Explanation:
Okay, so I need to find the least positive integer multiple of 30 that can be written using only the digits 0 and 2. Let me think about how to approach this.

First, let me recall that a number is a multiple of 30 if it is divisible by both 3 and 10. Since 30 is 3 times 10, right? So, for a number to be divisible by 10, it must end with a 0. That's straightforward. So whatever number I come up with has to end with a 0. Also, since it's a multiple of 3, the sum of its digits must be a multiple of 3.

But the number can only have digits 0 and 2. So, all digits are either 0 or 2. And it has to end with a 0. So, the last digit is 0. So, the number is something like 2...20, where there are some number of 2s followed by a 0.

So, the problem reduces to finding the smallest number composed of 2s and 0s, ending with 0, such that the sum of its digits is divisible by 3.

Let me think. Let's denote the number as having n digits. Since it ends with 0, the last digit is 0. The rest of the digits can be 0 or 2. But to make the number as small as possible, we want the fewest digits possible.

But since the number must be divisible by 3, the sum of digits must be divisible by 3. Let me think about the sum. Let's say there are k digits of 2 in the number. Since the last digit is 0, the rest of the digits (excluding the last 0) can be 2s and 0s. So, the total sum of digits is 2k. Therefore, 2k must be divisible by 3.

So, 2k ≡ 0 mod 3 ⇒ k ≡ 0 mod 3. Therefore, k must be a multiple of 3.

Therefore, the number of 2s in the number must be a multiple of 3.

So, the minimal number of 2s is 3. Therefore, the minimal number would be 2220? Let me check.

Wait, 2220. Let me check if that's divisible by 30.

First, it ends with 0, so divisible by 10. Sum of digits: 2+2+2+0 = 6, which is divisible by 3. Therefore, 2220 is divisible by 30.

But is there a smaller number? Let me check.

Wait, maybe there is a smaller number with fewer digits. Let me think.

The number must end with 0, so the minimal number of digits is 2? But a two-digit number ending with 0 would be 20. Let's check 20. Sum of digits is 2 + 0 = 2, which is not divisible by 3. So 20 is not divisible by 3.

Next, three-digit numbers. The number must end with 0. So possible numbers are 200, 220. Let's check their digit sums.

200: 2 + 0 + 0 = 2, not divisible by 3.

220: 2 + 2 + 0 = 4, not divisible by 3.

So no luck there.

Four-digit numbers. The number ends with 0. So possible numbers: 2000, 2020, 2200, 2220.

Check their digit sums:

2000: 2 + 0 + 0 + 0 = 2 → nope.

2020: 2 + 0 + 2 + 0 = 4 → nope.

2200: 2 + 2 + 0 + 0 = 4 → nope.

2220: 2 + 2 + 2 + 0 = 6 → yes! Divisible by 3.

So 2220 is a four-digit number.

But wait, is there a smaller number? Let me check if there is a number with more digits but smaller in value? Wait, no. Because numbers with more digits are larger. For example, a five-digit number would be larger than a four-digit number. So 2220 is the smallest so far.

But let me check if there is a smaller number with more digits but with leading zeros? Wait, no. Leading zeros are not allowed in numbers. So numbers can't have leading zeros. So the first digit must be 2.

Wait, but maybe there is a number with more digits but with fewer digits of 2? Wait, but earlier we saw that the number of 2s must be a multiple of 3. So if we have 3 digits of 2, that's the minimal. So 2220 is the minimal.

But let me think again. Maybe there is a number with more digits but with some zeros in between? For example, 2020? But we saw that 2020 has digit sum 4, which is not divisible by 3. Similarly, 20020? Wait, but that's a five-digit number. Let me check.

Wait, 20020: digits are 2, 0, 0, 2, 0. Sum is 2 + 0 + 0 + 2 + 0 = 4. Not divisible by 3.

How about 20220? Let's check. Sum is 2 + 0 + 2 + 2 + 0 = 6. Divisible by 3. So 20220 is a five-digit number. But 20220 is larger than 2220. So 2220 is smaller.

Alternatively, 22020? Sum is 2 + 2 + 0 + 2 + 0 = 6. Also divisible by 3. But again, 22020 is larger than 2220.

So 2220 is smaller.

But let me check if there is a number with more digits but smaller in value. Wait, for example, 200...0 with some digits. But if we have more digits, the number is longer, so it's larger. For example, 2000 is smaller than 2220? Wait, 2000 is 2000, which is less than 2220. But 2000's digit sum is 2, which is not divisible by 3. So 2000 is not acceptable.

Similarly, 2020 is 2020, which is less than 2220, but digit sum is 4. Not divisible by 3.

So, 2220 is the first one that works.

But wait, let me check if there is a number with more digits but with leading 2s and zeros arranged in a way that the number is smaller. For example, 2020 is smaller than 2220, but it doesn't work. How about 2002? Wait, but that ends with 2, not 0. So it's not divisible by 10.

So, numbers must end with 0. So, the last digit is 0. So, numbers like 20, 200, 2000, etc. But as we saw, the digit sums are not divisible by 3.

So, the next possible is 2220.

But let me think again. Maybe there is a number with more digits but with fewer digits? Wait, no. The number of digits is determined by the number of 2s and zeros.

Wait, but maybe there is a number with more digits but with some 2s and zeros arranged such that the number is smaller? For example, 2020 is smaller than 2220, but it doesn't work. How about 20020? That's 20020, which is larger than 2220.

Alternatively, 20220 is larger than 2220.

So, seems like 2220 is the smallest.

But let me check if there is a number with 3 digits of 2 and some zeros, but arranged in a way that the number is smaller. For example, 2020? But that has only two 2s. Wait, no. Wait, 2220 has three 2s.

Wait, if we have three 2s and one 0, the number is 2220. If we have three 2s and more zeros, like 2020? No, that has only two 2s. Wait, no. Wait, 2020 has two 2s.

Wait, so if we have three 2s, the minimal number would be 2220. If we have more 2s, like six 2s, then the number would be longer, so larger.

Alternatively, maybe arranging the digits with zeros in between? For example, 20220? But that's five digits.

But 2220 is four digits. So, 2220 is smaller.

Wait, but let me check if there is a number with three 2s and one 0, but arranged in a different way. For example, 2022? But that ends with 2, not 0. So not divisible by 10.

Or 2202? Ends with 2. Not good.

Or 2220. Ends with 0. So that's the only one.

So, seems like 2220 is the minimal.

But let me check if there is a smaller number with more digits. For example, maybe 200... something?

Wait, for example, 200... but with more digits. Let me think.

Suppose we have a number with 6 digits. Let's say 200220. Let me check the digit sum: 2 + 0 + 0 + 2 + 2 + 0 = 6. Divisible by 3. And it ends with 0. So 200220 is divisible by 30. But 200220 is larger than 2220.

Alternatively, 202020. Same digit sum. Also larger.

So, no.

Alternatively, maybe a number with 3 digits of 2 and some zeros, but arranged in a way that the number is smaller. But since 2220 is four digits, and any number with more digits is longer, so larger.

Wait, but what about numbers with leading zeros? But numbers can't have leading zeros. So, the first digit has to be 2.

So, the minimal number is 2220.

But let me check if there is a number with fewer digits. For example, 2220 is four digits. Is there a three-digit number? Let me check again.

Three-digit numbers ending with 0: 200, 210, 220. But digits can only be 0 and 2. So 200, 220.

Sum of digits for 200: 2, not divisible by 3.

Sum for 220: 4, not divisible by 3.

So no.

Two-digit numbers: 20, which is 2, not divisible by 3.

One-digit numbers: 0, but 0 is not positive.

So, yes, 2220 is the minimal.

Wait, but let me think again. Maybe there is a number with more digits but with smaller value? For example, 200... but with more digits. Wait, for example, 2000... but with more zeros. But 2000 is smaller than 2220? Wait, 2000 is 2000, which is less than 2220. But 2000 is not divisible by 3.

But if we can find a number with more digits but with digits 0 and 2, ending with 0, and sum of digits divisible by 3.

For example, let's think of numbers with 6 digits. Let me think.

If we have 6 digits, ending with 0. So, the first five digits can be 0s and 2s. The sum of digits is 2k, where k is the number of 2s. So, 2k must be divisible by 3. So k must be multiple of 3.

So, for example, if k=3, then the number has three 2s and two 0s (since total digits excluding the last 0 is 5 digits). Wait, total digits is 6. Last digit is 0, so first five digits have three 2s and two 0s.

So, the number would be something like 202020. Which is 202020. Let me check if that's smaller than 2220? No, 202020 is a six-digit number, which is way larger than 2220.

Alternatively, maybe arranging the digits to make the number smaller. For example, 200220. Which is 200220. Still larger than 2220.

Alternatively, 200022? Ends with 2, so not allowed.

So, no.

Alternatively, maybe with k=6? Then the number would have six 2s and one 0. So, the number would be something like 2222220. Which is even larger.

So, no.

Therefore, seems like 2220 is the minimal.

But let me check if there is a number with more digits but with leading zeros? No, leading zeros are not allowed.

Wait, but maybe there is a number with more digits but with fewer digits? No, more digits mean longer numbers, which are larger.

Wait, but maybe there is a number with digits arranged in a way that it's smaller? For example, 2020 is smaller than 2220, but it doesn't work.

Alternatively, 2002? Ends with 2, not 0.

So, no.

Alternatively, 200... with more digits.

Wait, let me think of another approach.

Since the number must be divisible by 30, it must be divisible by 10 and 3.

Divisible by 10: last digit is 0.

Divisible by 3: sum of digits is multiple of 3.

Digits are only 0 and 2.

So, the number is of the form 2...20, where there are some number of 2s followed by a 0.

But the number can have zeros in between. For example, 2020.

But the key is that the sum of digits is 2k, where k is the number of 2s.

So, 2k must be divisible by 3. Therefore, k must be divisible by 3.

Therefore, the minimal k is 3. Therefore, the minimal number is 2220.

But let me check if there is a number with k=3 but with more zeros, but arranged in a way that the number is smaller.

For example, 20220? Wait, that's five digits. But 20220 is larger than 2220.

Alternatively, 20022? Ends with 2, not 0.

Or 20202? Ends with 2.

Or 20002? Ends with 2.

So, no.

Alternatively, 20220 is five digits.

So, 2220 is four digits.

Therefore, 2220 is smaller.

Therefore, 2220 is the minimal.

But let me check if there is a number with k=3 and fewer digits.

Wait, k=3 means three 2s and one 0 (since the number must end with 0). So, the number has four digits.

So, the number is 2220.

Alternatively, if there are more zeros, but the number would have more digits.

So, for example, 20220 is five digits.

So, 2220 is the minimal.

But let me check if there is a number with k=3 but with leading zeros? No, leading zeros are not allowed.

Therefore, 2220 is the minimal.

But wait, let me check if there is a number with k=3 but with different arrangement. For example, 20220? But that's five digits.

Alternatively, 22020? Also five digits.

But 2220 is four digits.

So, 2220 is smaller.

Therefore, I think 2220 is the answer.

But let me check if there is a smaller number with more digits. For example, 200... but with more digits.

Wait, for example, 200... but with more digits. Let me think.

Suppose we have a number with 6 digits. Let me think of numbers with 3 2s and 3 0s, ending with 0.

So, the first five digits have three 2s and two 0s.

