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. Divisibility by 10 means the number must end with a 0. 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, the digits are restricted to 0 and 2. Let me note that down:

- Ends with 0 (for divisibility by 10)
- Sum of digits is a multiple of 3 (for divisibility by 3)
- Only digits 0 and 2 are allowed

So, the problem reduces to finding the smallest number composed of 0s and 2s that ends with 0 and has a digit sum divisible by 3.

Let me think about how to construct such a number. Since it has to end with 0, the last digit is fixed as 0. The rest of the digits can be 0s and 2s. But we need the sum of digits to be divisible by 3.

Let me consider the possible numbers. Since we want the smallest such number, we should aim for the smallest number of digits possible. Let me start checking from the smallest possible numbers.

First, let's think about numbers with 1 digit. The only possible numbers are 0 and 2. But 0 is not positive, and 2 is not divisible by 30. So, no luck here.

Next, numbers with 2 digits. The number must end with 0, so the first digit can be 2 or 0. But since it's a two-digit number, the first digit can't be 0. So, the only possibility is 20. Let's check if 20 is divisible by 30. 20 divided by 30 is 0.666..., so no. Also, the sum of digits is 2 + 0 = 2, which is not divisible by 3. So, 20 is out.

Moving on to three-digit numbers. The number must end with 0. So, the first two digits can be 0s and 2s. Let's list all possibilities:

- 200: sum is 2 + 0 + 0 = 2 → not divisible by 3
- 220: sum is 2 + 2 + 0 = 4 → not divisible by 3
- 020: but that's 20, which we already considered, and leading zeros aren't allowed. So, only 200 and 220. Neither works.

Four-digit numbers. Again, last digit is 0. The first three digits can be 0s and 2s. Let's think about possible numbers:

Possible numbers:

- 2000: sum is 2 → nope
- 2020: sum is 2 + 0 + 2 + 0 = 4 → nope
- 2200: sum is 2 + 2 + 0 + 0 = 4 → nope
- 2220: sum is 2 + 2 + 2 + 0 = 6 → 6 is divisible by 3! Wait, 2220. Let me check if that's divisible by 30.

Divisible by 10? Yes, ends with 0. Divisible by 3? Sum of digits is 6, which is divisible by 3. So, 2220 is divisible by 30. Is this the smallest? Let me check if there is a smaller number with fewer digits.

Wait, we checked 2-digit and 3-digit numbers, and they didn't work. So 4-digit numbers: 2220 is a candidate. But let me check if there is a smaller 4-digit number. For example, numbers like 2020, 2000, 2200, etc., but their digit sums are 4, 2, 4, which aren't divisible by 3. So 2220 is the only 4-digit number that works.

But wait, maybe there is a smaller number with more digits? Wait, no. Because 2220 is a 4-digit number. If there is a smaller number with more digits, it would have to have more digits, which would make it larger. For example, a 5-digit number would be larger than a 4-digit number. So 2220 is the smallest? Wait, but maybe there is a smaller number with more digits? Wait, no. For example, 2020 is smaller than 2220, but it doesn't satisfy the divisibility by 3. Similarly, 2000 is smaller but also doesn't work. So 2220 is the first one that works.

But let me check if there is a smaller number with more digits. Wait, for example, maybe a number with 5 digits that is smaller than 2220? But 2220 is 2220, and a 5-digit number would be at least 20000, which is way bigger. So no. So 2220 is the smallest? Wait, but let me think again.

Wait, maybe there is a number with 4 digits that is smaller than 2220? Let's see. The numbers with 4 digits starting with 2, then followed by digits. The smallest 4-digit number with digits 0 and 2 is 2000. Then 2000, 2020, 2200, 2220. So 2000 is 2000, which is smaller than 2220. But 2000's digit sum is 2, not divisible by 3. 2020 is 2+0+2+0=4, not divisible by 3. 2200 is 2+2+0+0=4, same. 2220 is 6. So yes, 2220 is the first one.

But wait, is there a number with more digits but smaller in value? For example, maybe 2020 is smaller than 2220, but it doesn't work. How about numbers with leading zeros? But numbers can't have leading zeros. So, no.

But let me think again. Maybe there is a number with more digits but with fewer digits? No, more digits would make the number larger. So 2220 is the smallest.

Wait, but let me check if there is a smaller number with digits 0 and 2, ending with 0, and digit sum divisible by 3. Let me think of numbers with digit sum 3, 6, 9, etc. Since the digits are only 0 and 2, the digit sum is 2 times the number of 2s. So, the sum must be a multiple of 3. Therefore, the number of 2s must be such that 2 * (number of 2s) is divisible by 3. Therefore, the number of 2s must be a multiple of 3. Because 2 and 3 are coprime.

So, the number of 2s must be divisible by 3. Therefore, the number of 2s is 3, 6, 9, etc.

So, for example, if we have 3 twos and the rest zeros, then the digit sum is 6, which is divisible by 3.

So, the minimal number of digits would be 3 twos and one zero (since the number must end with 0). Wait, but the number must end with 0, so the last digit is 0. Therefore, the number of digits is (number of 2s) + 1 (for the last 0).

So, if we have 3 twos, then the number has 3 + 1 = 4 digits. Which is exactly the case of 2220.

Alternatively, if we have 6 twos, then the number would have 6 + 1 = 7 digits. Which would be larger than 2220. So, 2220 is the minimal.

