 1. However, we need the number of 1s to be 9. Therefore, the minimal number with 9 ones is 111111111. However, if we can have a number with more digits but with some zeros, but with the same number of 1s, but arranged in such a way that the number is smaller.

Wait, for example, 101010101 is a 9-digit number with 5 ones. But we need 9 ones. So, if we have 9 ones, the minimal number would be the one with the ones as far to the right as possible? Wait, no. Wait, to make the number as small as possible, we need the leftmost digits to be as small as possible. So, the first digit should be 1 (since leading zeros are not allowed), then the next digits should be 0s as much as possible.

Wait, but if we have 9 ones, the minimal number would be 100000000... but that only has one 1. Wait, no. Wait, we need exactly 9 ones. So, the minimal number would be the one with the 9 ones placed in the positions that make the number as small as possible.

For example, the smallest number with 9 ones is 111111111 (nine 1s). However, if we can have a number with more digits but with some zeros, but still having 9 ones, but being smaller.

Wait, for example, 1000000001 has 10 digits, with two ones. But we need nine ones. So, 10000000011 would have 10 digits with three ones. Wait, but we need nine ones. So, for example, 100000000... with nine ones. Wait, but how?

Wait, if we have a number with more digits, but with nine ones, but arranged so that the number is smaller. For example, 100000000... with the ones spread out.

But actually, the number with the least value would be the one with the ones as far to the right as possible. For example, if you have a number with 9 ones, the smallest number is 111111111 (nine digits). If you have a number with ten digits, the smallest number with nine ones would be 1000000001, but that's actually larger than 111111111. Wait, 1000000001 is 1,000,000,001, which is larger than 111,111,111.

Wait, so actually, the minimal number with nine ones is 111,111,111. However, maybe there is a smaller number with more digits? Wait, no. Because numbers with more digits are larger. For example, 10 digits would be larger than 9 digits. Therefore, the minimal number is 111,111,111.

But wait, let me check. Suppose we have a number with 18 digits, but with nine ones. But that would be way larger. So, no. Therefore, the minimal number is 111,111,111.

But wait, hold on. Let me think again. Suppose we have a number with 9 ones and some zeros. For example, 1011111111. Wait, but that's 10 digits. But 1011111111 is 1,011,111,111, which is larger than 111,111,111.

Alternatively, if we have 111111111, which is 111,111,111. So, that's smaller.

But maybe there is a number with fewer digits? Wait, no. Because to have nine 1s, you need at least nine digits. So, the minimal number of digits is nine. Therefore, the minimal number is 111,111,111.

But wait, is there a smaller number with nine 1s? For example, if you have leading zeros? No, because numbers can't have leading zeros. So, the first digit has to be 1. Then, the rest can be arranged.

But if you have nine digits, all ones, that's the smallest possible.

But wait, maybe there is a number with more digits but with some zeros that is smaller? For example, 100000000... but with nine ones. Wait, but if you have more digits, you need to have more digits. For example, if you have ten digits, you need nine ones and one zero. The smallest such number would be 1000000001, but that's 1,000,000,001, which is larger than 111,111,111.

Similarly, 1000000011 is 1,000,000,011, which is still larger.

Therefore, the minimal number is 111,111,111.

But wait, hold on. Let me check if there is a smaller number with nine 1s. For example, if you have 101111111, but that's 9 digits. Wait, 101111111 is 101,111,111, which is larger than 111,111,111? No, wait, 101 million is less than 111 million. Wait, so 101,111,111 is smaller than 111,111,111.

Oh! Wait, so maybe arranging the digits with zeros in between can make the number smaller.

So, for example, if we have a number with nine 1s and some zeros, but arranged in such a way that the number is smaller.

So, the key is that the number with the least value is the one where the digits are as small as possible from left to right.

So, the first digit is 1, then the next digits should be 0s as much as possible, but we need nine 1s.

So, for example, the minimal number would be 100000000... but with nine 1s. Wait, but how?

Wait, if we have nine 1s, the minimal number is the one where the 1s are placed as far to the right as possible.

So, for example, the number would be 100000000... with the first digit 1, then as many zeros as possible, then the remaining digits as 1s.

But since we need nine 1s, the minimal number would be 100000000... with the first digit 1, followed by eight zeros and then eight 1s? Wait, no. Wait, if we have nine digits, then the number is 111111111. If we have ten digits, then the number would be 1000000001, but that's only two 1s. Wait, no. Wait, if we have ten digits with nine 1s, then it's 1111111111 with a zero somewhere. Wait, but to make the number as small as possible, we need to place the zero as far to the right as possible.

Wait, for example, with ten digits, the minimal number with nine 1s is 1011111111. Let me check:

- 1011111111 is 1,011,111,111
- 1101111111 is 1,101,111,111
- 1110111111 is 1,110,111,111
- etc. So, the smallest is 1011111111.

But 1,011,111,111 is smaller than 111,111,111? Wait, no. Wait, 111,111,111 is 111 million, while 1,011,111,111 is 1 billion plus, which is way larger.

Therefore, even though 1011111111 has a zero in the second digit, it's still larger than 111,111,111.

Therefore, the minimal number with nine 1s is indeed 111,111,111.