So, the number would be something like 200220.

But 200220 is 200,220. Which is larger than 2220.

Alternatively, 202020. Also larger.

So, no.

Alternatively, maybe with more 2s? For example, k=6. Then the number would have six 2s and one 0. So, 2222220. Which is way larger.

So, no.

Therefore, 2220 is the minimal.

But let me check if there is a number with k=3 but with leading zeros. But no, leading zeros are not allowed.

Alternatively, maybe there is a number with k=3 and some zeros in between, but arranged to make the number smaller.

For example, 20220 is five digits. But 2220 is four digits.

So, 2220 is smaller.

Therefore, I think 2220 is the answer.

But let me check if there is a number with k=3 and fewer digits.

Wait, k=3 means three 2s. Since the number must end with 0, the total number of digits is 3 + 1 = 4. So, 2220.

Therefore, there is no smaller number.

But let me check if there is a number with k=3 but with leading zeros. But no.

Alternatively, maybe there is a number with k=3 and some zeros in between, but with fewer digits?

Wait, no. If you have three 2s and one 0, you need four digits.

So, 2220 is the only four-digit number with three 2s and one 0.

Therefore, I think 2220 is the answer.

But let me check with another approach.

Let me think of numbers made up of 0s and 2s, ending with 0, and check divisibility by 3.

Start from the smallest possible numbers:

- 20: sum 2 → not divisible by 3.

- 200: sum 2 → nope.

- 2020: sum 4 → nope.

- 2000: sum 2 → nope.

- 20220: sum 6 → yes. But 20220 is larger than 2220.

- 20020: sum 4 → nope.

- 2200: sum 4 → nope.

- 22020: sum 6 → yes. But again, larger than 2220.

- 2220: sum 6 → yes.

So, 2220 is the first one that works.

Therefore, the answer is 2220.

But wait, let me check if there is a number between 2000 and 2220 that meets the criteria.

For example, numbers like 200... but with more digits.

Wait, numbers between 2000 and 2220 would be four-digit numbers starting with 2, then digits.

So, numbers from 2000 to 2220.

We already checked 2000, 2020, 2200, 2220.

Other numbers in between: 2002 (ends with 2, invalid), 2022 (ends with 2), 2000, 2002, 2002, 2020, 2022, 2020, 2022, 2020, 2200, 2202, 2202, 2200, 2220.

So, all numbers between 2000 and 2220 that end with 0 are 2000, 2020, 2200, 2220.

We checked those, and only 2220 has digit sum divisible by 3.

Therefore, 2220 is indeed the minimal.

Therefore, I think the answer is 2220.

But let me check once more.

Is there a number with more digits but smaller in value? For example, 200... but with more digits.

Wait, for example, 200... but with more digits. Let me think.

Suppose we have a number like 200...0 with more zeros. But 200...0 is 2 followed by some zeros. The digit sum is 2, which is not divisible by 3.

If we add more 2s. For example, 2020... but with more digits.

But for example, 2020 is 2020, which is less than 2220? No, 2020 is less than 2220. Wait, 2020 is 2020, which is less than 2220. But 2020's digit sum is 2 + 0 + 2 + 0 = 4, not divisible by 3.

So, if we can find a number between 2000 and 2220 that ends with 0, has digits only 0 and 2, and sum divisible by 3.

But the numbers are 2000, 2020, 2200, 2220.

As checked before, only 2220 works.

So, 2220 is the answer.

Therefore, I think the answer is 2220.

But let me check if there is a number with more digits but smaller in value. For example, 200... but with more digits.

Wait, for example, 20020 is 20020, which is larger than 2220.

Similarly, 2020 is 2020, which is less than 2220. But 2020 doesn't work.

So, no.

Alternatively, 200... with more digits.

But 200... with more digits would be larger.

Therefore, 2220 is the minimal.

Therefore, I think the answer is 2220.

**Final Answer**
The least positive integer multiple of 30 that can be written with only the digits 0 and 2 is \boxed{2220}.

To solve the problem of finding the **least positive integer multiple of 30** that can be written using **only the digits 0 and 2**, we need to consider the following:

---

### **Understanding the Requirements**

A number is a **multiple of 30** if it is divisible by both **3** and **10**.

- **Divisibility by 10** requires the number to **end in 0**.
- **Divisibility by 3** requires the **sum of its digits** to be divisible by 3.

Additionally, the number must be composed **only of the digits 0 and 2**.

---

### **Strategy**

Let’s denote the number as a sequence of digits made up of only 0s and 2s, ending in 0 (to satisfy divisibility by 10). Let’s denote the number of 2s in the number as $ k $. Since the last digit is 0, the total number of digits is $ k + 1 $.

The **sum of the digits** is $ 2k $, because each 2 contributes 2 to the sum, and the 0 contributes nothing. For the number to be divisible by 3, we require:

$$
2k \equiv 0 \pmod{3} \Rightarrow k \equiv 0 \pmod{3}
$$

So, the number of 2s must be a **multiple of 3**.

---

### **Finding the Smallest Valid Number**

We now look for the **smallest number** that satisfies:

- Ends in 0
- Composed only of 0s and 2s
- Has a number of 2s that is a multiple of 3

Let’s test the smallest such values:

- **k = 3** (smallest multiple of 3):
  The number has 3 twos and one zero. The smallest such number is **2220**.

Check:
- Ends in 0 ✅
- Digits are only 0 and 2 ✅
- Sum of digits: $ 2 + 2 + 2 + 0 = 6 $, which is divisible by 3 ✅

So, **2220** is a valid number.

---

### **Is There a Smaller Number?**

Let’s check smaller numbers:

- **20** → sum = 2 → not divisible by 3 ❌
- **200** → sum = 2 → not divisible by 3 ❌
- **2020** → sum = 4 → not divisible by 3 ❌
- **2200** → sum = 4 → not divisible by 3 ❌
- **2220** → sum = 6 → divisible by 3 ✅

All smaller numbers either:
- Don’t end in 0 (not divisible by 10), or
- Have digit sums not divisible by 3

Thus, **2220** is the **smallest** such number.

---

### **Conclusion**

The least positive integer multiple of 30 that can be written using only the digits 0 and 2 is:

$$
\boxed{2220}
$$

[Assistant]
Explanation:
Okay, so I need to find the least positive integer multiple of 30 that can be written using only the digits 0 and 2. Let me think about how to approach this.

First, let me recall that a number is a multiple of 30 if it is divisible by both 3 and 10. Since 30 is 3 times 10, right? So, for a number to be divisible by 10, it must end with a 0. That's straightforward. So whatever number I come up with has to end with a 0. Also, since it's a multiple of 3, the sum of its digits must be a multiple of 3.

But the number can only have digits 0 and 2. So, all digits are either 0 or 2. And it has to end with a 0. So, the last digit is 0. So, the number is something like 2...20, where there are some number of 2s followed by a 0.

So, the problem reduces to finding the smallest number composed of 2s and 0s, ending with 0, such that the sum of its digits is divisible by 3.

Let me think. Let's denote the number as having n digits. Since it ends with 0, the last digit is 0. The rest of the digits can be 0 or 2. But to make the number as small as possible, we want the fewest digits possible.

But since the number must be divisible by 3, the sum of digits must be divisible by 3. Let me think about the sum. Let's say there are k digits of 2 in the number. Since the last digit is 0, the rest of the digits (excluding the last 0) can be 2s and 0s. So, the total sum of digits is 2k. Therefore, 2k must be divisible by 3.

So, 2k ≡ 0 mod 3 ⇒ k ≡ 0 mod 3. Therefore, k must be a multiple of 3.

Therefore, the number of 2s in the number must be a multiple of 3.

So, the minimal number of 2s is 3. Therefore, the minimal number would be 2220? Let me check.

Wait, 2220. Let me check if that's divisible by 30.

First, it ends with 0, so divisible by 10. Sum of digits: 2+2+2+0 = 6, which is divisible by 3. Therefore, 2220 is divisible by 30.

But is there a smaller number? Let me check.

Wait, maybe there is a smaller number with fewer digits. Let me think.

The number must end with 0, so the minimal number of digits is 2? But a two-digit number ending with 0 would be 20. Let's check 20. Sum of digits is 2 + 0 = 2, which is not divisible by 3. So 20 is not divisible by 3.

Next, three-digit numbers. The number must end with 0. So possible numbers are 200, 220. Let's check their digit sums.

200: 2 + 0 + 0 = 2, not divisible by 3.

220: 2 + 2 + 0 = 4, not divisible by 3.

So no luck there.

Four-digit numbers. The number ends with 0. So possible numbers: 2000, 2020, 2200, 2220.

Check their digit sums:

2000: 2 + 0 + 0 + 0 = 2 → nope.

2020: 2 + 0 + 2 + 0 = 4 → nope.

2200: 2 + 2 + 0 + 0 = 4 → nope.

2220: 2 + 2 + 2 + 0 = 6 → yes! Divisible by 3.

So 2220 is a four-digit number.

But wait, is there a smaller number? Let me check if there is a number with more digits but smaller in value? Wait, no. Because numbers with more digits are larger. For example, a five-digit number would be larger than a four-digit number. So 2220 is the smallest so far.

But let me check if there is a smaller number with more digits but with leading zeros? Wait, no. Leading zeros are not allowed in numbers. So numbers can't have leading zeros. So the first digit must be 2.

Wait, but maybe there is a number with more digits but with fewer digits of 2? Wait, but earlier we saw that the number of 2s must be a multiple of 3. So if we have 3 digits of 2, that's the minimal. So 2220 is the minimal.

But let me think again. Maybe there is a number with more digits but with some zeros in between? For example, 2020? But we saw that 2020 has digit sum 4, which is not divisible by 3. Similarly, 20020? Wait, but that's a five-digit number. Let me check.

Wait, 20020: digits are 2, 0, 0, 2, 0. Sum is 2 + 0 + 0 + 2 + 0 = 4. Not divisible by 3.

How about 20220? Let's check. Sum is 2 + 0 + 2 + 2 + 0 = 6. Divisible by 3. So 20220 is a five-digit number. But 20220 is larger than 2220. So 2220 is smaller.

Alternatively, 22020? Sum is 2 + 2 + 0 + 2 + 0 = 6. Also divisible by 3. But again, 22020 is larger than 2220.

So 2220 is smaller.

But let me check if there is a number with more digits but smaller in value. Wait, for example, 200...0 with some digits. But if we have more digits, the number is longer, so it's larger. For example, 2000 is smaller than 2220? Wait, 2000 is 2000, which is less than 2220. But 2000's digit sum is 2, which is not divisible by 3. So 2000 is not acceptable.

Similarly, 2020 is 2020, which is less than 2220, but digit sum is 4. Not divisible by 3.

So, 2220 is the first one that works.

But wait, let me check if there is a number with more digits but with leading 2s and zeros arranged in a way that the number is smaller. For example, 2020 is smaller than 2220, but it doesn't work. How about 2002? Wait, but that ends with 2, not 0. So it's not divisible by 10.

So, numbers must end with 0. So, the last digit is 0. So, numbers like 20, 200, 2000, etc. But as we saw, the digit sums are not divisible by 3.

So, the next possible is 2220.

But let me think again. Maybe there is a number with more digits but with fewer digits? Wait, no. The number of digits is determined by the number of 2s and zeros.