But let me check if there is a way to have fewer digits. For example, if the number of 2s is 3, then the number of digits is 4. If we can have a number with fewer digits? Let's see.

If the number of 2s is 3, then the minimal number of digits is 4 (since one digit is 0). If we have fewer digits, say 3 digits. Then, the number of 2s would have to be 3, but then the number would be 222, but it doesn't end with 0. So, we need to have the last digit as 0. Therefore, the number of digits is at least 4.

Therefore, 2220 is the minimal.

Wait, but let me check if there is a number with 4 digits, with 3 twos and one zero, but arranged in a way that it's smaller than 2220. For example, 2022? Wait, but it has to end with 0. So, the last digit must be 0. Therefore, numbers like 2020, 2200, 2220.

So, 2020 is smaller than 2220, but its digit sum is 4. 2200 is also smaller than 2220, but digit sum is 4. 2220 is the only one with digit sum 6. Therefore, 2220 is the minimal.

But let me check if there is a number with more 2s but arranged in a way that the number is smaller. For example, if we have 6 twos and one zero, but that would be a 7-digit number, which is way bigger.

Alternatively, maybe there is a number with 3 twos and one zero, but arranged in a different way? But since the number must end with 0, the last digit is fixed. So, the first three digits can be arranged with 3 twos and 0? Wait, no. Wait, the total number of digits is 4. The last digit is 0, so the first three digits can be 2s and 0s. But if we need exactly 3 twos, then the first three digits must be 222, so the number is 2220. If we have fewer than 3 twos, then the digit sum would not be divisible by 3.

Wait, for example, if we have two twos, then the digit sum is 4, which is not divisible by 3. If we have one two, digit sum is 2, not divisible by 3. If we have zero twos, digit sum is 0, which is divisible by 3, but then the number would be 000...0, which is just 0, not positive.

Therefore, the minimal number of twos is 3, leading to the number 2220.

But let me check if there is a number with more digits but smaller in value. For example, maybe a number with 5 digits? Let's think. For example, 20020? Wait, but that ends with 0. Let me check the digit sum. 2 + 0 + 0 + 2 + 0 = 4, not divisible by 3. How about 20200? Sum is 2 + 0 + 2 + 0 + 0 = 4. Still 4. How about 22020? Sum is 2 + 2 + 0 + 2 + 0 = 6. That's divisible by 3. So 22020 is a 5-digit number. But 22020 is larger than 2220. So 2220 is still smaller.

Alternatively, 20020 is 20020, which is larger than 2220. So no.

Alternatively, maybe a number with 6 digits? For example, 200020? Sum is 2 + 0 + 0 + 0 + 2 + 0 = 4. Not good. 200200? Sum is 2 + 0 + 0 + 2 + 0 + 0 = 4. Still 4. 202020? Sum is 2 + 0 + 2 + 0 + 2 + 0 = 6. So 202020 is a 6-digit number. But again, it's larger than 2220.

So, seems like 2220 is the smallest.

But let me check if there is a number with 4 digits that is smaller than 2220. For example, numbers like 2020, 2000, 2200. But as we saw, their digit sums are 4, 2, 4. Not divisible by 3. So no.

Wait, but what about numbers with more than 3 twos? For example, 2220 is 3 twos. If we have 6 twos, but that would be longer.

Alternatively, maybe there is a number with 3 twos and one zero, but arranged in a different way? But since the number must end with 0, the only way to arrange 3 twos and one zero is 2220. Because if you put the zero in the middle, like 2202, but that doesn't end with 0. So, no.

Therefore, 2220 is the minimal.

But wait, let me check if there is a number with 4 digits, but with more zeros? For example, 2000, but that's sum 2. Not divisible by 3.

Alternatively, maybe there is a number with 4 digits where the digits are 2, 2, 2, 0, but arranged in a different way? But since the last digit must be 0, the first three digits are 2, 2, 2. So, 2220 is the only one.

Therefore, I think 2220 is the answer.

But let me check if there is a smaller number with more digits. For example, maybe a number with 5 digits that is smaller than 2220? But 2220 is a 4-digit number. Any 5-digit number is going to be larger than 2220. For example, 20000 is 20000, which is larger than 2220.

Wait, but 2000 is 2000, which is smaller than 2220, but it doesn't satisfy the divisibility by 3.

Alternatively, maybe there is a number with leading zeros? But numbers can't have leading zeros. So, no.

Therefore, I think 2220 is the minimal.

But let me check another approach. Since the number must be divisible by 30, which is 2*3*5. So, the number must be divisible by 2, 3, and 5. Divisible by 2 and 5 implies that it ends with 0, which we already considered. Divisible by 3 requires the sum of digits to be divisible by 3.

So, since digits are only 0 and 2, the sum is 2 times the number of 2s. Therefore, 2 * (number of 2s) must be divisible by 3. Therefore, number of 2s must be divisible by 3.

Therefore, the minimal number of 2s is 3, leading to the number with 3 twos and one zero (since it must end with 0). Therefore, the number is 2220.

Therefore, 2220 is the minimal.

But let me check if there is a smaller number with more 2s? For example, if we have 6 twos and one zero, but that would be 2222220, which is way bigger.

Alternatively, maybe there is a number with 3 twos and more zeros? But adding more zeros would make the number longer, hence larger.