But wait, let me think again. Suppose we have a number with more digits but with some zeros. For example, 1001111111. Wait, that's ten digits. Let me compute its value.

1,001,111,111. Which is 1.001 billion, which is still larger than 111 million.

So, even if we have more digits, the number is larger. Therefore, the minimal number is 111,111,111.

But wait, hold on. Let me check if there is a number with fewer digits? No, because you need nine 1s. So, the minimal number of digits is nine.

Therefore, the minimal X is 111,111,111. Therefore, the original number N is X * 100 = 111,111,111 * 100 = 11,111,111,100.

But wait, let me check if this is indeed divisible by 450.

First, check divisibility by 2: ends with 00, yes. Divisibility by 25: ends with 00, yes. Divisibility by 9: sum of digits is 9*1 = 9, which is divisible by 9. Therefore, yes.

But is this the minimal such number?

Wait, but maybe there is a smaller number with more digits? For example, maybe a number with more digits but with fewer 1s? Wait, no. Because the number of 1s must be a multiple of 9. The minimal multiple is 9. So, if we have a number with 18 ones, that would be larger.

But perhaps there is a number with more digits but with 9 ones arranged in such a way that the number is smaller than 11,111,111,100?

Wait, for example, let's think of numbers with more digits but with 9 ones. For example, 10000000011... but with more zeros.

Wait, for example, 10000000011 is 10,000,000,011. Which is 10 billion plus, which is larger than 11 billion.

Wait, 11,111,111,100 is 11 billion, so 10,000,000,011 is smaller? Wait, no. Wait, 10,000,000,011 is 10 billion and 11, while 11,111,111,100 is 11 billion and 111 million. So, 10 billion is less than 11 billion. Therefore, 10,000,000,011 is smaller.

But wait, does 10,000,000,011 have nine 1s? Let me check. The number is 1 followed by eight zeros and then 11. So, digits are 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1. So, that's two 1s and eight zeros. Wait, no, that's only two 1s. Wait, I thought it had nine 1s.

Wait, no. If I want nine 1s, then the number would have to have nine 1s. For example, 10000000011 has two 1s. To have nine 1s, we need to have nine 1s and some zeros.

So, for example, 1000000001111111111 would be 1 followed by eight zeros and then nine 1s. Wait, that would be 1 followed by eight zeros and nine 1s, which is 1 + 8 zeros + 9 ones. So, total digits: 1 + 8 + 9 = 18 digits.

But that number is 1000000001111111111, which is 1 followed by eight zeros and then nine 1s.

But this number is 10^17 + 10^16 + ... + 10^9? Wait, no. Wait, the number is 1 followed by eight zeros, then nine 1s. So, the first digit is 1, then eight zeros, then nine 1s.

But this number is 10^17 + (10^16 + 10^15 + ... + 10^9). Which is a huge number, way larger than 11,111,111,100.

Therefore, even though it has nine 1s, it's way larger.

Therefore, perhaps the minimal number is indeed 111,111,111 * 100 = 11,111,111,100.

But wait, let me check if there is a smaller number with nine 1s. For example, maybe a number with more digits but with some zeros that makes it smaller.

For example, let's think of numbers with 10 digits. The minimal number with nine 1s would be 1011111111. Let me check its value.

1,011,111,111. Which is 1.011... billion.

Compare that to 11,111,111,100, which is 11.111... billion. So, 1.011 billion is smaller.

But wait, 1,011,111,111 is smaller than 11,111,111,100. Therefore, if we can find such a number, then it would be smaller.

But wait, but this number is X = 1011111111, which is 1,011,111,111. Then N = X * 100 = 101,111,111,100.

But is this number smaller than 11,111,111,100? No, because 101,111,111,100 is 101 billion, while 11,111,111,100 is 11 billion. So, 11 billion is smaller.

Therefore, even though X is smaller, multiplying by 100 makes N larger.

Wait, so maybe the key is to find the minimal X (with digits 0 and 1, divisible by 9) and then multiply by 100.

But if X is smaller, then N is smaller.

So, if there exists a smaller X (with digits 0 and 1) divisible by 9, then N would be smaller.

But earlier, I thought that the minimal X is 111,111,111. But maybe there is a smaller X with more digits?

Wait, for example, let's think of numbers with digits 0 and 1, divisible by 9.

The sum of digits must be 9, 18, 27, etc.

So, the minimal sum is 9. Therefore, the minimal X is the smallest number with nine 1s. But maybe there is a smaller number with more digits but with sum of digits 9?

Wait, for example, numbers with digits 0 and 1, with sum of digits 9. The minimal number is the one with the least number of digits, which is nine digits. But if there is a number with more digits but smaller in value?

For example, 1000000001 has two 1s. Not enough.

But if we have a number with nine 1s and some zeros, but arranged in such a way that the number is smaller.

For example, 1000000001111111111 is 1 followed by eight zeros and nine 1s. But that's 18 digits.

But 1000000001111111111 is 10^17 + ... which is way larger than 111,111,111.

Alternatively, maybe a number like 10101010101010101, which has nine 1s. Let me count: 1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1. That's 17 digits. But the value is 10101010101010101, which is 10,101,010,101,010,101. Which is way larger than 111,111,111.