Wait, but maybe there is a number with more digits but with some 2s and zeros arranged such that the number is smaller? For example, 2020 is smaller than 2220, but it doesn't work. How about 20020? That's 20020, which is larger than 2220.

Alternatively, 20220 is larger than 2220.

So, seems like 2220 is the smallest.

But let me check if there is a number with 3 digits of 2 and some zeros, but arranged in a way that the number is smaller. For example, 2020? But that has only two 2s. Wait, no. Wait, 2220 has three 2s.

Wait, if we have three 2s and one 0, the number is 2220. If we have three 2s and more zeros, like 2020? No, that has only two 2s. Wait, no. Wait, 2020 has two 2s.

Wait, so if we have three 2s, the minimal number would be 2220. If we have more 2s, like six 2s, then the number would be longer, so larger.

Alternatively, maybe arranging the digits with zeros in between? For example, 20220? But that's five digits.

But 2220 is four digits. So, 2220 is smaller.

Wait, but let me check if there is a number with three 2s and one 0, but arranged in a different way. For example, 2022? But that ends with 2, not 0. So not divisible by 10.

Or 2202? Ends with 2. Not good.

Or 2220. Ends with 0. So that's the only one.

So, seems like 2220 is the minimal.

But let me check if there is a smaller number with more digits. For example, maybe 200... something?

Wait, for example, 200... but with more digits. Let me think.

Suppose we have a number with 6 digits. Let's say 200220. Let me check the digit sum: 2 + 0 + 0 + 2 + 2 + 0 = 6. Divisible by 3. And it ends with 0. So 200220 is divisible by 30. But 200220 is larger than 2220.

Alternatively, 202020. Same digit sum. Also larger.

So, no.

Alternatively, maybe a number with 3 digits of 2 and some zeros, but arranged in a way that the number is smaller. But since 2220 is four digits, and any number with more digits is longer, so larger.

Wait, but what about numbers with leading zeros? But numbers can't have leading zeros. So, the first digit has to be 2.

So, the minimal number is 2220.

But let me check if there is a number with fewer digits. For example, 2220 is four digits. Is there a three-digit number? Let me check again.

Three-digit numbers ending with 0: 200, 210, 220. But digits can only be 0 and 2. So 200, 220.

Sum of digits for 200: 2, not divisible by 3.

Sum for 220: 4, not divisible by 3.

So no.

Two-digit numbers: 20, which is 2, not divisible by 3.

One-digit numbers: 0, but 0 is not positive.

So, yes, 2220 is the minimal.

Wait, but let me think again. Maybe there is a number with more digits but with smaller value? For example, 200... but with more digits. Wait, for example, 2000... but with more zeros. But 2000 is smaller than 2220? Wait, 2000 is 2000, which is less than 2220. But 2000 is not divisible by 3.

But if we can find a number with more digits but with digits 0 and 2, ending with 0, and sum of digits divisible by 3.

For example, let's think of numbers with 6 digits. Let me think.

If we have 6 digits, ending with 0. So, the first five digits can be 0s and 2s. The sum of digits is 2k, where k is the number of 2s. So, 2k must be divisible by 3. So k must be multiple of 3.

So, for example, if k=3, then the number has three 2s and two 0s (since total digits excluding the last 0 is 5 digits). Wait, total digits is 6. Last digit is 0, so first five digits have three 2s and two 0s.

So, the number would be something like 202020. Which is 202020. Let me check if that's smaller than 2220? No, 202020 is a six-digit number, which is way larger than 2220.

Alternatively, maybe arranging the digits to make the number smaller. For example, 200220. Which is 200220. Still larger than 2220.

Alternatively, 200022? Ends with 2, so not allowed.

So, no.

Alternatively, maybe with k=6? Then the number would have six 2s and one 0. So, the number would be something like 2222220. Which is even larger.

So, no.

Therefore, seems like 2220 is the minimal.

But let me check if there is a number with more digits but with leading zeros? No, leading zeros are not allowed.

Wait, but maybe there is a number with more digits but with fewer digits? No, more digits mean longer numbers, which are larger.

Wait, but maybe there is a number with digits arranged in a way that it's smaller? For example, 2020 is smaller than 2220, but it doesn't work.

Alternatively, 2002? Ends with 2, not 0.

So, no.

Alternatively, 200... with more digits.

Wait, let me think of another approach.

Since the number must be divisible by 30, it must be divisible by 10 and 3.

Divisible by 10: last digit is 0.

Divisible by 3: sum of digits is multiple of 3.

Digits are only 0 and 2.

So, the number is of the form 2...20, where there are some number of 2s followed by a 0.

But the number can have zeros in between. For example, 2020.

But the key is that the sum of digits is 2k, where k is the number of 2s.

So, 2k must be divisible by 3. Therefore, k must be divisible by 3.

Therefore, the minimal k is 3. Therefore, the minimal number is 2220.

But let me check if there is a number with k=3 but with more zeros, but arranged in a way that the number is smaller.

For example, 20220? Wait, that's five digits. But 20220 is larger than 2220.

Alternatively, 20022? Ends with 2, not 0.

Or 20202? Ends with 2.

Or 20002? Ends with 2.

So, no.

Alternatively, 20220 is five digits.

So, 2220 is four digits.

Therefore, 2220 is smaller.

Therefore, 2220 is the minimal.

But let me check if there is a number with k=3 and fewer digits.

Wait, k=3 means three 2s and one 0 (since the number must end with 0). So, the number has four digits.

So, the number is 2220.

Alternatively, if there are more zeros, but the number would have more digits.

So, for example, 20220 is five digits.

So, 2220 is the minimal.

But let me check if there is a number with k=3 but with leading zeros? No, leading zeros are not allowed.

Therefore, 2220 is the minimal.

But wait, let me check if there is a number with k=3 but with different arrangement. For example, 20220? But that's five digits.

Alternatively, 22020? Also five digits.

But 2220 is four digits.

So, 2220 is smaller.

Therefore, I think 2220 is the answer.

But let me check if there is a smaller number with more digits. For example, 200... but with more digits.

Wait, for example, 200... but with more digits. Let me think.

Suppose we have a number with 6 digits. Let me think of numbers with 3 2s and 3 0s, ending with 0.

So, the first five digits have three 2s and two 0s.

So, the number would be something like 200220.

But 200220 is 200,220. Which is larger than 2220.

Alternatively, 202020. Also larger.

So, no.

Alternatively, maybe with more 2s? For example, k=6. Then the number would have six 2s and one 0. So, 2222220. Which is way larger.

So, no.

Therefore, 2220 is the minimal.

But let me check if there is a number with k=3 but with leading zeros. But no, leading zeros are not allowed.

Alternatively, maybe there is a number with k=3 and some zeros in between, but arranged to make the number smaller.

For example, 20220 is five digits. But 2220 is four digits.

So, 2220 is smaller.

Therefore, I think 2220 is the answer.

But let me check if there is a number with k=3 and fewer digits.

Wait, k=3 means three 2s. Since the number must end with 0, the total number of digits is 3 + 1 = 4. So, 2220.

Therefore, there is no smaller number.

But let me check if there is a number with k=3 but with leading zeros. But no.

Alternatively, maybe there is a number with k=3 and some zeros in between, but with fewer digits?

Wait, no. If you have three 2s and one 0, you need four digits.

So, 2220 is the only four-digit number with three 2s and one 0.

Therefore, I think 2220 is the answer.

But let me check with another approach.

Let me think of numbers made up of 0s and 2s, ending with 0, and check divisibility by 3.

Start from the smallest possible numbers:

- 20: sum 2 → not divisible by 3.

- 200: sum 2 → nope.

- 2020: sum 4 → nope.

- 2000: sum 2 → nope.

- 20220: sum 6 → yes. But 20220 is larger than 2220.

- 20020: sum 4 → nope.

- 2200: sum 4 → nope.

- 22020: sum 6 → yes. But again, larger than 2220.

- 2220: sum 6 → yes.

So, 2220 is the first one that works.

Therefore, the answer is 2220.

But wait, let me check if there is a number between 2000 and 2220 that meets the criteria.

For example, numbers like 200... but with more digits.

Wait, numbers between 2000 and 2220 would be four-digit numbers starting with 2, then digits.

So, numbers from 2000 to 2220.

We already checked 2000, 2020, 2200, 2220.

Other numbers in between: 2002 (ends with 2, invalid), 2022 (ends with 2), 2000, 2002, 2002, 2020, 2022, 2020, 2022, 2020, 2200, 2202, 2202, 2200, 2220.

So, all numbers between 2000 and 2220 that end with 0 are 2000, 2020, 2200, 2220.

We checked those, and only 2220 has digit sum divisible by 3.

Therefore, 2220 is indeed the minimal.

Therefore, I think the answer is 2220.

But let me check once more.

Is there a number with more digits but smaller in value? For example, 200... but with more digits.

Wait, for example, 200... but with more digits. Let me think.

Suppose we have a number like 200...0 with more zeros. But 200...0 is 2 followed by some zeros. The digit sum is 2, which is not divisible by 3.

If we add more 2s. For example, 2020... but with more digits.

But for example, 2020 is 2020, which is less than 2220? No, 2020 is less than 2220. Wait, 2020 is 2020, which is less than 2220. But 2020's digit sum is 2 + 0 + 2 + 0 = 4, not divisible by 3.

So, if we can find a number between 2000 and 2220 that ends with 0, has digits only 0 and 2, and sum divisible by 3.

But the numbers are 2000, 2020, 2200, 2220.

As checked before, only 2220 works.

So, 2220 is the answer.

Therefore, I think the answer is 2220.

But let me check if there is a number with more digits but smaller in value. For example, 200... but with more digits.

Wait, for example, 20020 is 20020, which is larger than 2220.

Similarly, 2020 is 2020, which is less than 2220. But 2020 doesn't work.

So, no.

Alternatively, 200... with more digits.

But 200... with more digits would be larger.

Therefore, 2220 is the minimal.

Therefore, I think the answer is 2220.

**Final Answer**
The least positive integer multiple of 30 that can be written with only the digits 0 and 2 is \boxed{2220}.

To solve the problem of finding the **least positive integer multiple of 30** that can be written using **only the digits 0 and 2**, we need to consider the following:

---

### **Understanding the Requirements**

A number is a **multiple of 30** if it is divisible by both **3** and **10**.

- **Divisibility by 10** requires the number to **end in 0**.
- **Divisibility by 3** requires the **sum of its digits** to be divisible by 3.

Additionally, the number must be composed **only of the digits 0 and 2**.

---

### **Strategy**

Let’s denote the number as a sequence of digits made up of only 0s and 2s, ending in 0 (to satisfy divisibility by 10). Let’s denote the number of 2s in the number as $ k $. Since the last digit is 0, the total number of digits is $ k + 1 $.

The **sum of the digits** is $ 2k $, because each 2 contributes 2 to the sum, and the 0 contributes nothing. For the number to be divisible by 3, we require:

$$
2k \equiv 0 \pmod{3} \Rightarrow k \equiv 0 \pmod{3}
$$

So, the number of 2s must be a **multiple of 3**.

---

### **Finding the Smallest Valid Number**

We now look for the **smallest number** that satisfies:

- Ends in 0
- Composed only of 0s and 2s
- Has a number of 2s that is a multiple of 3

Let’s test the smallest such values:

- **k = 3** (smallest multiple of 3):
  The number has 3 twos and one zero. The smallest such number is **2220**.