Therefore, 2220 is the minimal.

But let me check if there is a number with 3 twos and one zero, but arranged in a way that it's smaller. For example, 2022? But that doesn't end with 0. If we have to end with 0, then the last digit is 0, so the number is 2220.

Alternatively, 2020? But that has only two 2s.

So, no.

Therefore, I think the answer is 2220.

But wait, let me check if there is a number with 3 twos and one zero, but with leading zeros? But leading zeros are not allowed.

Alternatively, maybe there is a number with 3 twos and one zero, but with more digits? For example, 20020? But that has two 2s. Wait, no. 20020 has two 2s.

Wait, if I have 3 twos and one zero, the total digits are 4. So, 2220. If I have more digits, like 5 digits, then I need to have 3 twos and two zeros. For example, 20200, which is 2+0+2+0+0=4, not divisible by 3. Or 20020, same thing. Or 22000, sum is 2+2+0+0+0=4. Not good.

Alternatively, 22200, which is 2+2+2+0+0=6. So, 22200 is a 5-digit number. But 22200 is larger than 2220.

So, 2220 is still the smallest.

Therefore, I think the answer is 2220.

But let me check if there is a smaller number with more digits. For example, maybe 2020? No, sum is 4. 2000? Sum is 2. 2200? Sum is 4. 2220? Sum is 6.

So, yes, 2220 is the first one.

Alternatively, maybe there is a number with 3 twos and one zero, but with some other arrangement? But since the last digit must be zero, the first three digits can be 2, 2, 2. So, 2220.

Alternatively, if the number is 2022, but that doesn't end with 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 their divisibility by 3.

Start from the smallest:

- 20: sum 2 → no
- 200: sum 2 → no
- 2000: sum 2 → no
- 20000: sum 2 → no
- 200000: sum 2 → no
- 2000000: sum 2 → no
- 20000000: sum 2 → no
- 200000000: sum 2 → no
- 2000000000: sum 2 → no
- 200...0 (any number of zeros): sum is 2, not divisible by 3.

Then numbers with two 2s:

- 220: sum 4 → no
- 2020: sum 4 → no
- 20020: sum 4 → no
- 200020: sum 4 → no
- 2000020: sum 4 → no
- etc. All these have sum 4, not divisible by 3.

Then numbers with three 2s:

- 2220: sum 6 → yes! Divisible by 3. So 2220 is the first one.

Therefore, 2220 is the answer.

But let me check if there is a number with three 2s and more zeros, but smaller than 2220. For example, 20220? But that's 20220, which is larger than 2220.

Alternatively, 200220? Also larger.

So, no.

Therefore, 2220 is the minimal.

But just to be thorough, let me check numbers with three 2s and one zero, but arranged differently. Since the number must end with 0, the first three digits can be:

- 222 → 2220
- 220 → 2200 (but that has only two 2s)
- 202 → 2020 (only two 2s)
- 022 → 0220, which is 220, which is a three-digit number, but we already checked that.

So, only 2220 is valid.

Therefore, I think the answer is 2220.

But let me check if there is a number with more digits but smaller. For example, 2020 is smaller than 2220, but it doesn't work. 2000 is smaller, but doesn't work.

Alternatively, maybe numbers with more 2s but arranged in a way that the number is smaller? For example, 2020 is smaller than 2220, but it doesn't work.

Alternatively, 20020 is larger than 2220.

Therefore, 2220 is indeed the smallest.

Wait, but let me check if there is a number with 3 twos and one zero, but with leading zeros? But numbers can't have leading zeros.

Therefore, I think I can conclude that the least positive integer multiple of 30 that can be written with only the digits 0 and 2 is 2220.

But just to be 100% sure, let me check if 2220 is indeed divisible by 30.

Divide 2220 by 30: 2220 ÷ 30 = 74. So, 30 × 74 = 2220. Yes, that's correct.

Therefore, 2220 is a multiple of 30, and it's composed only of 0s and 2s.

Is there any smaller number? Let me think again.

Suppose someone says 2020? No, because 2020 ÷ 30 = 67.333... Not integer.

What about 20020? 20020 ÷ 30 = 667.333... Not integer.

What about 22020? 22020 ÷ 30 = 734. So, that's an integer. But 22020 is larger than 2220.

So, 2220 is smaller.

Therefore, I think 2220 is the correct answer.

**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**. This gives us two key conditions:

1. **Divisibility by 10**: The number must **end in 0**.
2. **Divisibility by 3**: The **sum of its digits** must be divisible by 3.

Additionally, the number can **only contain the digits 0 and 2**.

---

### **Strategy**

We aim to find the **smallest** such number. Since the number must end in 0, we can focus on numbers that end in 0 and are composed only of 0s and 2s.

Let’s consider the **sum of digits**. Since the digits are only 0 and 2, the sum of the digits is simply **2 × (number of 2s)**. For the number to be divisible by 3, this sum must be a multiple of 3.

So, the number of 2s must be a **multiple of 3** (since 2 and 3 are coprime). The **smallest such number of 2s** is **3**, which gives a digit sum of **6**, divisible by 3.

