<Question>
What is the least positive integer multiple of 30 that can be written with only the digits 0 and 2?
</Question>

<think>
<Plan_1>
Okay, so I need to find the smallest positive integer multiple of 30 that can be written using only the digits 0 and 2. Let me start by breaking down the problem step by step.

<prompt_1.1> Understand the problem requirements and identify the key aspects that need to be considered. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let's recall that a multiple of 30 must satisfy the divisibility rules for 30. Since 30 is 3 multiplied by 10, the number has to be divisible by both 3 and 10. Divisibility by 10 is straightforward: the number must end with a 0. That means whatever number we come up with has to have a 0 as its last digit. 

Next, divisibility by 3 requires that the sum of the digits is a multiple of 3. Since the only digits we can use are 0 and 2, the sum of the digits will be 2 times the number of 2s in the number. So, 2 times the count of 2s must be divisible by 3. Let's denote the number of 2s as n. Then, 2n ≡ 0 mod 3. Solving for n, we find that n must be a multiple of 3 because 2 and 3 are coprime. So, the number of 2s in the number has to be 3, 6, 9, etc.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Our task is to find the smallest such number. Since we want the smallest number, we should aim for the least number of digits possible. The more digits a number has, the larger it is, assuming we're dealing with numbers of the same length. However, since leading zeros are not allowed, the length is fixed once we determine the number of digits.

Let's start with the smallest possible number of digits. The number must end with 0, so the last digit is fixed. The rest of the digits can be 2s or 0s. But we also need the sum of the digits to be a multiple of 3. Let's start with the smallest possible numbers and check them.

<prompt_2.1> Start with the smallest possible numbers and check them for the required conditions. </prompt_2.1>

Let me verify that 2220 is indeed divisible by 30.

<prompt_2.2> Verify that 2220 is divisible by 30. </prompt_2.2>

But wait, is there a smaller number? Let's check numbers with fewer digits. The smallest possible multiple of 30 is 30 itself, but 30 contains a digit 3, which is not allowed. The next one is 60, which has a 6, also not allowed. 90 has a 9, which isn't permitted. 120 has a 1, which is not allowed. 150 has a 1 and a 5. 180 has an 8. 210 has a 1. 240 has a 4. 270 has a 7. 300 has a 3. All of these have digits other than 0 and 2. 

So moving to four-digit numbers. The smallest four-digit number ending with 0 and composed of 2s and 0s is 2000. Let's check the sum of digits: 2 + 0 + 0 + 0 = 2.
Not divisible by 3. Next would be 2020: sum is 2 + 0 + 2 + 0 = 4. Not divisible by 3. Then 2002: sum is 2 + 0 + 0 + 2 = 4. Also not. Then 2200: sum is 2 + 2 + 0 + 0 = 4. Still not. Then 2220: sum is 6, which works. 

But wait, maybe there's a four-digit number with three 2s and one 0. But since the last digit must be 0, the other digits can be 2s. So 2220 is the only four-digit number ending with 0 and having three 2s. So that's the first candidate.
</Plan_2>

<Execution_2>
<execution_2.1>
Case 1: Two-digit number. The number must end with 0, so the possibilities are 20, 00, but 00 is not a positive integer. 20 is a candidate. Let's check the sum of digits: 2 + 0 = 2. 2 is not divisible by 3, so 20 is not divisible by 3. Therefore, 20 is not a solution.

Case 2: Three-digit number. The number must end with 0. So the possible numbers are 220, 202, and 022 (but 022 is not a three-digit number). Let's check their digit sums. 

- 220: 2 + 2 + 0 = 4. Not divisible by 3.
- 202: 2 + 0 + 2 = 4. Also not divisible by 3.
</execution_2.1>

<execution_2.2>
2220: The digits are two 2s, one 2, and a 0. Wait, no, let's count the digits. 2220 has three 2s and one 0. So the sum of the digits is 2 + 2 + 2 + 0 = 6, which is divisible by 3. Since it ends with a 0, it's divisible by 10. Therefore, it's divisible by 30.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2:
But let me check if there's a smaller three-digit number. Wait, three-digit numbers can't have more than three digits. The sum of digits needed is a multiple of 3. For three-digit numbers ending with 0, the first two digits can be 2s or 0s. The maximum sum would be 2 + 2 + 0 = 4, which is less than 6.
So three-digit numbers can't have a digit sum of 6. Therefore, the smallest possible number must have four digits.