Check:
- Ends in 0 ✅
- Digits are only 0 and 2 ✅
- Sum of digits: $ 2 + 2 + 2 + 0 = 6 $, which is divisible by 3 ✅

So, **2220** is a valid number.

---

### **Is There a Smaller Number?**

Let’s check smaller numbers:

- **20** → sum = 2 → not divisible by 3 ❌
- **200** → sum = 2 → not divisible by 3 ❌
- **2020** → sum = 4 → not divisible by 3 ❌
- **2200** → sum = 4 → not divisible by 3 ❌
- **2220** → sum = 6 → divisible by 3 ✅

All smaller numbers either:
- Don’t end in 0 (not divisible by 10), or
- Have digit sums not divisible by 3

Thus, **2220** is the **smallest** such number.

---

### **Conclusion**

The least positive integer multiple of 30 that can be written using only the digits 0 and 2 is:

$$
\boxed{2220}
$$

Solution: Let $M$ be the least positive multiple of 30 that can be written with only the digits 0 and 2. First, $M$ is a multiple of 10, so its units digit must be 0. $M$ is also a multiple of 3, which means the sum of its digits must be a multiple of 3. Therefore, we must take at least  three 2's. Since $M$ is minimal, we take exactly three 2's and do not have any additional 0's: $M=\boxed{2220}$.


Explanation:
Okay, so I need to find the smallest possible value of a positive real number a such that all the roots of the cubic equation x³ + a x² + a x + 1 = 0 are real. Hmm, let me think about how to approach this.

First, I remember that for a cubic equation to have all real roots, certain conditions must be satisfied. Maybe I can use Vieta's formulas or some other properties of polynomials. Let me recall Vieta's formulas for a cubic equation. If the roots are r, s, t, then:

r + s + t = -a (since the coefficient of x² is a, and Vieta says sum of roots is -coefficient of x² divided by coefficient of x³, which is -a/1 = -a)

rs + rt + st = a (sum of products of roots two at a time is coefficient of x divided by coefficient of x³, which is a/1 = a)

rst = -1 (product of roots is -constant term divided by coefficient of x³, which is -1/1 = -1)

So, given that all roots are real, we need to find the minimal a > 0 such that these conditions hold.

But how do I relate these to find the minimal a? Maybe I can use some inequalities or consider the nature of the roots.

Alternatively, maybe I can factor the cubic equation. Let me try to factor it. Let me see if there's a rational root. By Rational Root Theorem, possible rational roots are ±1. Let me test x = -1:

Plugging x = -1 into the equation: (-1)^3 + a(-1)^2 + a(-1) + 1 = -1 + a - a + 1 = 0. Oh! So x = -1 is a root for any a. That's interesting. So the cubic can be factored as (x + 1)(quadratic). Let me perform the division or factorization.

Let me do polynomial division or use synthetic division. Let me use synthetic division with root -1.

Coefficients of the cubic: 1 (x³), a (x²), a (x), 1 (constant).

Set up synthetic division:

-1 | 1   a   a   1
    |     -1  -(a - 1)   -(a + 1)
    ----------------------------
      1  a - 1   - (a - 1) + a   1 - (a + 1)

Wait, let me do that step by step. The coefficients are 1, a, a, 1.

Bring down the 1.

Multiply by -1: 1 * (-1) = -1. Add to next coefficient: a + (-1) = a - 1.

Multiply by -1: (a - 1)(-1) = -a + 1. Add to next coefficient: a + (-a + 1) = 1.

Multiply by -1: 1 * (-1) = -1. Add to last coefficient: 1 + (-1) = 0. Perfect, so the division works.

Therefore, the cubic factors as (x + 1)(x² + (a - 1)x + 1). So the original equation is (x + 1)(x² + (a - 1)x + 1) = 0.

Therefore, the roots are x = -1 and the roots of x² + (a - 1)x + 1 = 0. Since we need all roots to be real, the quadratic must have real roots. Therefore, discriminant must be non-negative.

Discriminant D of quadratic is [ (a - 1) ]² - 4 * 1 * 1 ≥ 0.

So, (a - 1)² - 4 ≥ 0 ⇒ (a - 1)² ≥ 4 ⇒ |a - 1| ≥ 2 ⇒ a - 1 ≥ 2 or a - 1 ≤ -2 ⇒ a ≥ 3 or a ≤ -1.

But since a is a positive real number, a ≥ 3 or a ≤ -1. But a is positive, so only a ≥ 3. Therefore, the minimal value of a is 3? Wait, but hold on. Wait, but the problem says "all the roots are real". So if a ≥ 3, then quadratic has real roots, so all roots of the cubic are real. But is that the case?

Wait, but if a = 3, then quadratic is x² + (3 - 1)x + 1 = x² + 2x + 1 = (x + 1)^2. So the roots are -1, -1, -1. So triple root at -1. So all roots are real. So that's okay. But if a is less than 3, say a = 2, then discriminant is (2 - 1)^2 - 4 = 1 - 4 = -3 < 0, so quadratic has complex roots, so original equation would have one real root (-1) and two complex roots. Therefore, not all real. Therefore, the minimal a is 3? But wait, the problem says "all the roots are real", so the minimal a is 3? But wait, maybe there's a case where even if discriminant is negative, but... no, because if discriminant is negative, quadratic has complex roots. Therefore, the only way for all roots to be real is discriminant ≥ 0, so a ≥ 3. Therefore, minimal a is 3. But wait, the problem says "Find the smallest possible value of a." So answer is 3? But let me check again.

Wait, but maybe there's another way where the quadratic has real roots but with some other conditions? Wait, but the discriminant is the only condition for quadratic to have real roots. So if discriminant is non-negative, quadratic has real roots. Therefore, if a ≥ 3, then quadratic has real roots, so original equation has three real roots. Therefore, minimal a is 3. But wait, the problem says "all the roots are real", so maybe there's a case where even if discriminant is negative, but... no, because then two roots are complex. Therefore, the answer is 3? But I need to check if there's a possibility that even with discriminant negative, but maybe the cubic has three real roots? Wait, no. If the quadratic has complex roots, then the cubic has one real root (x = -1) and two complex roots. Therefore, not all real. Therefore, the minimal a is 3. But wait, the problem says "Find the smallest possible value of a." So answer is 3? But I have a feeling that maybe there's something wrong here. Let me check with a = 3. Then equation is x³ + 3x² + 3x + 1 = 0. Which factors as (x + 1)^3 = 0. So triple root at -1. All real. If a is less than 3, say a = 2, then equation is x³ + 2x² + 2x + 1 = 0. Which factors as (x + 1)(x² + x + 1) = 0. The quadratic x² + x + 1 has discriminant 1 - 4 = -3 < 0, so complex roots. Therefore, only one real root. Therefore, not all real. If a = 3, all real. If a > 3, then quadratic has two distinct real roots. So for example, a = 4: quadratic is x² + 3x + 1, which has roots [-3 ± sqrt(9 - 4)]/2 = [-3 ± sqrt(5)]/2, which are real. So then original equation has three real roots: -1, [-3 + sqrt(5)]/2, [-3 - sqrt(5)]/2. Therefore, all real. Therefore, seems like minimal a is 3. But the problem says "Find the smallest possible value of a." So answer is 3? But maybe I need to check if there's another case where the cubic has three real roots with different multiplicities? Wait, but if discriminant of quadratic is zero, then quadratic has a double root. So when a = 3, quadratic is (x + 1)^2, so the cubic has a triple root. So that's allowed. So maybe 3 is the minimal value. But I have a doubt because sometimes problems like this have more nuanced answers. Let me think again.

Alternatively, maybe there's another approach. Let me consider the cubic equation x³ + a x² + a x + 1 = 0. Since we know that x = -1 is a root, as we saw earlier, so we can write it as (x + 1)(x² + (a - 1)x + 1) = 0. So the other two roots are roots of x² + (a - 1)x + 1 = 0. Let me denote them as r and s. Then, by Vieta's formula for quadratic:

r + s = -(a - 1) = 1 - a

and rs = 1.

So, since rs = 1, the product of the roots is 1. Also, since they are real, discriminant must be non-negative. So discriminant D = (a - 1)^2 - 4 ≥ 0 ⇒ a ≥ 3 or a ≤ -1. Since a is positive, a ≥ 3. Therefore, minimal a is 3. Therefore, answer is 3. But maybe there's a mistake here? Let me check with another method.

Alternatively, maybe using calculus. Let me consider the function f(x) = x³ + a x² + a x + 1. For all roots to be real, the function must have three real critical points? Wait, no. Wait, for a cubic to have three real roots, it must have two critical points (local maximum and minimum) and the function must cross the x-axis three times. Therefore, the discriminant of the cubic must be positive. But maybe I can compute the discriminant of the cubic?

Alternatively, since we already factored it into (x + 1)(quadratic), maybe we can use the discriminant of the cubic. But maybe that's more complicated. Alternatively, since we have already factored it, and the quadratic must have real roots, so discriminant ≥ 0. Therefore, leading to a ≥ 3. Therefore, answer is 3. But maybe I need to check if there's a case where even if discriminant is negative, but the cubic still has three real roots? But no, because if quadratic has complex roots, then the cubic has only one real root. Therefore, the answer is 3. But maybe the problem is more complicated? Let me check with another approach.

Suppose that all roots are real. Let me denote them as r, s, t. Then, as before, r + s + t = -a, rs + rt + st = a, rst = -1. Also, since all roots are real, maybe I can use inequalities. For example, using AM ≥ GM or something else. Let me think.

But rst = -1. Since a is positive, and r + s + t = -a < 0. So the sum of roots is negative. Also, rs + rt + st = a > 0. So the sum of products is positive. Also, product is negative. So, since product is negative, there must be either one negative root and two positive roots or three negative roots. But if all roots are negative, then their sum would be negative, which is okay. But if there are two positive roots and one negative root, then their sum could be negative if the negative root is large in magnitude. Let me check.

But given that rst = -1. If all roots are negative, then their product is negative? Wait, if all roots are negative, then product is negative times negative times negative = negative. Which matches rst = -1. So that's possible. Alternatively, if there are two positive roots and one negative root, then product is positive * positive * negative = negative, which also matches. So both cases are possible.

But let me think about the case where all roots are negative. Let me denote them as -p, -q, -r, where p, q, r are positive real numbers. Then:

Sum: -p - q - r = -a ⇒ p + q + r = a

Sum of products: (-p)(-q) + (-p)(-r) + (-q)(-r) = pq + pr + qr = a

Product: (-p)(-q)(-r) = -pqr = -1 ⇒ pqr = 1

So, we have p + q + r = a, pq + pr + qr = a, pqr = 1. So, we need to find the minimal a such that there exist positive real numbers p, q, r with these properties.