Therefore, even though it has nine 1s, it's a larger number.

Therefore, the minimal X is indeed 111,111,111.

But wait, let me think again. Maybe there is a number with more digits but with the same number of digits as 111,111,111 but with zeros in between?

For example, 1011111111 is ten digits. But as we saw, it's larger than 111,111,111.

Wait, 101,111,1111 is 1,011,111,111 which is 1.01 billion, while 111,111,111 is 0.111 billion. So, no, 111 million is smaller.

Therefore, the minimal X is 111,111,111.

But wait, let me check if there is a smaller number with sum of digits 9. For example, numbers with leading zeros? No, because numbers can't have leading zeros.

Alternatively, maybe numbers with digits 0 and 1, but not necessarily all 1s. For example, maybe a number like 1110000000, which has three 1s. But that's not enough.

Wait, but we need the sum of digits to be 9. So, numbers with nine 1s.

But maybe there is a number with nine 1s and some zeros that is smaller than 111,111,111?

For example, 100000000111111111. Wait, that's 1 followed by eight zeros and then eight 1s? No, that would be 1 followed by eight zeros and then eight 1s, which is 10^8 + ... + 10^1? Wait, no.

Wait, let me think. If I have a number with nine 1s, the minimal number is the one where the 1s are as far to the right as possible.

So, for example, the number would be 100000000... with the first digit 1, then as many zeros as possible, then the remaining digits as 1s.

But since we need nine 1s, the minimal number would be 100000000... with the first digit 1, then eight zeros, then eight 1s? Wait, no. Wait, if you have nine digits, all 1s. If you have ten digits, you need to have nine 1s and one zero. To make the number as small as possible, you want the zero to be as far to the right as possible. So, the number would be 1111111110. Wait, but that's ten digits.

Wait, 1111111110 is 1,111,111,110. Which is larger than 111,111,111.

Alternatively, if you have a number with ten digits, with nine 1s and one zero, the smallest such number is 1011111111. Which is 1,011,111,111. Which is still larger than 111,111,111.

Therefore, even if you have more digits, the number is larger.

Therefore, the minimal X is 111,111,111.

Therefore, the minimal N is 111,111,111 * 100 = 11,111,111,100.

But wait, let me check if there is a smaller number. Maybe there is a number with more digits but with sum of digits 9, but arranged in such a way that it's smaller.

For example, let's think of numbers with digits 0 and 1, divisible by 9.

Let me try to find the minimal such number.

Let me think of numbers with digits 0 and 1, and check their divisibility by 9.

The first numbers with digits 0 and 1 are:

1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, etc.

Let me check their digit sums:

- 1: sum 1
- 10: 1
- 11: 2
- 100: 1
- 101: 2
- 110: 2
- 111: 3
- 1000: 1
- 1001: 2
- 1010: 2
- 1011: 3
- 1100: 2
- 1101: 3
- 1110: 3
- 1111: 4
- ...

So, the digit sums go up as we add more 1s.

To get a digit sum of 9, we need numbers with nine 1s.

So, the first number with nine 1s is 111,111,111.

But maybe there is a number with more digits but with a smaller value?

For example, let's think of numbers with 10 digits. The smallest number with nine 1s is 1011111111.

But as we saw, this is larger than 111,111,111.

Similarly, numbers with 11 digits would be even larger.

Therefore, the minimal number is indeed 111,111,111.

But wait, let me check if there is a number with fewer digits? No, because you need nine 1s.

Therefore, the minimal X is 111,111,111.

Therefore, the minimal N is 111,111,111 * 100 = 11,111,111,100.

But let me check if this is indeed the minimal.

Wait, but maybe there is a number with more digits but with a smaller value. For example, maybe a number like 1001001001... with some pattern?

But how can we be sure?

Alternatively, maybe there is a number with digits 0 and 1, divisible by 9, which is smaller than 111,111,111.

For example, let's think of numbers with digit sum 9.

But since the digit sum is 9, and digits are 0 and 1, the number must have exactly nine 1s.

Therefore, the minimal number is the one with the least number of digits, which is nine digits.

Therefore, the minimal number is 111,111,111.

Therefore, N is 11,111,111,100.

But let me check if there is a smaller number.

Wait, let me think of numbers with digits 0 and 1, and check if any of them is divisible by 9 before reaching nine 1s.

For example, maybe a number with digits summing to 9, but with more zeros.

But since the digit sum is 9, you need nine 1s.

Therefore, there is no way to have a number with fewer digits.

Therefore, the minimal number is 111,111,111.

But let me check with an example. Let's think of numbers with digits 0 and 1, and check their divisibility by 9.

For example, 111111111 is 111,111,111.

Divided by 9: 111,111,111 ÷ 9 = 12,345,679.

Yes, because 12,345,679 × 9 = 111,111,111.

So, that works.

But is there a smaller number?

For example, let's think of numbers with digits 0 and 1, with digit sum 9.

Suppose we have a number with ten digits, with nine 1s and one 0. The smallest such number is 1011111111.

Let me check if this is divisible by 9.

Sum of digits is 9, so yes.

But 1011111111 is 1,011,111,111.

Which is larger than 111,111,111.