---

### **Constructing the Smallest Valid Number**

We now need to construct the **smallest number** using **3 twos and one zero**, with the **zero at the end** (to satisfy divisibility by 10). The only such number is:

- **2220**

Let’s verify:

- **Ends in 0** ✅
- **Digits are only 0 and 2** ✅
- **Sum of digits = 2 + 2 + 2 + 0 = 6**, which is divisible by 3 ✅
- **Divisible by 30** ✅ (since it's divisible by both 3 and 10)

---

### **Checking for Smaller Numbers**

We can check smaller numbers with fewer digits:

- **20** → sum = 2 → not divisible by 3
- **200** → sum = 2 → not divisible by 3
- **2000** → sum = 2 → not divisible by 3
- **220** → sum = 4 → not divisible by 3
- **2020** → sum = 4 → not divisible by 3
- **2200** → sum = 4 → not divisible by 3

All of these fail the divisibility by 3 condition. The next candidate is **2220**, which satisfies all the conditions and is the **smallest such number**.

---

### **Conclusion**

After analyzing all possible combinations and verifying the conditions, 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. Divisibility by 10 means the number must end with a 0. 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, the digits are restricted to 0 and 2. Let me note that down:

- Ends with 0 (for divisibility by 10)
- Sum of digits is a multiple of 3 (for divisibility by 3)
- Only digits 0 and 2 are allowed

So, the problem reduces to finding the smallest number composed of 0s and 2s that ends with 0 and has a digit sum divisible by 3.

Let me think about how to construct such a number. Since it has to end with 0, the last digit is fixed as 0. The rest of the digits can be 0s and 2s. But we need the sum of digits to be divisible by 3.

Let me consider the possible numbers. Since we want the smallest such number, we should aim for the smallest number of digits possible. Let me start checking from the smallest possible numbers.

First, let's think about numbers with 1 digit. The only possible numbers are 0 and 2. But 0 is not positive, and 2 is not divisible by 30. So, no luck here.

Next, numbers with 2 digits. The number must end with 0, so the first digit can be 2 or 0. But since it's a two-digit number, the first digit can't be 0. So, the only possibility is 20. Let's check if 20 is divisible by 30. 20 divided by 30 is 0.666..., so no. Also, the sum of digits is 2 + 0 = 2, which is not divisible by 3. So, 20 is out.

Moving on to three-digit numbers. The number must end with 0. So, the first two digits can be 0s and 2s. Let's list all possibilities:

- 200: sum is 2 + 0 + 0 = 2 → not divisible by 3
- 220: sum is 2 + 2 + 0 = 4 → not divisible by 3
- 020: but that's 20, which we already considered, and leading zeros aren't allowed. So, only 200 and 220. Neither works.

Four-digit numbers. Again, last digit is 0. The first three digits can be 0s and 2s. Let's think about possible numbers:

Possible numbers:

- 2000: sum is 2 → nope
- 2020: sum is 2 + 0 + 2 + 0 = 4 → nope
- 2200: sum is 2 + 2 + 0 + 0 = 4 → nope
- 2220: sum is 2 + 2 + 2 + 0 = 6 → 6 is divisible by 3! Wait, 2220. Let me check if that's divisible by 30.

Divisible by 10? Yes, ends with 0. Divisible by 3? Sum of digits is 6, which is divisible by 3. So, 2220 is divisible by 30. Is this the smallest? Let me check if there is a smaller number with fewer digits.

Wait, we checked 2-digit and 3-digit numbers, and they didn't work. So 4-digit numbers: 2220 is a candidate. But let me check if there is a smaller 4-digit number. For example, numbers like 2020, 2000, 2200, etc., but their digit sums are 4, 2, 4, which aren't divisible by 3. So 2220 is the only 4-digit number that works.

But wait, maybe there is a smaller number with more digits? Wait, no. Because 2220 is a 4-digit number. If there is a smaller number with more digits, it would have to have more digits, which would make it larger. For example, a 5-digit number would be larger than a 4-digit number. So 2220 is the smallest? Wait, but maybe there is a smaller number with more digits? Wait, no. For example, 2020 is smaller than 2220, but it doesn't satisfy the divisibility by 3. Similarly, 2000 is smaller but also doesn't work. So 2220 is the first one that works.

But let me check if there is a smaller number with more digits. Wait, for example, maybe a number with 5 digits that is smaller than 2220? But 2220 is 2220, and a 5-digit number would be at least 20000, which is way bigger. So no. So 2220 is the smallest? Wait, but let me think again.

Wait, maybe there is a number with 4 digits that is smaller than 2220? Let's see. The numbers with 4 digits starting with 2, then followed by digits. The smallest 4-digit number with digits 0 and 2 is 2000. Then 2000, 2020, 2200, 2220. So 2000 is 2000, which is smaller than 2220. But 2000's digit sum is 2, not divisible by 3. 2020 is 2+0+2+0=4, not divisible by 3. 2200 is 2+2+0+0=4, same. 2220 is 6. So yes, 2220 is the first one.

But wait, is there a number with more digits but smaller in value? For example, maybe 2020 is smaller than 2220, but it doesn't work. How about numbers with leading zeros? But numbers can't have leading zeros. So, no.

But let me think again. Maybe there is a number with more digits but with fewer digits? No, more digits would make the number larger. So 2220 is the smallest.