Alternatively, if there are two positive roots and one negative root. Let me suppose that two roots are positive and one is negative. Let me denote them as p, q, -r, where p, q, r are positive. Then:

Sum: p + q - r = -a ⇒ r = p + q + a

Sum of products: pq + (-pr) + (-qr) = pq - r(p + q) = a

Product: p q (-r) = -pqr = -1 ⇒ pqr = 1

So, from the first equation, r = p + q + a. Substitute into the second equation:

pq - (p + q + a)(p + q) = a

Let me denote s = p + q. Then, r = s + a. Then, the second equation becomes:

pq - s(s + a) = a ⇒ pq = s(s + a) + a

Also, from the product equation: p q r = 1 ⇒ pq (s + a) = 1 ⇒ pq = 1 / (s + a)

Therefore, combining the two expressions for pq:

s(s + a) + a = 1 / (s + a)

Multiply both sides by (s + a):

[s(s + a) + a] (s + a) = 1

Let me expand this:

s(s + a)^2 + a(s + a) = 1

This seems complicated. Maybe there's a better way. Alternatively, maybe considering symmetry. Let me think.

But maybe the case where all roots are negative is easier. Let me consider that case first. So, p + q + r = a, pq + pr + qr = a, pqr = 1. Let me think about this system. Maybe using inequalities. For example, by AM ≥ GM:

(p + q + r)/3 ≥ (pqr)^{1/3} ⇒ a/3 ≥ 1^{1/3} ⇒ a ≥ 3. Equality holds when p = q = r. So if p = q = r, then p + q + r = 3p = a, pq + pr + qr = 3p² = a, and pqr = p³ = 1 ⇒ p = 1. Therefore, a = 3. Therefore, in this case, the minimal a is 3. Therefore, if all roots are negative and equal, then a = 3. Therefore, this suggests that the minimal a is 3. Also, since in the case of two positive roots and one negative root, maybe we can get a smaller a? But according to the previous equations, maybe not. Let me check.

Alternatively, maybe the minimal a occurs when two roots are equal? Let me think. For example, in the case of the quadratic, when discriminant is zero, which is when a = 3, we have a double root. So maybe that's the minimal case. But let me check if there's a case where even with a < 3, the cubic has three real roots. But earlier, we saw that if a < 3, then quadratic has complex roots, so only one real root. Therefore, not all real. Therefore, the minimal a is 3. Therefore, answer is 3. But let me check with another approach.

Alternatively, using calculus. Let me consider the function f(x) = x³ + a x² + a x + 1. For it to have three real roots, it must have a local maximum and a local minimum, and the function must cross the x-axis three times. Therefore, the derivative f’(x) = 3x² + 2a x + a. For there to be two critical points, discriminant of derivative must be positive. So discriminant D’ = (2a)^2 - 4 * 3 * a = 4a² - 12a = 4a(a - 3). For D’ > 0, we need 4a(a - 3) > 0. Since a > 0, this implies a - 3 > 0 ⇒ a > 3. Therefore, when a > 3, the derivative has two real critical points, so the function has a local maximum and a local minimum. For the cubic to have three real roots, the function must cross the x-axis three times, which requires that the local maximum is above the x-axis and the local minimum is below the x-axis. Therefore, we need f(x) evaluated at the local maximum to be positive and at the local minimum to be negative. However, when a = 3, the discriminant of the derivative is zero, so there is only one critical point (a point of inflection). Therefore, the function is increasing or decreasing throughout? Wait, if discriminant of derivative is zero, then the derivative has a repeated root. Let me check for a = 3, f’(x) = 3x² + 6x + 3 = 3(x² + 2x + 1) = 3(x + 1)^2. Therefore, derivative is always non-negative, so function is non-decreasing. Therefore, the function is increasing everywhere, so it can have only one real root. Wait, but earlier we saw that when a = 3, the equation has a triple root at x = -1. So, even though the derivative is zero at x = -1, the function is increasing everywhere except at x = -1 where it has a horizontal tangent. Therefore, the function touches the x-axis at x = -1 and is increasing on both sides. Therefore, it only has one real root (with multiplicity three). Therefore, for a = 3, the function has one real root (with multiplicity three). Therefore, for a > 3, the function has two critical points, and if the function crosses the x-axis three times, then there are three real roots. Therefore, the condition for three real roots is that the local maximum is positive and the local minimum is negative. Therefore, we need to compute f at the critical points and set them to have opposite signs.

But since we already have that when a ≥ 3, the quadratic factor has real roots, which would imply that the cubic has three real roots. However, when a = 3, the quadratic factor is (x + 1)^2, so the cubic has a triple root. Therefore, maybe the condition for three real roots is a ≥ 3. However, according to the calculus approach, for a > 3, there are two critical points, and the function can have three real roots. But when a = 3, there is only one critical point, but the function still has three real roots (with multiplicity). Therefore, the minimal a is 3. Therefore, answer is 3. However, let me check with the calculus approach for a > 3. Let me suppose a > 3. Then, the derivative f’(x) = 3x² + 2a x + a. Let me find the critical points. The critical points are at x = [-2a ± sqrt(4a² - 12a)] / 6 = [-2a ± 2sqrt(a² - 3a)] / 6 = [-a ± sqrt(a² - 3a)] / 3.

Let me denote x1 = [-a - sqrt(a² - 3a)] / 3 and x2 = [-a + sqrt(a² - 3a)] / 3. These are the critical points. Now, to have three real roots, f(x1) and f(x2) must have opposite signs. Let me compute f(x1) and f(x2). However, this seems complicated. Alternatively, since we already have that when a ≥ 3, the quadratic factor has real roots, which would imply that the cubic has three real roots. Therefore, maybe the calculus approach is redundant here. Therefore, the minimal a is 3.

But let me check with a specific example. Let me take a = 3. Then, the equation is x³ + 3x² + 3x + 1 = 0, which factors as (x + 1)^3 = 0. So, triple root at -1. All real. If I take a slightly larger than 3, say a = 4. Then, the equation is x³ + 4x² + 4x + 1 = 0. Which factors as (x + 1)(x² + 3x + 1) = 0. The quadratic x² + 3x + 1 has roots [-3 ± sqrt(9 - 4)] / 2 = [-3 ± sqrt(5)] / 2. These are real numbers. Therefore, three real roots. Therefore, for a ≥ 3, all roots are real. Therefore, the minimal a is 3. Therefore, answer is 3. But why does the calculus approach suggest that for a > 3, there are two critical points? Because when a = 3, the derivative has a double root, so only one critical point. Therefore, the function is increasing everywhere, but still has three real roots (with multiplicity). Therefore, the minimal value of a is 3.

But let me check if there's a case where even with a < 3, the cubic has three real roots. Suppose a = 2. Then, equation is x³ + 2x² + 2x + 1 = 0. As before, factors as (x + 1)(x² + x + 1) = 0. The quadratic has complex roots, so only one real root. Therefore, not all real. If a = 1, equation is x³ + x² + x + 1 = 0. Which factors as (x + 1)(x² + 0x + 1) = 0. Again, complex roots. Therefore, only one real root. If a approaches 0, equation approaches x³ + 0 + 0 + 1 = x³ + 1 = 0, which has one real root -1 and two complex roots. Therefore, no. Therefore, seems like a must be at least 3. Therefore, answer is 3. Therefore, I think the minimal value of a is 3.

But wait, the problem says "all the roots are real". So maybe there's another case where the cubic has three real roots with different multiplicities? For example, a double root and a simple root? But in that case, discriminant of the quadratic would be zero, which occurs when a = 3. Which gives a triple root. So that's covered. Therefore, I think the answer is 3. Therefore, I think the minimal a is 3. Therefore, the answer is \boxed{3}.

Wait, but I just recalled that sometimes for cubics, even if the discriminant of the quadratic is negative, but maybe the cubic has three real roots? But no, because if the quadratic has complex roots, then the cubic has only one real root. Therefore, the answer is 3. Therefore, I think the answer is 3. Therefore, I think the answer is \boxed{3}.

**Final Answer**
\boxed{3}

To determine the smallest possible value of the positive real number $ a $ such that all the roots of the cubic polynomial

$$
x^3 + ax^2 + ax + 1 = 0
$$

are real, we begin by analyzing the structure of the polynomial.

---

### Step 1: Factor the Polynomial

We observe that $ x = -1 $ is a root of the polynomial. To verify:

$$
(-1)^3 + a(-1)^2 + a(-1) + 1 = -1 + a - a + 1 = 0
$$

So, $ x = -1 $ is indeed a root. We can factor the polynomial as:

$$
x^3 + ax^2 + ax + 1 = (x + 1)(x^2 + (a - 1)x + 1)
$$

This factorization is confirmed by polynomial division or synthetic division.

---

### Step 2: Analyze the Quadratic Factor

The remaining roots of the cubic are the roots of the quadratic:

$$
x^2 + (a - 1)x + 1 = 0
$$

For all roots of the cubic to be real, the quadratic must also have real roots. This happens when the **discriminant** of the quadratic is non-negative:

$$
D = (a - 1)^2 - 4 \geq 0
$$

Solving this inequality:

$$
(a - 1)^2 \geq 4 \Rightarrow |a - 1| \geq 2
$$

This gives two cases:

- $ a - 1 \geq 2 \Rightarrow a \geq 3 $
- $ a - 1 \leq -2 \Rightarrow a \leq -1 $

Since $ a $ is a **positive real number**, we discard the second case. Thus, the condition becomes:

$$
a \geq 3
$$

---

### Step 3: Determine the Minimal Value of $ a $

We now check whether $ a = 3 $ satisfies the condition that all roots are real.

Substitute $ a = 3 $ into the quadratic:

$$
x^2 + (3 - 1)x + 1 = x^2 + 2x + 1 = (x + 1)^2
$$

So the roots of the quadratic are both $ x = -1 $, and the original cubic becomes:

$$
(x + 1)^3 = 0
$$

This has a **triple root** at $ x = -1 $, which is real. Therefore, all roots are real when $ a = 3 $.

Now, we check whether a smaller value of $ a $, say $ a = 2 $, would result in all real roots. For $ a = 2 $, the quadratic becomes:

$$
x^2 + (2 - 1)x + 1 = x^2 + x + 1
$$

The discriminant is:

$$
D = 1^2 - 4(1)(1) = 1 - 4 = -3 < 0
$$

So the quadratic has **complex roots**, and the cubic has only one real root $ x = -1 $, with two complex roots. Therefore, **not all roots are real** when $ a < 3 $.

---

### Step 4: Conclusion

The smallest value of $ a $ for which all roots of the cubic are real is:

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest possible value of a positive real number a such that all the roots of the cubic equation x³ + a x² + a x + 1 = 0 are real. Hmm, let me think about how to approach this.

First, I remember that for a cubic equation to have all real roots, certain conditions must be satisfied. Maybe I can use Vieta's formulas or some other properties of polynomials. Let me recall Vieta's formulas for a cubic equation. If the roots are r, s, t, then:

r + s + t = -a (since the coefficient of x² is a, and Vieta says sum of roots is -coefficient of x² divided by coefficient of x³, which is -a/1 = -a)

rs + rt + st = a (sum of products of roots two at a time is coefficient of x divided by coefficient of x³, which is a/1 = a)

rst = -1 (product of roots is -constant term divided by coefficient of x³, which is -1/1 = -1)

So, given that all roots are real, we need to find the minimal a > 0 such that these conditions hold.

But how do I relate these to find the minimal a? Maybe I can use some inequalities or consider the nature of the roots.

Alternatively, maybe I can factor the cubic equation. Let me try to factor it. Let me see if there's a rational root. By Rational Root Theorem, possible rational roots are ±1. Let me test x = -1:

Plugging x = -1 into the equation: (-1)^3 + a(-1)^2 + a(-1) + 1 = -1 + a - a + 1 = 0. Oh! So x = -1 is a root for any a. That's interesting. So the cubic can be factored as (x + 1)(quadratic). Let me perform the division or factorization.