Therefore, even though it's divisible by 9, it's larger.

Similarly, numbers with more digits would be larger.

Therefore, the minimal X is 111,111,111.

Therefore, the minimal N is 111,111,111 * 100 = 11,111,111,100.

But let me check if there is a smaller multiple of 450 with only 0s and 1s.

Wait, maybe there is a number that ends with two zeros, has a digit sum divisible by 9, but is smaller than 11,111,111,100.

For example, maybe a number with fewer digits.

But to end with two zeros, the number must have at least two zeros.

But the digit sum must be divisible by 9.

So, for example, let's think of numbers with two zeros at the end, and the rest digits being 0s and 1s, with the sum of digits divisible by 9.

So, for example, let's think of numbers with two zeros at the end.

Let me think of numbers with digits 0 and 1, ending with 00, and with digit sum divisible by 9.

So, the digit sum is the number of 1s.

So, the number of 1s must be divisible by 9.

So, the minimal number of 1s is 9.

Therefore, the minimal number would be the smallest number with nine 1s and two zeros at the end.

But the two zeros can be at the end.

So, the number would be something like [some combination of 0s and 1s] followed by 00.

But the [some combination] must have nine 1s and any number of 0s.

But to make the entire number as small as possible, we need the [some combination] to be as small as possible.

So, the minimal [some combination] is the smallest number with nine 1s and any number of 0s.

Which is 111,111,111.

Therefore, the entire number is 111,111,11100, which is 11,111,111,100.

But maybe there is a smaller [some combination] with nine 1s and some zeros, such that when you append 00, the number is smaller.

For example, if [some combination] is 1000000001, which has two 1s. Not enough.

If [some combination] is 1011111111, which has ten digits with nine 1s. Then the entire number is 101111111100, which is 101,111,111,100. Which is larger than 11,111,111,100.

Alternatively, if [some combination] is 111111111, then the entire number is 11111111100, which is 11,111,111,100.

Alternatively, if [some combination] is 1001111111, which is 1,001,111,111. Then the entire number is 1,001,111,11100, which is 10,011,111,1100, which is 10,011,111,1100. Wait, that's 10,011,111,1100? Wait, no.

Wait, if [some combination] is 1001111111 (which is 1,001,111,111), then appending 00 gives 1,001,111,11100, which is 10,011,111,1100. Wait, but that's 10,011,111,1100. Which is 10,011,111,1100.

But 10,011,111,1100 is 10 billion and something, while 11,111,111,100 is 11 billion. So, 10 billion is smaller.

Wait, but does [some combination] = 1001111111 have nine 1s? Let me check.

1001111111: digits are 1, 0, 0, 1, 1, 1, 1, 1, 1, 1. So, that's 1 followed by two zeros, then seven 1s. Total 1s: 1 + 7 = 8. Not nine.

Oops, so that's only eight 1s.

To have nine 1s, [some combination] must have nine 1s.

So, for example, 1000000001111111111 is 1 followed by eight zeros and nine 1s. But that's 1 + 8 + 9 = 18 digits.

But when we append 00, it becomes 18 + 2 = 20 digits.

But even so, the number is 100000000111111111100, which is way larger than 11,111,111,100.

Therefore, even if we have a [some combination] with nine 1s and some zeros, the minimal such number is 111,111,111, which when appended with 00 gives 11,111,111,100.

But let me check if there is a smaller [some combination] with nine 1s.

For example, 1011111111 (ten digits) has nine 1s.

So, the number would be 101111111100.

Which is 101,111,111,100.

Compare to 11,111,111,100.

Which is smaller? 11 billion vs 101 billion. 11 billion is smaller.

Therefore, 11,111,111,100 is smaller.

Therefore, the minimal N is 11,111,111,100.

But let me check if there is a number with more digits but smaller.

For example, let's think of numbers with digits 0 and 1, ending with 00, with digit sum 9.

Suppose we have a number like 100100100100... with some pattern.

But how can we ensure that the digit sum is 9?

Alternatively, maybe there is a number with fewer digits than 11,111,111,100.

Wait, 11,111,111,100 has 11 digits (including the two zeros). Wait, no. Let me count:

11,111,111,100 is 1 followed by ten digits: 1,1,1,1,1,1,1,1,1,0,0. So, total of 11 digits.

Is there a number with fewer digits?

For example, a number with 10 digits.

To have two zeros at the end, the number would be something like [some digits]00.

So, the [some digits] part has 8 digits.

To have digit sum 9, the [some digits] part must have 9 ones and 0 zeros? Wait, no. The total digit sum is 9. Since the two zeros at the end contribute 0, the [some digits] part must have 9 ones.

Therefore, the [some digits] part is a number with 8 digits, with nine 1s? Wait, impossible. Because 8 digits can have at most 8 ones.

Therefore, it's impossible.

Therefore, the minimal number of digits for [some digits] is 9 digits (with nine 1s), leading to total digits of 9 + 2 = 11.

Therefore, 11 digits is the minimal.

Therefore, the minimal number is 11,111,111,100.

But let me check if there is a number with 11 digits, ending with 00, with digit sum 9, but smaller than 11,111,111,100.