Wait, but let me check if there is a smaller number with digits 0 and 2, ending with 0, and digit sum divisible by 3. Let me think of numbers with digit sum 3, 6, 9, etc. Since the digits are only 0 and 2, the digit sum is 2 times the number of 2s. So, the sum must be a multiple of 3. Therefore, the number of 2s must be such that 2 * (number of 2s) is divisible by 3. Therefore, the number of 2s must be a multiple of 3. Because 2 and 3 are coprime.

So, the number of 2s must be divisible by 3. Therefore, the number of 2s is 3, 6, 9, etc.

So, for example, if we have 3 twos and the rest zeros, then the digit sum is 6, which is divisible by 3.

So, the minimal number of digits would be 3 twos and one zero (since the number must end with 0). Wait, but the number must end with 0, so the last digit is 0. Therefore, the number of digits is (number of 2s) + 1 (for the last 0).

So, if we have 3 twos, then the number has 3 + 1 = 4 digits. Which is exactly the case of 2220.

Alternatively, if we have 6 twos, then the number would have 6 + 1 = 7 digits. Which would be larger than 2220. So, 2220 is the minimal.

But let me check if there is a way to have fewer digits. For example, if the number of 2s is 3, then the number of digits is 4. If we can have a number with fewer digits? Let's see.

If the number of 2s is 3, then the minimal number of digits is 4 (since one digit is 0). If we have fewer digits, say 3 digits. Then, the number of 2s would have to be 3, but then the number would be 222, but it doesn't end with 0. So, we need to have the last digit as 0. Therefore, the number of digits is at least 4.

Therefore, 2220 is the minimal.

Wait, but let me check if there is a number with 4 digits, with 3 twos and one zero, but arranged in a way that it's smaller than 2220. For example, 2022? Wait, but it has to end with 0. So, the last digit must be 0. Therefore, numbers like 2020, 2200, 2220.

So, 2020 is smaller than 2220, but its digit sum is 4. 2200 is also smaller than 2220, but digit sum is 4. 2220 is the only one with digit sum 6. Therefore, 2220 is the minimal.

But let me check if there is a number with more 2s but arranged in a way that the number is smaller. For example, if we have 6 twos and one zero, but that would be a 7-digit number, which is way bigger.

Alternatively, maybe there is a number with 3 twos and one zero, but arranged in a different way? But since the number must end with 0, the last digit is fixed. So, the first three digits can be arranged with 3 twos and 0? Wait, no. Wait, the total number of digits is 4. The last digit is 0, so the first three digits can be 2s and 0s. But if we need exactly 3 twos, then the first three digits must be 222, so the number is 2220. If we have fewer than 3 twos, then the digit sum would not be divisible by 3.

Wait, for example, if we have two twos, then the digit sum is 4, which is not divisible by 3. If we have one two, digit sum is 2, not divisible by 3. If we have zero twos, digit sum is 0, which is divisible by 3, but then the number would be 000...0, which is just 0, not positive.

Therefore, the minimal number of twos is 3, leading to the number 2220.

But let me check if there is a number with more digits but smaller in value. For example, maybe a number with 5 digits? Let's think. For example, 20020? Wait, but that ends with 0. Let me check the digit sum. 2 + 0 + 0 + 2 + 0 = 4, not divisible by 3. How about 20200? Sum is 2 + 0 + 2 + 0 + 0 = 4. Still 4. How about 22020? Sum is 2 + 2 + 0 + 2 + 0 = 6. That's divisible by 3. So 22020 is a 5-digit number. But 22020 is larger than 2220. So 2220 is still smaller.

Alternatively, 20020 is 20020, which is larger than 2220. So no.

Alternatively, maybe a number with 6 digits? For example, 200020? Sum is 2 + 0 + 0 + 0 + 2 + 0 = 4. Not good. 200200? Sum is 2 + 0 + 0 + 2 + 0 + 0 = 4. Still 4. 202020? Sum is 2 + 0 + 2 + 0 + 2 + 0 = 6. So 202020 is a 6-digit number. But again, it's larger than 2220.

So, seems like 2220 is the smallest.

But let me check if there is a number with 4 digits that is smaller than 2220. For example, numbers like 2020, 2000, 2200. But as we saw, their digit sums are 4, 2, 4. Not divisible by 3. So no.

Wait, but what about numbers with more than 3 twos? For example, 2220 is 3 twos. If we have 6 twos, but that would be longer.

Alternatively, maybe there is a number with 3 twos and one zero, but arranged in a different way? But since the number must end with 0, the only way to arrange 3 twos and one zero is 2220. Because if you put the zero in the middle, like 2202, but that doesn't end with 0. So, no.

Therefore, 2220 is the minimal.

But wait, let me check if there is a number with 4 digits, but with more zeros? For example, 2000, but that's sum 2. Not divisible by 3.

Alternatively, maybe there is a number with 4 digits where the digits are 2, 2, 2, 0, but arranged in a different way? But since the last digit must be 0, the first three digits are 2, 2, 2. So, 2220 is the only one.

Therefore, I think 2220 is the answer.

But let me check if there is a smaller number with more digits. For example, maybe a number with 5 digits that is smaller than 2220? But 2220 is a 4-digit number. Any 5-digit number is going to be larger than 2220. For example, 20000 is 20000, which is larger than 2220.

Wait, but 2000 is 2000, which is smaller than 2220, but it doesn't satisfy the divisibility by 3.

Alternatively, maybe there is a number with leading zeros? But numbers can't have leading zeros. So, no.