Let me do polynomial division or use synthetic division. Let me use synthetic division with root -1.

Coefficients of the cubic: 1 (x³), a (x²), a (x), 1 (constant).

Set up synthetic division:

-1 | 1   a   a   1
    |     -1  -(a - 1)   -(a + 1)
    ----------------------------
      1  a - 1   - (a - 1) + a   1 - (a + 1)

Wait, let me do that step by step. The coefficients are 1, a, a, 1.

Bring down the 1.

Multiply by -1: 1 * (-1) = -1. Add to next coefficient: a + (-1) = a - 1.

Multiply by -1: (a - 1)(-1) = -a + 1. Add to next coefficient: a + (-a + 1) = 1.

Multiply by -1: 1 * (-1) = -1. Add to last coefficient: 1 + (-1) = 0. Perfect, so the division works.

Therefore, the cubic factors as (x + 1)(x² + (a - 1)x + 1). So the original equation is (x + 1)(x² + (a - 1)x + 1) = 0.

Therefore, the roots are x = -1 and the roots of x² + (a - 1)x + 1 = 0. Since we need all roots to be real, the quadratic must have real roots. Therefore, discriminant must be non-negative.

Discriminant D of quadratic is [ (a - 1) ]² - 4 * 1 * 1 ≥ 0.

So, (a - 1)² - 4 ≥ 0 ⇒ (a - 1)² ≥ 4 ⇒ |a - 1| ≥ 2 ⇒ a - 1 ≥ 2 or a - 1 ≤ -2 ⇒ a ≥ 3 or a ≤ -1.

But since a is a positive real number, a ≥ 3 or a ≤ -1. But a is positive, so only a ≥ 3. Therefore, the minimal value of a is 3? Wait, but hold on. Wait, but the problem says "all the roots are real". So if a ≥ 3, then quadratic has real roots, so all roots of the cubic are real. But is that the case?

Wait, but if a = 3, then quadratic is x² + (3 - 1)x + 1 = x² + 2x + 1 = (x + 1)^2. So the roots are -1, -1, -1. So triple root at -1. So all roots are real. So that's okay. But if a is less than 3, say a = 2, then discriminant is (2 - 1)^2 - 4 = 1 - 4 = -3 < 0, so quadratic has complex roots, so original equation would have one real root (-1) and two complex roots. Therefore, not all real. Therefore, the minimal a is 3? But wait, the problem says "all the roots are real", so the minimal a is 3? But wait, maybe there's a case where even if discriminant is negative, but... no, because if discriminant is negative, quadratic has complex roots. Therefore, the only way for all roots to be real is discriminant ≥ 0, so a ≥ 3. Therefore, minimal a is 3. But wait, the problem says "Find the smallest possible value of a." So answer is 3? But let me check again.

Wait, but maybe there's another way where the quadratic has real roots but with some other conditions? Wait, but the discriminant is the only condition for quadratic to have real roots. So if discriminant is non-negative, quadratic has real roots. Therefore, if a ≥ 3, then quadratic has real roots, so original equation has three real roots. Therefore, minimal a is 3. But wait, the problem says "all the roots are real", so maybe there's a case where even if discriminant is negative, but... no, because then two roots are complex. Therefore, the answer is 3? But I need to check if there's a possibility that even with discriminant negative, but maybe the cubic has three real roots? Wait, no. If the quadratic has complex roots, then the cubic has one real root (x = -1) and two complex roots. Therefore, not all real. Therefore, the minimal a is 3. But wait, the problem says "Find the smallest possible value of a." So answer is 3? But I have a feeling that maybe there's something wrong here. Let me check with a = 3. Then equation is x³ + 3x² + 3x + 1 = 0. Which factors as (x + 1)^3 = 0. So triple root at -1. All real. If a is less than 3, say a = 2, then equation is x³ + 2x² + 2x + 1 = 0. Which factors as (x + 1)(x² + x + 1) = 0. The quadratic x² + x + 1 has discriminant 1 - 4 = -3 < 0, so complex roots. Therefore, only one real root. Therefore, not all real. If a = 3, all real. If a > 3, then quadratic has two distinct real roots. So for example, a = 4: quadratic is x² + 3x + 1, which has roots [-3 ± sqrt(9 - 4)]/2 = [-3 ± sqrt(5)]/2, which are real. So then original equation has three real roots: -1, [-3 + sqrt(5)]/2, [-3 - sqrt(5)]/2. Therefore, all real. Therefore, seems like minimal a is 3. But the problem says "Find the smallest possible value of a." So answer is 3? But maybe I need to check if there's another case where the cubic has three real roots with different multiplicities? Wait, but if discriminant of quadratic is zero, then quadratic has a double root. So when a = 3, quadratic is (x + 1)^2, so the cubic has a triple root. So that's allowed. So maybe 3 is the minimal value. But I have a doubt because sometimes problems like this have more nuanced answers. Let me think again.

Alternatively, maybe there's another approach. Let me consider the cubic equation x³ + a x² + a x + 1 = 0. Since we know that x = -1 is a root, as we saw earlier, so we can write it as (x + 1)(x² + (a - 1)x + 1) = 0. So the other two roots are roots of x² + (a - 1)x + 1 = 0. Let me denote them as r and s. Then, by Vieta's formula for quadratic:

r + s = -(a - 1) = 1 - a

and rs = 1.

So, since rs = 1, the product of the roots is 1. Also, since they are real, discriminant must be non-negative. So discriminant D = (a - 1)^2 - 4 ≥ 0 ⇒ a ≥ 3 or a ≤ -1. Since a is positive, a ≥ 3. Therefore, minimal a is 3. Therefore, answer is 3. But maybe there's a mistake here? Let me check with another method.

Alternatively, maybe using calculus. Let me consider the function f(x) = x³ + a x² + a x + 1. For all roots to be real, the function must have three real critical points? Wait, no. Wait, for a cubic to have three real roots, it must have two critical points (local maximum and minimum) and the function must cross the x-axis three times. Therefore, the discriminant of the cubic must be positive. But maybe I can compute the discriminant of the cubic?

Alternatively, since we already factored it into (x + 1)(quadratic), maybe we can use the discriminant of the cubic. But maybe that's more complicated. Alternatively, since we have already factored it, and the quadratic must have real roots, so discriminant ≥ 0. Therefore, leading to a ≥ 3. Therefore, answer is 3. But maybe I need to check if there's a case where even if discriminant is negative, but the cubic still has three real roots? But no, because if quadratic has complex roots, then the cubic has only one real root. Therefore, the answer is 3. But maybe the problem is more complicated? Let me check with another approach.

Suppose that all roots are real. Let me denote them as r, s, t. Then, as before, r + s + t = -a, rs + rt + st = a, rst = -1. Also, since all roots are real, maybe I can use inequalities. For example, using AM ≥ GM or something else. Let me think.

But rst = -1. Since a is positive, and r + s + t = -a < 0. So the sum of roots is negative. Also, rs + rt + st = a > 0. So the sum of products is positive. Also, product is negative. So, since product is negative, there must be either one negative root and two positive roots or three negative roots. But if all roots are negative, then their sum would be negative, which is okay. But if there are two positive roots and one negative root, then their sum could be negative if the negative root is large in magnitude. Let me check.

But given that rst = -1. If all roots are negative, then their product is negative? Wait, if all roots are negative, then product is negative times negative times negative = negative. Which matches rst = -1. So that's possible. Alternatively, if there are two positive roots and one negative root, then product is positive * positive * negative = negative, which also matches. So both cases are possible.

But let me think about the case where all roots are negative. Let me denote them as -p, -q, -r, where p, q, r are positive real numbers. Then:

Sum: -p - q - r = -a ⇒ p + q + r = a

Sum of products: (-p)(-q) + (-p)(-r) + (-q)(-r) = pq + pr + qr = a

Product: (-p)(-q)(-r) = -pqr = -1 ⇒ pqr = 1

So, we have p + q + r = a, pq + pr + qr = a, pqr = 1. So, we need to find the minimal a such that there exist positive real numbers p, q, r with these properties.

Alternatively, if there are two positive roots and one negative root. Let me suppose that two roots are positive and one is negative. Let me denote them as p, q, -r, where p, q, r are positive. Then:

Sum: p + q - r = -a ⇒ r = p + q + a

Sum of products: pq + (-pr) + (-qr) = pq - r(p + q) = a

Product: p q (-r) = -pqr = -1 ⇒ pqr = 1

So, from the first equation, r = p + q + a. Substitute into the second equation:

pq - (p + q + a)(p + q) = a

Let me denote s = p + q. Then, r = s + a. Then, the second equation becomes:

pq - s(s + a) = a ⇒ pq = s(s + a) + a

Also, from the product equation: p q r = 1 ⇒ pq (s + a) = 1 ⇒ pq = 1 / (s + a)

Therefore, combining the two expressions for pq:

s(s + a) + a = 1 / (s + a)

Multiply both sides by (s + a):

[s(s + a) + a] (s + a) = 1

Let me expand this:

s(s + a)^2 + a(s + a) = 1

This seems complicated. Maybe there's a better way. Alternatively, maybe considering symmetry. Let me think.

But maybe the case where all roots are negative is easier. Let me consider that case first. So, p + q + r = a, pq + pr + qr = a, pqr = 1. Let me think about this system. Maybe using inequalities. For example, by AM ≥ GM:

(p + q + r)/3 ≥ (pqr)^{1/3} ⇒ a/3 ≥ 1^{1/3} ⇒ a ≥ 3. Equality holds when p = q = r. So if p = q = r, then p + q + r = 3p = a, pq + pr + qr = 3p² = a, and pqr = p³ = 1 ⇒ p = 1. Therefore, a = 3. Therefore, in this case, the minimal a is 3. Therefore, if all roots are negative and equal, then a = 3. Therefore, this suggests that the minimal a is 3. Also, since in the case of two positive roots and one negative root, maybe we can get a smaller a? But according to the previous equations, maybe not. Let me check.

Alternatively, maybe the minimal a occurs when two roots are equal? Let me think. For example, in the case of the quadratic, when discriminant is zero, which is when a = 3, we have a double root. So maybe that's the minimal case. But let me check if there's a case where even with a < 3, the cubic has three real roots. But earlier, we saw that if a < 3, then quadratic has complex roots, so only one real root. Therefore, not all real. Therefore, the minimal a is 3. Therefore, answer is 3. But let me check with another approach.