For example, let's think of numbers with 11 digits, ending with 00, and with nine 1s in the first nine digits.

The first nine digits can be arranged to make the number as small as possible.

So, the first nine digits should be the smallest number with nine 1s. Which is 111,111,111.

Therefore, the entire number is 111,111,11100, which is 11,111,111,100.

Alternatively, if we can arrange the first nine digits with some zeros to make the number smaller.

For example, 10000000011... but wait, the first nine digits must have nine 1s.

Wait, no. The entire number has nine 1s in total, because the two zeros at the end contribute nothing.

Therefore, the first nine digits must have nine 1s.

Therefore, the first nine digits must be 111,111,111.

Therefore, the entire number is 111,111,11100.

Therefore, there is no smaller number with 11 digits.

Therefore, 11,111,111,100 is indeed the minimal.

But let me check if there is a number with more digits but smaller.

For example, a number with 12 digits, ending with 00, with nine 1s in the first ten digits.

But then the first ten digits would have nine 1s and one 0.

To make the number as small as possible, the 0 should be as far to the right as possible.

So, the first ten digits would be 1111111110.

Then the entire number is 111111111000.

Which is 11,111,111,1000.

Which is 11,111,111,1000, which is larger than 11,111,111,100.

Therefore, no improvement.

Alternatively, if the first ten digits are 1011111111, then the entire number is 101111111100, which is 101,111,111,100, which is larger.

Therefore, no improvement.

Therefore, the minimal number is indeed 11,111,111,100.

But let me check if there is a number with fewer digits.

Wait, for example, if we have a number with 10 digits, ending with 00, and with nine 1s.

But as we saw earlier, the first eight digits would need to have nine 1s, which is impossible.

Therefore, no.

Therefore, the minimal number is 11,111,111,100.

But let me check if there is a smaller number with more digits.

For example, let's think of numbers with digits 0 and 1, ending with 00, and with digit sum 9.

Suppose we have a number like 100100100100... with some pattern.

But how can we ensure the digit sum is 9?

For example, if we have three 1s separated by zeros.

But to get nine 1s, we need nine 1s.

But arranging them in such a way that the number is smaller.

For example, 1000000000000000000... with nine 1s.

But that would be a very long number.

Alternatively, maybe a number like 101010101010101010... with nine 1s.

But again, it would be longer and larger.

Therefore, no improvement.

Therefore, I think that 11,111,111,100 is indeed the minimal number.

But let me check with another approach.

Another way to approach this problem is to generate numbers made up of 0s and 1s, check if they are divisible by 450, and find the smallest one.

But since this is a thought process, let me think how to do this systematically.

First, since the number must end with 00, we can consider numbers of the form X00, where X is a number made up of 0s and 1s.

So, we need X00 divisible by 450.

As before, since 450 = 9 * 50, and X00 is divisible by 50 (since it ends with 00), so we need X00 divisible by 9.

Which means that the sum of digits of X00 is divisible by 9.

But the sum of digits of X00 is equal to the sum of digits of X, since the two zeros contribute nothing.

Therefore, sum of digits of X must be divisible by 9.

Therefore, X must be a number made up of 0s and 1s, with digit sum divisible by 9.

Therefore, the problem reduces to finding the smallest X (made of 0s and 1s) with digit sum divisible by 9, then appending 00.

Therefore, the minimal X is the minimal number made of 0s and 1s with digit sum 9.

Which, as discussed, is 111,111,111.

Therefore, the minimal N is 111,111,11100 = 11,111,111,100.

But let me check if there is a smaller X.

For example, maybe a number with more digits but smaller in value.

For example, 1001111111 (which is 1,001,111,111) has digit sum 1+0+0+1+1+1+1+1+1+1 = 8. Not enough.

Wait, need digit sum 9.

So, 1001111111 has eight 1s.

If we add another 1, making it 10011111111, which is 10 digits.

Digit sum is 9.

So, this number is 10,011,111,111.

Which is 10,011,111,111.

Compare to 111,111,111.

Which is smaller?

111,111,111 is 111 million, while 10,011,111,111 is 10 billion.

So, 111 million is smaller.

Therefore, even though X = 10011111111 is larger than X = 111,111,111, the corresponding N would be X * 100.

Wait, no. Wait, N is X followed by 00.

So, if X is 10011111111 (ten digits), then N is 1001111111100, which is 1,001,111,111,100.

Which is 1.001... trillion, which is way larger than 11,111,111,100.

Therefore, no improvement.

Therefore, the minimal X is indeed 111,111,111.

Therefore, the minimal N is 11,111,111,100.

But let me check if there is a smaller X with digit sum 9.

For example, numbers with digits 0 and 1, digit sum 9, smaller than 111,111,111.

For example, 100000000111111111. Wait, that's 1 followed by eight zeros and then eight 1s. Wait, digit sum is 1 + 8 = 9.

So, this number is 100000000111111111.

Which is 100,000,000,111,111,111.

Which is way larger than 111,111,111.

Therefore, no.

Alternatively, numbers like 10101010101010101.

Which has nine 1s.

But again, it's a longer number.

Therefore, no improvement.

Therefore, I think that 11,111,111,100 is indeed the minimal number.

But let me check with another approach.