Therefore, I think 2220 is the minimal.

But let me check another approach. Since the number must be divisible by 30, which is 2*3*5. So, the number must be divisible by 2, 3, and 5. Divisible by 2 and 5 implies that it ends with 0, which we already considered. Divisible by 3 requires the sum of digits to be divisible by 3.

So, since digits are only 0 and 2, the sum is 2 times the number of 2s. Therefore, 2 * (number of 2s) must be divisible by 3. Therefore, number of 2s must be divisible by 3.

Therefore, the minimal number of 2s is 3, leading to the number with 3 twos and one zero (since it must end with 0). Therefore, the number is 2220.

Therefore, 2220 is the minimal.

But let me check if there is a smaller number with more 2s? For example, if we have 6 twos and one zero, but that would be 2222220, which is way bigger.

Alternatively, maybe there is a number with 3 twos and more zeros? But adding more zeros would make the number longer, hence larger.

Therefore, 2220 is the minimal.

But let me check if there is a number with 3 twos and one zero, but arranged in a way that it's smaller. For example, 2022? But that doesn't end with 0. If we have to end with 0, then the last digit is 0, so the number is 2220.

Alternatively, 2020? But that has only two 2s.

So, no.

Therefore, I think the answer is 2220.

But wait, let me check if there is a number with 3 twos and one zero, but with leading zeros? But leading zeros are not allowed.

Alternatively, maybe there is a number with 3 twos and one zero, but with more digits? For example, 20020? But that has two 2s. Wait, no. 20020 has two 2s.

Wait, if I have 3 twos and one zero, the total digits are 4. So, 2220. If I have more digits, like 5 digits, then I need to have 3 twos and two zeros. For example, 20200, which is 2+0+2+0+0=4, not divisible by 3. Or 20020, same thing. Or 22000, sum is 2+2+0+0+0=4. Not good.

Alternatively, 22200, which is 2+2+2+0+0=6. So, 22200 is a 5-digit number. But 22200 is larger than 2220.

So, 2220 is still the smallest.

Therefore, I think the answer is 2220.

But let me check if there is a smaller number with more digits. For example, maybe 2020? No, sum is 4. 2000? Sum is 2. 2200? Sum is 4. 2220? Sum is 6.

So, yes, 2220 is the first one.

Alternatively, maybe there is a number with 3 twos and one zero, but with some other arrangement? But since the last digit must be zero, the first three digits can be 2, 2, 2. So, 2220.

Alternatively, if the number is 2022, but that doesn't end with 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 their divisibility by 3.

Start from the smallest:

- 20: sum 2 → no
- 200: sum 2 → no
- 2000: sum 2 → no
- 20000: sum 2 → no
- 200000: sum 2 → no
- 2000000: sum 2 → no
- 20000000: sum 2 → no
- 200000000: sum 2 → no
- 2000000000: sum 2 → no
- 200...0 (any number of zeros): sum is 2, not divisible by 3.

Then numbers with two 2s:

- 220: sum 4 → no
- 2020: sum 4 → no
- 20020: sum 4 → no
- 200020: sum 4 → no
- 2000020: sum 4 → no
- etc. All these have sum 4, not divisible by 3.

Then numbers with three 2s:

- 2220: sum 6 → yes! Divisible by 3. So 2220 is the first one.

Therefore, 2220 is the answer.

But let me check if there is a number with three 2s and more zeros, but smaller than 2220. For example, 20220? But that's 20220, which is larger than 2220.

Alternatively, 200220? Also larger.

So, no.

Therefore, 2220 is the minimal.

But just to be thorough, let me check numbers with three 2s and one zero, but arranged differently. Since the number must end with 0, the first three digits can be:

- 222 → 2220
- 220 → 2200 (but that has only two 2s)
- 202 → 2020 (only two 2s)
- 022 → 0220, which is 220, which is a three-digit number, but we already checked that.

So, only 2220 is valid.

Therefore, I think the answer is 2220.

But let me check if there is a number with more digits but smaller. For example, 2020 is smaller than 2220, but it doesn't work. 2000 is smaller, but doesn't work.

Alternatively, maybe numbers with more 2s but arranged in a way that the number is smaller? For example, 2020 is smaller than 2220, but it doesn't work.

Alternatively, 20020 is larger than 2220.

Therefore, 2220 is indeed the smallest.

Wait, but let me check if there is a number with 3 twos and one zero, but with leading zeros? But numbers can't have leading zeros.

Therefore, I think I can conclude that the least positive integer multiple of 30 that can be written with only the digits 0 and 2 is 2220.

But just to be 100% sure, let me check if 2220 is indeed divisible by 30.

Divide 2220 by 30: 2220 ÷ 30 = 74. So, 30 × 74 = 2220. Yes, that's correct.

Therefore, 2220 is a multiple of 30, and it's composed only of 0s and 2s.

Is there any smaller number? Let me think again.

Suppose someone says 2020? No, because 2020 ÷ 30 = 67.333... Not integer.

What about 20020? 20020 ÷ 30 = 667.333... Not integer.

What about 22020? 22020 ÷ 30 = 734. So, that's an integer. But 22020 is larger than 2220.

So, 2220 is smaller.

Therefore, I think 2220 is the correct answer.