Alternatively, using calculus. Let me consider the function f(x) = x³ + a x² + a x + 1. For it to have three real roots, it must have a local maximum and a local minimum, and the function must cross the x-axis three times. Therefore, the derivative f’(x) = 3x² + 2a x + a. For there to be two critical points, discriminant of derivative must be positive. So discriminant D’ = (2a)^2 - 4 * 3 * a = 4a² - 12a = 4a(a - 3). For D’ > 0, we need 4a(a - 3) > 0. Since a > 0, this implies a - 3 > 0 ⇒ a > 3. Therefore, when a > 3, the derivative has two real critical points, so the function has a local maximum and a local minimum. For the cubic to have three real roots, the function must cross the x-axis three times, which requires that the local maximum is above the x-axis and the local minimum is below the x-axis. Therefore, we need f(x) evaluated at the local maximum to be positive and at the local minimum to be negative. However, when a = 3, the discriminant of the derivative is zero, so there is only one critical point (a point of inflection). Therefore, the function is increasing or decreasing throughout? Wait, if discriminant of derivative is zero, then the derivative has a repeated root. Let me check for a = 3, f’(x) = 3x² + 6x + 3 = 3(x² + 2x + 1) = 3(x + 1)^2. Therefore, derivative is always non-negative, so function is non-decreasing. Therefore, the function is increasing everywhere, so it can have only one real root. Wait, but earlier we saw that when a = 3, the equation has a triple root at x = -1. So, even though the derivative is zero at x = -1, the function is increasing everywhere except at x = -1 where it has a horizontal tangent. Therefore, the function touches the x-axis at x = -1 and is increasing on both sides. Therefore, it only has one real root (with multiplicity three). Therefore, for a = 3, the function has one real root (with multiplicity three). Therefore, for a > 3, the function has two critical points, and if the function crosses the x-axis three times, then there are three real roots. Therefore, the condition for three real roots is that the local maximum is positive and the local minimum is negative. Therefore, we need to compute f at the critical points and set them to have opposite signs.

But since we already have that when a ≥ 3, the quadratic factor has real roots, which would imply that the cubic has three real roots. However, when a = 3, the quadratic factor is (x + 1)^2, so the cubic has a triple root. Therefore, maybe the condition for three real roots is a ≥ 3. However, according to the calculus approach, for a > 3, there are two critical points, and the function can have three real roots. But when a = 3, there is only one critical point, but the function still has three real roots (with multiplicity). Therefore, the minimal a is 3. Therefore, answer is 3. However, let me check with the calculus approach for a > 3. Let me suppose a > 3. Then, the derivative f’(x) = 3x² + 2a x + a. Let me find the critical points. The critical points are at x = [-2a ± sqrt(4a² - 12a)] / 6 = [-2a ± 2sqrt(a² - 3a)] / 6 = [-a ± sqrt(a² - 3a)] / 3.

Let me denote x1 = [-a - sqrt(a² - 3a)] / 3 and x2 = [-a + sqrt(a² - 3a)] / 3. These are the critical points. Now, to have three real roots, f(x1) and f(x2) must have opposite signs. Let me compute f(x1) and f(x2). However, this seems complicated. Alternatively, since we already have that when a ≥ 3, the quadratic factor has real roots, which would imply that the cubic has three real roots. Therefore, maybe the calculus approach is redundant here. Therefore, the minimal a is 3.

But let me check with a specific example. Let me take a = 3. Then, the equation is x³ + 3x² + 3x + 1 = 0, which factors as (x + 1)^3 = 0. So, triple root at -1. All real. If I take a slightly larger than 3, say a = 4. Then, the equation is x³ + 4x² + 4x + 1 = 0. Which factors as (x + 1)(x² + 3x + 1) = 0. The quadratic x² + 3x + 1 has roots [-3 ± sqrt(9 - 4)] / 2 = [-3 ± sqrt(5)] / 2. These are real numbers. Therefore, three real roots. Therefore, for a ≥ 3, all roots are real. Therefore, the minimal a is 3. Therefore, answer is 3. But why does the calculus approach suggest that for a > 3, there are two critical points? Because when a = 3, the derivative has a double root, so only one critical point. Therefore, the function is increasing everywhere, but still has three real roots (with multiplicity). Therefore, the minimal value of a is 3.

But let me check if there's a case where even with a < 3, the cubic has three real roots. Suppose a = 2. Then, equation is x³ + 2x² + 2x + 1 = 0. As before, factors as (x + 1)(x² + x + 1) = 0. The quadratic has complex roots, so only one real root. Therefore, not all real. If a = 1, equation is x³ + x² + x + 1 = 0. Which factors as (x + 1)(x² + 0x + 1) = 0. Again, complex roots. Therefore, only one real root. If a approaches 0, equation approaches x³ + 0 + 0 + 1 = x³ + 1 = 0, which has one real root -1 and two complex roots. Therefore, no. Therefore, seems like a must be at least 3. Therefore, answer is 3. Therefore, I think the minimal value of a is 3.

But wait, the problem says "all the roots are real". So maybe there's another case where the cubic has three real roots with different multiplicities? For example, a double root and a simple root? But in that case, discriminant of the quadratic would be zero, which occurs when a = 3. Which gives a triple root. So that's covered. Therefore, I think the answer is 3. Therefore, I think the minimal a is 3. Therefore, the answer is \boxed{3}.

Wait, but I just recalled that sometimes for cubics, even if the discriminant of the quadratic is negative, but maybe the cubic has three real roots? But no, because if the quadratic has complex roots, then the cubic has only one real root. Therefore, the answer is 3. Therefore, I think the answer is 3. Therefore, I think the answer is \boxed{3}.

**Final Answer**
\boxed{3}

To determine the smallest possible value of the positive real number $ a $ such that all the roots of the cubic polynomial

$$
x^3 + ax^2 + ax + 1 = 0
$$

are real, we begin by analyzing the structure of the polynomial.

---

### Step 1: Factor the Polynomial

We observe that $ x = -1 $ is a root of the polynomial. To verify:

$$
(-1)^3 + a(-1)^2 + a(-1) + 1 = -1 + a - a + 1 = 0
$$

So, $ x = -1 $ is indeed a root. We can factor the polynomial as:

$$
x^3 + ax^2 + ax + 1 = (x + 1)(x^2 + (a - 1)x + 1)
$$

This factorization is confirmed by polynomial division or synthetic division.

---

### Step 2: Analyze the Quadratic Factor

The remaining roots of the cubic are the roots of the quadratic:

$$
x^2 + (a - 1)x + 1 = 0
$$

For all roots of the cubic to be real, the quadratic must also have real roots. This happens when the **discriminant** of the quadratic is non-negative:

$$
D = (a - 1)^2 - 4 \geq 0
$$

Solving this inequality:

$$
(a - 1)^2 \geq 4 \Rightarrow |a - 1| \geq 2
$$

This gives two cases:

- $ a - 1 \geq 2 \Rightarrow a \geq 3 $
- $ a - 1 \leq -2 \Rightarrow a \leq -1 $

Since $ a $ is a **positive real number**, we discard the second case. Thus, the condition becomes:

$$
a \geq 3
$$

---

### Step 3: Determine the Minimal Value of $ a $

We now check whether $ a = 3 $ satisfies the condition that all roots are real.

Substitute $ a = 3 $ into the quadratic:

$$
x^2 + (3 - 1)x + 1 = x^2 + 2x + 1 = (x + 1)^2
$$

So the roots of the quadratic are both $ x = -1 $, and the original cubic becomes:

$$
(x + 1)^3 = 0
$$

This has a **triple root** at $ x = -1 $, which is real. Therefore, all roots are real when $ a = 3 $.

Now, we check whether a smaller value of $ a $, say $ a = 2 $, would result in all real roots. For $ a = 2 $, the quadratic becomes:

$$
x^2 + (2 - 1)x + 1 = x^2 + x + 1
$$

The discriminant is:

$$
D = 1^2 - 4(1)(1) = 1 - 4 = -3 < 0
$$

So the quadratic has **complex roots**, and the cubic has only one real root $ x = -1 $, with two complex roots. Therefore, **not all roots are real** when $ a < 3 $.

---

### Step 4: Conclusion

The smallest value of $ a $ for which all roots of the cubic are real is:

$$
\boxed{3}
$$

Solution:
Note that $x = -1$ is always a root of $x^3 + ax^2 + ax + 1 = 0,$ so we can factor out $x + 1,$ to get
\[(x + 1) (x^2 + (a - 1) x + 1) = 0.\]The quadratic factor has real roots if and only if its discriminant is nonnegative:
\[(a - 1)^2 - 4 \ge 0.\]This reduces to $a^2 - 2a - 3 \ge 0,$ which factors as $(a + 1)(a - 3) \ge 0.$  The smallest positive value that satisfies this inequality is $\boxed{3}.$


Explanation:
Okay, so I need to evaluate the expression (1 + 2i)6 - 3i. Let me think about how to approach this. First, I remember that when you have a complex number multiplied by a real number, you distribute that real number to both the real and imaginary parts. Then, after that multiplication, I need to subtract 3i from the result. Let me write that down step by step.

Starting with the first part: (1 + 2i) multiplied by 6. So, distributing the 6, that should be 6 times 1 plus 6 times 2i. Let me calculate each term separately. 6 times 1 is straightforward, that's just 6. Then 6 times 2i... Well, 6 times 2 is 12, so that term is 12i. So putting those together, the first part simplifies to 6 + 12i.

Now, the original expression is that result minus 3i. So, I need to subtract 3i from 6 + 12i. Let me write that out: (6 + 12i) - 3i. To subtract these, I can combine like terms. The real part is just 6, and the imaginary parts are 12i and -3i. So combining those, 12i - 3i is 9i. Therefore, the entire expression simplifies to 6 + 9i.

Wait, let me check if I did that correctly. Let me go through each step again. First, multiplying (1 + 2i) by 6. Yes, 1*6 is 6, and 2i*6 is 12i. Then subtracting 3i. So 12i - 3i is indeed 9i. So the final result is 6 + 9i.

Hmm, is there another way to approach this? Maybe by considering the entire expression as a complex number. Let me think. The original expression is (1 + 2i) multiplied by 6, then subtract 3i. Alternatively, I could think of it as 6*(1 + 2i) - 3i. Which is exactly what I did.

Alternatively, maybe I can factor out something? Let me see. If I factor out an i from the terms involving i, but I don't think that's necessary here. Let me try. The expression after multiplying out is 6 + 12i - 3i. So combining the imaginary terms, 12i - 3i is 9i, so 6 + 9i.

Wait, maybe I can check my answer by converting to rectangular form? But that's already what I did. Alternatively, maybe plug into a different form? Let me think. If I consider complex numbers as vectors, multiplying by 6 scales the vector, and then subtracting 3i is like subtracting a vector in the imaginary direction.

But maybe I can verify with another method. Let me compute the original expression again.

First, (1 + 2i) * 6. Let me compute that again. 1*6 = 6, 2i*6 = 12i. So 6 + 12i. Then subtract 3i. So 6 + 12i - 3i. Which is 6 + (12i - 3i) = 6 + 9i.

Yes, that seems consistent.

Alternatively, maybe I can think of the entire expression as 6*(1 + 2i) - 3i. Let me compute 6*(1 + 2i) first. Which is 6*1 + 6*2i = 6 + 12i. Then subtract 3i. So 6 + 12i - 3i. Which is 6 + 9i.

Alternatively, maybe I can write the entire expression as 6*1 + 6*2i - 3i. Which is 6 + 12i - 3i. Then combine like terms.

So, 6 is the real part, and 12i - 3i is 9i. So the result is 6 + 9i.

I think that's correct. Let me check if there's any chance I made a mistake in the arithmetic. For example, 6*2i is 12i? Yes, because 6*2 is 12. Then subtracting 3i, so 12i - 3i is 9i. Yes.

Alternatively, if I had messed up the multiplication, like if I thought 6*2i was 6i or something, but no, 6*2 is 12.