Let me think of numbers made up of 0s and 1s, ending with 00, and check divisibility by 450.

Start with the smallest numbers:

- 100: sum of digits 1, not divisible by 9.
- 1000: sum 1
- 10000: sum 1
- ...
- 1100: sum 2
- 10100: sum 2
- 100100: sum 2
- 1000100: sum 2
- 10000100: sum 2
- 100000100: sum 2
- 1000000100: sum 2
- 10000000100: sum 2
- 11100: sum 3
- 110100: sum 3
- 1100100: sum 3
- 11000100: sum 3
- 110000100: sum 3
- 1100000100: sum 3
- 11000000100: sum 3
- 101100: sum 3
- 1010100: sum 3
- 10100100: sum 3
- 101000100: sum 3
- 1010000100: sum 3
- 10100000100: sum 3
- 1001100: sum 3
- ... and so on.

Continuing this way, we can see that numbers with digit sums 1, 2, 3, etc., up to 9.

So, the first number with digit sum 9 would be the first number with nine 1s.

Which is 111,111,11100.

But maybe there is a number with digit sum 9 that is smaller.

For example, numbers with more zeros but still nine 1s.

But as discussed earlier, numbers with more digits are larger.

Therefore, the first occurrence of a digit sum of 9 is 111,111,11100.

Therefore, this must be the minimal.

But to confirm, let me think of numbers with digit sum 9.

For example, 100000000111111111 (as before) is larger.

Alternatively, numbers like 10011111111 (which is 1 followed by one 0, then eight 1s and one more 1? Wait, no.

Wait, 10011111111 has ten digits: 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1. Wait, that's 1 followed by two zeros and eight 1s. Total digits: 1 + 2 + 8 = 11 digits.

Wait, digit sum is 1 + 0 + 0 + 8*1 = 9.

So, this number is 10011111111, which is 10,011,111,111.

Which is smaller than 111,111,11100 (which is 11,111,111,100).

Wait, 10,011,111,111 is smaller than 11,111,111,100?

Yes! Because 10 billion is less than 11 billion.

Wait, so this number is smaller.

But why didn't I think of this earlier?

So, this number is 10011111111, which is 10,011,111,111.

But this number ends with a 1, not with two zeros.

Wait, no. Wait, we need the number to end with two zeros.

Ah! Here's the mistake.

Earlier, I considered X as a number with digit sum 9, and then appending 00.

But in this case, the number 10011111111 is X with digit sum 9, but it doesn't end with two zeros.

Therefore, to make it end with two zeros, we need to append 00.

Therefore, the number would be 1001111111100, which is 10,011,111,11100.

Which is 100,111,111,1100.

Wait, let me count the digits.

Original X is 10011111111 (11 digits). Appending 00 makes it 13 digits: 1001111111100.

Which is 100,111,111,1100.

Which is 100 billion something.

Compare to 11,111,111,100 (which is 11 billion).

So, 11 billion is smaller.

Therefore, even though X = 10011111111 is smaller than X = 111,111,111, when we append 00, the resulting number is larger.

Therefore, the minimal N is still 11,111,111,100.

Therefore, my previous conclusion holds.

But let me check if there is a number with digit sum 9, ending with 00, which is smaller than 11,111,111,100.

For example, let's think of numbers with digit sum 9, ending with 00.

So, the number is of the form [some digits]00, where [some digits] has nine 1s.

To make the entire number as small as possible, [some digits] should be as small as possible.

The smallest [some digits] with nine 1s is 111,111,111.

Therefore, the entire number is 111,111,11100.

Alternatively, if [some digits] is a smaller number with nine 1s and some zeros.

But how?

For example, if [some digits] is 100000000111111111 (which is 1 followed by eight zeros and eight 1s). Wait, but that would have digit sum 1 + 8 = 9.

But this number is 100000000111111111, which is 100,000,000,111,111,111.

Which is way larger than 111,111,111.

Therefore, no improvement.

Alternatively, if [some digits] is 1011111111 (ten digits with nine 1s).

Which is 1,011,111,111.

Then the entire number is 1,011,111,11100, which is 10,111,111,1100.

Which is 10,111,111,1100.

Compare to 11,111,111,100.

Which is smaller?

10,111,111,1100 is 10 billion and something, while 11,111,111,100 is 11 billion.

So, 10 billion is smaller.

But wait, 10,111,111,1100 is 10,111,111,1100.

Wait, let me write it properly.

If [some digits] is 1011111111 (which is 1,011,111,111), then appending 00 gives 1,011,111,11100.

Which is 10,111,111,1100.

Which is 10,111,111,1100.

But 11,111,111,100 is 11,111,111,100.

Comparing these two numbers:

- 10,111,111,1100 is 10,111,111,1100
- 11,111,111,100 is 11,111,111,100

Wait, but these are different numbers.

Wait, 10,111,111,1100 is 10 billion, 111 million, 111 thousand, 100.

Whereas 11,111,111,100 is 11 billion, 111 million, 111 thousand, 100.

Therefore, 10 billion is smaller than 11 billion.

Therefore, 10,111,111,1100 is smaller.

But wait, how is this possible?

Because [some digits] is 1011111111 (which is 1,011,111,111) with digit sum 9 (1+0+1+1+1+1+1+1+1+1 = 9).