**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**. This gives us two key conditions:

1. **Divisibility by 10**: The number must **end in 0**.
2. **Divisibility by 3**: The **sum of its digits** must be divisible by 3.

Additionally, the number can **only contain the digits 0 and 2**.

---

### **Strategy**

We aim to find the **smallest** such number. Since the number must end in 0, we can focus on numbers that end in 0 and are composed only of 0s and 2s.

Let’s consider the **sum of digits**. Since the digits are only 0 and 2, the sum of the digits is simply **2 × (number of 2s)**. For the number to be divisible by 3, this sum must be a multiple of 3.

So, the number of 2s must be a **multiple of 3** (since 2 and 3 are coprime). The **smallest such number of 2s** is **3**, which gives a digit sum of **6**, divisible by 3.

---

### **Constructing the Smallest Valid Number**

We now need to construct the **smallest number** using **3 twos and one zero**, with the **zero at the end** (to satisfy divisibility by 10). The only such number is:

- **2220**

Let’s verify:

- **Ends in 0** ✅
- **Digits are only 0 and 2** ✅
- **Sum of digits = 2 + 2 + 2 + 0 = 6**, which is divisible by 3 ✅
- **Divisible by 30** ✅ (since it's divisible by both 3 and 10)

---

### **Checking for Smaller Numbers**

We can check smaller numbers with fewer digits:

- **20** → sum = 2 → not divisible by 3
- **200** → sum = 2 → not divisible by 3
- **2000** → sum = 2 → not divisible by 3
- **220** → sum = 4 → not divisible by 3
- **2020** → sum = 4 → not divisible by 3
- **2200** → sum = 4 → not divisible by 3

All of these fail the divisibility by 3 condition. The next candidate is **2220**, which satisfies all the conditions and is the **smallest such number**.

---

### **Conclusion**

After analyzing all possible combinations and verifying the conditions, 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. Therefore, the cubic 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, if discriminant is negative, quadratic has complex roots, so cubic would have one real root and two complex roots. Therefore, to have all roots real, discriminant must be non-negative, 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 maybe 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 all roots of cubic are real. Therefore, minimal a is 3. But let me check with 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. So that's okay. If a is less than 3, then quadratic has complex roots, so cubic has one real root and two complex roots. Therefore, the minimal a is 3. But wait, the problem says "all the roots are real", so maybe there's another case where even if discriminant is negative, but... no, that's not possible. So maybe answer is 3? But I have a feeling that maybe there's something wrong here. Wait, let me check again.

Wait, but maybe there's another way to have all roots real? For example, maybe if the quadratic has a double root? But even so, discriminant is zero. So when discriminant is zero, quadratic has a double root. So for a = 3, discriminant is zero. So that's the minimal a? But maybe there's another case where the cubic has three real roots, but not necessarily with x = -1 as a root? Wait, but we already factored the cubic as (x + 1)(quadratic). So if x = -1 is a root, then the other roots come from quadratic. Therefore, if we want all roots real, quadratic must have real roots. Therefore, discriminant must be non-negative. Therefore, a ≥ 3. Therefore, minimal a is 3. But maybe there's a mistake here? Let me check with a different approach.

Alternatively, maybe I can use 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 the cubic as (x + 1)(x² + (a - 1)x + 1), then the discriminant of the cubic is the discriminant of the quadratic times something? Wait, maybe not. Alternatively, since we already have one real root, and the other two roots depend on the quadratic. Therefore, the condition is that quadratic has real roots, so discriminant ≥ 0. Therefore, leading to a ≥ 3. Therefore, answer is 3. But maybe I need to check if there are other possibilities where the cubic has three real roots without x = -1 being a root? But since we have x = -1 as a root for any a, so if the quadratic has real roots, then all roots are real. If quadratic has complex roots, then only one real root. Therefore, the only way for all roots to be real is discriminant of quadratic ≥ 0, so a ≥ 3. Therefore, minimal a is 3. But maybe I need to check if there's a case where even if discriminant is negative, but... no. So answer is 3? But I have a doubt because sometimes problems like this have more nuanced answers. Let me check with a = 3. Then equation is (x + 1)^3 = 0. All roots are -1. So that's okay. If a is slightly more than 3, say a = 4, then quadratic is x² + 3x + 1, which has roots [-3 ± sqrt(9 - 4)]/2 = [-3 ± sqrt(5)]/2, which are real. So that's okay. If a is less than 3, say a = 2, quadratic is x² + x + 1, discriminant 1 - 4 = -3 < 0. So complex roots. Therefore, only one real root. Therefore, seems like a must be ≥ 3. Therefore, minimal a is 3. But maybe the problem is trickier? Let me think again.

Wait, but maybe there's another way to have all roots real? For example, if the quadratic has two real roots, but maybe with some multiplicities? But even so, discriminant must be non-negative. So same condition. Therefore, answer is 3. But maybe I need to check with another method. Let me try using calculus. Let me consider the cubic function f(x) = x³ + a x² + a x + 1. To have three real roots, the function must have a local maximum and a local minimum, and the function must cross the x-axis three times. Therefore, first, find the critical points by taking derivative:

f'(x) = 3x² + 2a x + a.