Another check: Let me think of the complex number (1 + 2i) multiplied by 6. If I consider 1 + 2i as a point in the complex plane, multiplying by 6 scales it by 6. So the original point is (1, 2), scaling by 6 would make it (6, 12). Then subtracting 3i is subtracting 3 from the imaginary part, so subtracting (0, 3), which would result in (6, 12 - 3) = (6, 9), which corresponds to 6 + 9i.

Yes, that matches. So that's another way to visualize it.

Alternatively, if I use algebraic manipulation. Let me write the original expression again:

(1 + 2i) * 6 - 3i

First, multiply out the terms:

= 1*6 + 2i*6 - 3i

= 6 + 12i - 3i

Combine like terms:

= 6 + (12i - 3i)

= 6 + 9i

So, same result.

I think that's solid. I don't see any mistakes here. Maybe I can check with decimal numbers? Let me think. Suppose I have a complex number, say, 1 + 2i. Multiply by 6: 6*(1 + 2i) = 6 + 12i. Then subtract 3i: 6 + 12i - 3i = 6 + 9i.

Alternatively, if I convert to polar form? Although that might be more complicated, but let me try.

First, the original complex number is 1 + 2i. Its magnitude is sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5). The angle is arctangent(2/1) = arctan(2). Multiplying by 6 would scale the magnitude by 6, so the new magnitude is 6*sqrt(5), and the angle remains arctan(2). Then subtracting 3i. Hmm, subtracting 3i is equivalent to adding a complex number -3i, which has magnitude 3 and angle -π/2.

But combining these two complex numbers in polar form would be complicated. Maybe not the best approach here.

Alternatively, maybe convert the final answer to polar form? Let me see. The result is 6 + 9i. Its magnitude is sqrt(6^2 + 9^2) = sqrt(36 + 81) = sqrt(117) = 3*sqrt(13). The angle is arctangent(9/6) = arctangent(3/2). But I don't know if that helps me verify anything.

But since I already have the rectangular form, and I can check that the arithmetic is correct, maybe that's sufficient.

Alternatively, maybe plug into a calculator? But since I don't have a calculator here, I need to rely on my own calculations.

Wait, let me think again. If I have (1 + 2i) multiplied by 6, that's 6 + 12i. Then subtract 3i. So 6 + 12i - 3i. So 6 + 9i.

Yes, that's straightforward.

Alternatively, maybe I can think of the entire expression as 6*(1 + 2i) - 3i. Let me factor out something? For example, can I factor out a 3? Let me see. 6 is 3*2, and 3i is 3i. So maybe:

= 3*(2*(1 + 2i)) - 3i

= 3*(2 + 4i) - 3i

= 3*2 + 3*4i - 3i

= 6 + 12i - 3i

= 6 + 9i

Same result.

Alternatively, maybe factor out 3 from the entire expression? Let me see:

Original expression: 6*(1 + 2i) - 3i.

If I factor out 3, I get 3*[2*(1 + 2i) - i]. Let me compute inside the brackets:

2*(1 + 2i) - i = 2 + 4i - i = 2 + 3i.

Then multiply by 3: 3*(2 + 3i) = 6 + 9i.

Same answer.

So that's another way to look at it.

Therefore, regardless of the method, I keep getting 6 + 9i.

I think that's solid. I don't see any mistakes here.

Therefore, the final answer is 6 + 9i.

**Final Answer**
The value of the expression is \boxed{6 + 9i}.

To evaluate the expression $(1 + 2i)6 - 3i$, we proceed step by step, applying the rules of complex number arithmetic.

---

### Step 1: Distribute the Multiplication

We begin by distributing the scalar $6$ across the complex number $1 + 2i$:

$$
(1 + 2i) \cdot 6 = 1 \cdot 6 + 2i \cdot 6 = 6 + 12i
$$

---

### Step 2: Subtract $3i$

Now, subtract $3i$ from the result:

$$
(6 + 12i) - 3i = 6 + (12i - 3i) = 6 + 9i
$$

---

### Final Result

After simplifying, the expression evaluates to:

$$
\boxed{6 + 9i}
$$

[Assistant]
Explanation:
Okay, so I need to evaluate the expression (1 + 2i)6 - 3i. Let me think about how to approach this. First, I remember that when you have a complex number multiplied by a real number, you distribute that real number to both the real and imaginary parts. Then, after that multiplication, I need to subtract 3i from the result. Let me write that down step by step.

Starting with the first part: (1 + 2i) multiplied by 6. So, distributing the 6, that should be 6 times 1 plus 6 times 2i. Let me calculate each term separately. 6 times 1 is straightforward, that's just 6. Then 6 times 2i... Well, 6 times 2 is 12, so that term is 12i. So putting those together, the first part simplifies to 6 + 12i.

Now, the original expression is that result minus 3i. So, I need to subtract 3i from 6 + 12i. Let me write that out: (6 + 12i) - 3i. To subtract these, I can combine like terms. The real part is just 6, and the imaginary parts are 12i and -3i. So combining those, 12i - 3i is 9i. Therefore, the entire expression simplifies to 6 + 9i.

Wait, let me check if I did that correctly. Let me go through each step again. First, multiplying (1 + 2i) by 6. Yes, 1*6 is 6, and 2i*6 is 12i. Then subtracting 3i. So 12i - 3i is indeed 9i. So the final result is 6 + 9i.

Hmm, is there another way to approach this? Maybe by considering the entire expression as a complex number. Let me think. The original expression is (1 + 2i) multiplied by 6, then subtract 3i. Alternatively, I could think of it as 6*(1 + 2i) - 3i. Which is exactly what I did.

Alternatively, maybe I can factor out something? Let me see. If I factor out an i from the terms involving i, but I don't think that's necessary here. Let me try. The expression after multiplying out is 6 + 12i - 3i. So combining the imaginary terms, 12i - 3i is 9i, so 6 + 9i.

Wait, maybe I can check my answer by converting to rectangular form? But that's already what I did. Alternatively, maybe plug into a different form? Let me think. If I consider complex numbers as vectors, multiplying by 6 scales the vector, and then subtracting 3i is like subtracting a vector in the imaginary direction.

But maybe I can verify with another method. Let me compute the original expression again.

First, (1 + 2i) * 6. Let me compute that again. 1*6 = 6, 2i*6 = 12i. So 6 + 12i. Then subtract 3i. So 6 + 12i - 3i. Which is 6 + (12i - 3i) = 6 + 9i.

Yes, that seems consistent.

Alternatively, maybe I can think of the entire expression as 6*(1 + 2i) - 3i. Let me compute 6*(1 + 2i) first. Which is 6*1 + 6*2i = 6 + 12i. Then subtract 3i. So 6 + 12i - 3i. Which is 6 + 9i.

Alternatively, maybe I can write the entire expression as 6*1 + 6*2i - 3i. Which is 6 + 12i - 3i. Then combine like terms.

So, 6 is the real part, and 12i - 3i is 9i. So the result is 6 + 9i.

I think that's correct. Let me check if there's any chance I made a mistake in the arithmetic. For example, 6*2i is 12i? Yes, because 6*2 is 12. Then subtracting 3i, so 12i - 3i is 9i. Yes.

Alternatively, if I had messed up the multiplication, like if I thought 6*2i was 6i or something, but no, 6*2 is 12.

Another check: Let me think of the complex number (1 + 2i) multiplied by 6. If I consider 1 + 2i as a point in the complex plane, multiplying by 6 scales it by 6. So the original point is (1, 2), scaling by 6 would make it (6, 12). Then subtracting 3i is subtracting 3 from the imaginary part, so subtracting (0, 3), which would result in (6, 12 - 3) = (6, 9), which corresponds to 6 + 9i.

Yes, that matches. So that's another way to visualize it.

Alternatively, if I use algebraic manipulation. Let me write the original expression again:

(1 + 2i) * 6 - 3i

First, multiply out the terms:

= 1*6 + 2i*6 - 3i

= 6 + 12i - 3i

Combine like terms:

= 6 + (12i - 3i)

= 6 + 9i

So, same result.

I think that's solid. I don't see any mistakes here. Maybe I can check with decimal numbers? Let me think. Suppose I have a complex number, say, 1 + 2i. Multiply by 6: 6*(1 + 2i) = 6 + 12i. Then subtract 3i: 6 + 12i - 3i = 6 + 9i.

Alternatively, if I convert to polar form? Although that might be more complicated, but let me try.

First, the original complex number is 1 + 2i. Its magnitude is sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5). The angle is arctangent(2/1) = arctan(2). Multiplying by 6 would scale the magnitude by 6, so the new magnitude is 6*sqrt(5), and the angle remains arctan(2). Then subtracting 3i. Hmm, subtracting 3i is equivalent to adding a complex number -3i, which has magnitude 3 and angle -π/2.

But combining these two complex numbers in polar form would be complicated. Maybe not the best approach here.

Alternatively, maybe convert the final answer to polar form? Let me see. The result is 6 + 9i. Its magnitude is sqrt(6^2 + 9^2) = sqrt(36 + 81) = sqrt(117) = 3*sqrt(13). The angle is arctangent(9/6) = arctangent(3/2). But I don't know if that helps me verify anything.

But since I already have the rectangular form, and I can check that the arithmetic is correct, maybe that's sufficient.

Alternatively, maybe plug into a calculator? But since I don't have a calculator here, I need to rely on my own calculations.

Wait, let me think again. If I have (1 + 2i) multiplied by 6, that's 6 + 12i. Then subtract 3i. So 6 + 12i - 3i. So 6 + 9i.

Yes, that's straightforward.

Alternatively, maybe I can think of the entire expression as 6*(1 + 2i) - 3i. Let me factor out something? For example, can I factor out a 3? Let me see. 6 is 3*2, and 3i is 3i. So maybe:

= 3*(2*(1 + 2i)) - 3i

= 3*(2 + 4i) - 3i

= 3*2 + 3*4i - 3i

= 6 + 12i - 3i

= 6 + 9i

Same result.

Alternatively, maybe factor out 3 from the entire expression? Let me see:

Original expression: 6*(1 + 2i) - 3i.

If I factor out 3, I get 3*[2*(1 + 2i) - i]. Let me compute inside the brackets:

2*(1 + 2i) - i = 2 + 4i - i = 2 + 3i.

Then multiply by 3: 3*(2 + 3i) = 6 + 9i.

Same answer.

So that's another way to look at it.

Therefore, regardless of the method, I keep getting 6 + 9i.

I think that's solid. I don't see any mistakes here.

Therefore, the final answer is 6 + 9i.

**Final Answer**
The value of the expression is \boxed{6 + 9i}.

To evaluate the expression $(1 + 2i)6 - 3i$, we proceed step by step, applying the rules of complex number arithmetic.

---

### Step 1: Distribute the Multiplication

We begin by distributing the scalar $6$ across the complex number $1 + 2i$:

$$
(1 + 2i) \cdot 6 = 1 \cdot 6 + 2i \cdot 6 = 6 + 12i
$$

---

### Step 2: Subtract $3i$

Now, subtract $3i$ from the result:

$$
(6 + 12i) - 3i = 6 + (12i - 3i) = 6 + 9i
$$

---

### Final Result

After simplifying, the expression evaluates to:

$$
\boxed{6 + 9i}
$$

Solution: Distribute the factor of 6 and simplify to obtain $(1+2i)6-3i=6+12i-3i=\boxed{6+9i}$.