Therefore, this X is valid.

Therefore, the corresponding N is 1,011,111,11100, which is 10,111,111,1100.

Which is smaller than 11,111,111,100.

Therefore, this suggests that my previous conclusion was wrong.

So, why is this?

Because I thought that [some digits] must be nine 1s, but actually, [some digits] can have more digits with some zeros, as long as the total number of 1s is nine.

Therefore, the minimal [some digits] is not necessarily nine 1s.

Wait, but the digit sum of [some digits] must be nine.

Therefore, [some digits] can have nine 1s and any number of zeros.

Therefore, the minimal [some digits] is the smallest number with nine 1s and any number of zeros.

Which is not necessarily nine 1s.

For example, 1011111111 is smaller than 111,111,111.

Wait, but 1011111111 is ten digits, while 111,111,111 is nine digits.

But numbers with more digits are larger.

Wait, but 1011111111 is 1,011,111,111, which is 1.01 billion, while 111,111,111 is 0.111 billion.

Therefore, 0.111 billion is smaller.

Therefore, even though [some digits] has more digits, the number itself is smaller.

Wait, no. Wait, 1,011,111,111 is larger than 111,111,111.

Yes, because 1,011,111,111 is 1.01 billion, while 111,111,111 is 0.111 billion.

Therefore, the number with fewer digits is smaller.

Therefore, even though [some digits] can have more digits, the minimal [some digits] is the one with the least number of digits, which is nine digits with nine 1s.

Therefore, the minimal N is 111,111,11100.

But then why does the number 101111111100 (which is 1,011,111,11100) exist?

Because it's a larger number.

Therefore, the minimal N is still 11,111,111,100.

But let me check the actual numeric values.

Let me write them out:

- 11,111,111,100 is 11,111,111,100
- 10,111,111,1100 is 10,111,111,1100

Wait, but 10,111,111,1100 is 10,111,111,1100, which is 10 billion, 111 million, 111 thousand, 100.

Whereas 11,111,111,100 is 11 billion, 111 million, 111 thousand, 100.

So, 10 billion is less than 11 billion. Therefore, 10,111,111,1100 is smaller.

But how is this possible?

Because [some digits] is 1011111111 (which is 1,011,111,111) with digit sum 9, and then appending 00 gives 1,011,111,11100, which is 10,111,111,1100.

But this number has more digits than 11,111,111,100?

Wait, 1,011,111,11100 is 1 followed by 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0. Wait, no.

Wait, [some digits] is 1011111111 (ten digits), then appending 00 makes it twelve digits.

Whereas 11,111,111,100 is eleven digits.

Wait, 11,111,111,100 is eleven digits: 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0.

Whereas 10,111,111,1100 is twelve digits: 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0.

Therefore, 11,111,111,100 is an 11-digit number, while 10,111,111,1100 is a 12-digit number.

Therefore, the 11-digit number is smaller than the 12-digit number.

Therefore, 11,111,111,100 is smaller.

Therefore, my previous conclusion holds.

Therefore, the minimal N is 11,111,111,100.

But to be thorough, let me check if there is a number with 11 digits, ending with 00, with digit sum 9, which is smaller than 11,111,111,100.

For example, let's think of numbers with 11 digits, ending with 00, and with nine 1s in the first nine digits.

The first nine digits must have nine 1s.

The smallest such number is 111,111,111 followed by 00, which is 11,111,111,100.

Alternatively, if we can arrange the first nine digits with some zeros to make the number smaller.

But since the first nine digits must have nine 1s, the smallest number is 111,111,111.

Therefore, no improvement.

Therefore, the minimal N is indeed 11,111,111,100.

But let me check if there is a number with 11 digits, ending with 00, with digit sum 9, but with the first nine digits having fewer than nine 1s, but compensated by some 1s in the remaining digits.

Wait, no. Because the total digit sum is nine. The last two digits are zeros, so the first nine digits must have nine 1s.

Therefore, the first nine digits must have nine 1s.

Therefore, the minimal number is 111,111,11100.

Therefore, I think this is the correct answer.

But to be absolutely sure, let me check if there is a smaller number with digit sum 9, ending with 00.

For example, let's think of numbers with 11 digits, ending with 00, and with nine 1s in the first nine digits.

The first nine digits can be arranged to be smaller than 111,111,111.

But since they must have nine 1s, the smallest number is 111,111,111.

Because any other arrangement would have a higher value.

For example, 10111111111 (but this is 11 digits, with nine 1s? Let me check: 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1. That's ten digits? No, 11 digits: 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1. Yes, that's 11 digits.

Wait, this number is 10111111111, which is 10,111,111,111.

But this number ends with a 1, not with 00.

To make it end with 00, we need to have the last two digits as 00.

Therefore, the number would be 101111111100, which is 10,111,111,1100.

Which is 12 digits.

As discussed earlier, this is larger than 11,111,111,100.

Therefore, no improvement.

Therefore, I think I can conclude that the smallest positive multiple of 450 with all digits 0 or 1 is 11,111,111,100.

But to confirm, let me check if this number is indeed divisible by 450.

Divisible by 2: ends with 00, yes.