Wait, but hold on. Let's verify again.

<prompt_3.1> Verify the solution by checking if there is a smaller number with four digits. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
If we have a three-digit number, the maximum sum of digits (other than 0s) would be 2 + 2 + 0 = 4. So, as I thought, not enough. So three-digit numbers are out. Then four-digit numbers. The smallest four-digit number ending with 0 is 2000, but as we saw, the digit sum is 2. Then 2020: sum 4, 2200: sum 4, 2220: sum 6. So 2220 is the first four-digit number that meets the criteria. 

But is there a four-digit number with three 2s and one 0 that is smaller than 2220? Let's think. If the number ends with 0, the first three digits must be 2s. So 2220 is the only one. If we try to place the 0 earlier, like 2020 or 2200, but those have only two 2s, which gives a digit sum of 4, not divisible by 3. So 2220 seems to be the smallest four-digit number.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let me check if there's a five-digit number that's smaller in value but has more digits. Wait, no. A five-digit number is larger than a four-digit number, so even if the digits are smaller, the overall number is bigger. For example, 10000 is larger than 9999, but in our case, the numbers are composed of 2s and 0s, so 2220 (four digits) is smaller than any five-digit number like 20000 or 22200.

But let me confirm if 2220 is indeed the minimal. Let's see if there's a four-digit number with three 2s and a 0 that is smaller than 2220. The thousands place must be 2 (since it's a four-digit number), then the hundreds, tens, and units places. The units place has to be 0. So the thousands digit is 2, the units digit is 0, and the hundreds and tens digits can be 2s.

<prompt_4.1> Confirm if 2220 is the minimal solution by checking all possibilities. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
So the number would be 2220. If we try to have the hundreds digit as 0, then the number would be 2020, but the sum of digits is 2 + 0 + 2 + 0 = 4, which isn't divisible by 3. Similarly, 2200 has a digit sum of 4. So only 2220 meets the digit sum requirement. 

Therefore, 2220 is the number. But let me check once more if there is a smaller number with more digits but smaller in value. Wait, no. For example, 20220 is a five-digit number, but it's larger than 2220. So 2220 is the smallest.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But just to be thorough, let's check if there's a number with three 2s and one 0 in a different arrangement that's smaller. For example, 2020, but as mentioned, the digit sum is 4. 2200: sum 4. 2002: ends with 2, which is invalid because it needs to end with 0. So no. Therefore, 2220 is the only four-digit number that fits.

Wait another thought: maybe a number with six digits? For example, 222000. But 2220 is already four digits, which is smaller. So even if there's a six-digit number with three 2s and three 0s, it would be larger than 2220.

Therefore, after considering all possibilities, 2220 seems to be the smallest multiple of 30 that can be written with only the digits 0 and 2.

But let me verify once again. Let's list the multiples of 30 and check each one:

<prompt_5.1> Re-verify the solution by listing multiples of 30 and checking each one. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
30: 30 (has a 3) → invalid
60: 60 (has a 6) → invalid
90: 90 (has a 9) → invalid
120: 120 (has 1 and 2) → invalid (1 is not allowed)
150: 150 (1 and 5) → invalid
180: 180 (1 and 8) → invalid
210: 210 (2, 1, 0) → invalid (1 is not allowed)
240: 240 (2, 4, 0) → invalid (4 is not allowed)
270: 270 (2, 7, 0) → invalid (7 is not allowed)
300: 300 (3 and 0s) → invalid
330: 330 (3s and 0) → invalid
360: 360 (3, 6, 0) → invalid
... and so on. 

All these numbers have digits other than 0 and 2. The next multiple would be 390, 420, etc. , all of which have invalid digits. 

So moving to numbers composed of 2s and 0s. The smallest such number divisible by 30. Since it must end with 0, let's consider numbers like 20, 220, 2220, etc. 

20: sum of digits 2 → not divisible by 3
220: sum 4 → not divisible by 3
2220: sum 6 → divisible by 3. So yes, 2220 is the first one.
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, the least positive integer multiple of 30 that can be written with only the digits 0 and 2 is $\boxed{2220}$.
</Final_answer>
</think>