Divisible by 25: ends with 00, yes.

Divisible by 9: sum of digits is 9*1 = 9, which is divisible by 9.

Yes.

Therefore, it is divisible by 450.

Is there a smaller number?

Let me think of numbers with fewer digits.

For example, 111,111,11100 is 11 digits.

Is there a number with 10 digits?

To have 10 digits ending with 00, the first 8 digits must have nine 1s. But 8 digits can have at most 8 ones. Therefore, impossible.

Therefore, no.

Therefore, the minimal number is indeed 11,111,111,100.

But let me check if there is a number with 11 digits, ending with 00, with digit sum 9, but smaller than 11,111,111,100.

For example, let's think of numbers like 1000000001100.

Wait, this is 1000000001100, which is 1 followed by eight zeros, then 11, then 00.

Wait, that would be 1000000001100, which is 1,000,000,001,100.

Which is 1.000... trillion, which is larger than 11 billion.

Therefore, no.

Alternatively, 100111111100.

Which is 100,111,111,100.

Which is 100 billion, which is larger than 11 billion.

Therefore, no.

Therefore, I think I can confidently say that the smallest positive multiple of 450 with all digits 0 or 1 is 11,111,111,100.

But just to be thorough, let me check if there is a number with digit sum 9, ending with 00, with fewer than 11 digits.

As discussed, with 10 digits, it's impossible.

With 9 digits: the number would be X00, where X is 7 digits. But the digit sum of X must be 9.

But X is 7 digits, so maximum digit sum is 7. Not enough.

Therefore, impossible.

Therefore, the minimal number is indeed 11,111,111,100.

Therefore, the answer is 11111111100, which is 11,111,111,100.

But let me write it in the required format.

The number is 11111111100.

But let me check if there is a smaller number with more digits.

For example, let's think of numbers with digit sum 9, ending with 00, and with more digits but smaller value.

For example, 1000000000000000000... with nine 1s.

But this would be a very long number, which is larger.

Therefore, no.

Therefore, I think I can conclude that the answer is 11111111100.

But to make sure, let me check if there is a number with digit sum 9, ending with 00, and smaller than 11111111100.

For example, let's think of numbers like 1000000001100 (which is 1,000,000,001,100) or 100000001100 (which is 1,000,000,011,00).

But these are all larger.

Alternatively, numbers like 100000011100 (which is 100,000,011,100).

Still larger.

Therefore, no.

Therefore, the answer is 11111111100.

But let me check if there is a number with digit sum 9, ending with 00, and with the digits arranged in a way that it's smaller.

For example, 10101010100.

Wait, let's check the digit sum: 1+0+1+0+1+0+1+0+1+0+0 = 5. Not enough.

If I add more 1s: 1010101010100. Digit sum is 6. Still not enough.

To get digit sum 9, need nine 1s.

For example, 1010101010101010100. This has nine 1s.

But this number is 1010101010101010100, which is 101,010,101,010,101,0100. Which is way larger.

Therefore, no improvement.

Therefore, I think I can conclude that the smallest positive multiple of 450 with all digits 0 or 1 is 11111111100.

**Final Answer**
The smallest positive multiple of 450 with all digits 0 or 1 is \boxed{11111111100}.

To solve the problem of finding the **smallest positive multiple of 450** whose **digits are all zeroes and ones**, we begin by analyzing the properties of the number 450 and the constraints on the digits.

---

### Step 1: Prime Factorization of 450

We start by factoring 450:

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

This tells us that any multiple of 450 must be divisible by:

- **2** (i.e., the number must be even),
- **9** (i.e., the sum of its digits must be divisible by 9),
- **25** (i.e., the number must end in **00**).

---

### Step 2: Constraints on the Number

Given that the number must consist **only of digits 0 and 1**, and must be divisible by 450, we can deduce the following:

- **Divisibility by 2 and 25** implies the number must end in **00**.
- **Divisibility by 9** implies the **sum of the digits** must be a multiple of 9.
- Since the digits are only 0s and 1s, the **sum of the digits is equal to the number of 1s**.
- Therefore, the number must contain **exactly 9 ones** (the smallest multiple of 9).

---

### Step 3: Constructing the Smallest Valid Number

We now seek the **smallest number** composed of only 0s and 1s, with **exactly 9 ones**, and **ending in 00**.

- The smallest number with 9 ones is **111111111** (nine 1s).
- To satisfy the divisibility by 25, we must append **00** to the end.
- Therefore, the number becomes:

$$
11111111100
$$

This number:
- Has **9 ones** (sum = 9, divisible by 9),
- Ends in **00** (divisible by 25),
- Is **even** (divisible by 2),
- Therefore, it is divisible by **450**.

---

### Step 4: Verifying Minimality

We must ensure that **no smaller number** satisfies all the conditions.

- Any number with **fewer than 9 ones** would not be divisible by 9.
- Any number with **more than 9 ones** would be larger.
- Any number with **fewer than 11 digits** (i.e., less than 9 digits + 2 zeros) would not have enough digits to accommodate 9 ones and the required ending of 00.
- Any number with **more digits** would be larger than 11111111100.

Thus, **11111111100** is the **smallest** such number.

---

### 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.