Set derivative equal to zero: 3x² + 2a x + a = 0. For there to be two critical points, discriminant must be positive. Discriminant D' = (2a)^2 - 4 * 3 * a = 4a² - 12a = 4a(a - 3). Therefore, discriminant is positive when 4a(a - 3) > 0. Since a is positive, this is when a - 3 > 0 ⇒ a > 3. Therefore, when a > 3, there are two critical points. When a = 3, discriminant is zero, so one critical point (a double root). When a < 3, no critical points. Wait, but if there are no critical points, then the function is strictly increasing or decreasing? Since leading coefficient is positive, as x approaches infinity, f(x) approaches infinity, and as x approaches negative infinity, f(x) approaches negative infinity. Therefore, if there are no critical points, the function is strictly increasing, so it crosses the x-axis only once. Therefore, for a < 3, only one real root. For a = 3, derivative is 3x² + 6x + 3 = 3(x² + 2x + 1) = 3(x + 1)^2, so derivative is zero only at x = -1. Therefore, the function has a horizontal tangent at x = -1, but since it's a double root in the derivative, does that mean the function has a point of inflection? Wait, but if the derivative has a double root, then the function has a stationary point at x = -1, but since the derivative doesn't change sign there (since it's a square), the function is increasing everywhere? Wait, let me check. For a = 3, f'(x) = 3(x + 1)^2 ≥ 0 for all x, so the function is non-decreasing. Therefore, it can have only one real root (since it's strictly increasing except at x = -1 where derivative is zero). But when a = 3, the original equation is (x + 1)^3 = 0, so triple root at x = -1. So even though the derivative is zero there, the function still only crosses the x-axis once (with multiplicity three). Therefore, for a = 3, there is only one real root (with multiplicity three). Wait, but the problem says "all the roots are real". If we count multiplicity, then yes, all roots are real. But if we consider distinct roots, then there is only one distinct root. However, the problem says "all the roots are real", which usually includes multiplicity. So in that case, a = 3 is acceptable. However, if the problem requires three distinct real roots, then a must be greater than 3. But the problem doesn't specify distinct roots, just that all roots are real. Therefore, a = 3 is acceptable. Therefore, the minimal value of a is 3. But let me check again with the previous approach where we factored the cubic. If a = 3, then quadratic factor is x² + 2x + 1 = (x + 1)^2, so the roots are -1, -1, -1. So all real. Therefore, answer is 3. But maybe the problem expects a different answer? Wait, let me check with another approach. Let me suppose that all roots are real. Let me denote them as r, s, t. Then, as per Vieta:

r + s + t = -a

rs + rt + st = a

rst = -1

Since all roots are real, and a is positive. Let me try to find relations between these equations. Let me consider that 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. Let me think about possible roots. Since the product rst = -1, so either one negative and two positive roots, or three negative roots. But if all roots are negative, then their sum would be negative, which matches r + s + t = -a < 0. Also, rs + rt + st would be positive because product of two negatives is positive, and sum of positive terms. Alternatively, if there is one negative and two positive roots, then the product rst would be negative (since one negative and two positive), which matches rst = -1. So both cases are possible.

But let me check if there are constraints. Let me suppose that all roots are negative. Let me set r = -p, s = -q, t = -k, where p, q, k are positive real numbers. Then:

Sum: -p - q - k = -a ⇒ p + q + k = a

Sum of products: (-p)(-q) + (-p)(-k) + (-q)(-k) = pq + pk + qk = a

Product: (-p)(-q)(-k) = -p q k = -1 ⇒ p q k = 1

So we have p + q + k = a, pq + pk + qk = a, and p q k = 1, with p, q, k > 0.

So we need to find the minimal a such that there exist positive real numbers p, q, k with these three equations. So this seems like a symmetric system. Maybe we can assume that p = q = k? Let me check. If p = q = k, then:

3p = a

3p² = a

p³ = 1 ⇒ p = 1

Therefore, 3(1) = a ⇒ a = 3. So that gives a = 3. So that's the case where all roots are equal to -1. Which matches our previous result. So that's one case. But maybe there are other cases where p, q, k are not equal, but still satisfy the equations, leading to a smaller a? Let me check.

Suppose that two variables are equal, say p = q ≠ k. Let me set p = q. Then:

From p + p + k = a ⇒ 2p + k = a

From p² + 2pk = a

From p² k = 1

So we have three equations:

1) 2p + k = a

2) p² + 2pk = a

3) p² k = 1

Let me try to solve these equations. From equations 1 and 2, since both equal to a:

2p + k = p² + 2pk ⇒ 2p + k = p² + 2pk ⇒ Let me rearrange:

p² + 2pk - 2p - k = 0 ⇒ p² - 2p + 2pk - k = 0 ⇒ p² - 2p + k(2p - 1) = 0

Hmm, not sure. Let me try to express k from equation 1: k = a - 2p. Then plug into equation 2:

p² + 2p(a - 2p) = a ⇒ p² + 2pa - 4p² = a ⇒ -3p² + 2pa = a ⇒ -3p² + 2pa - a = 0 ⇒ Let me write it as:

-3p² + 2pa - a = 0 ⇒ Multiply both sides by -1:

3p² - 2pa + a = 0

Now, from equation 3: p² k = 1. Since k = a - 2p, then p²(a - 2p) = 1 ⇒ a p² - 2p³ = 1.

So now, we have two equations:

Equation A: 3p² - 2pa + a = 0

Equation B: a p² - 2p³ = 1

Let me try to solve these equations. From equation A: 3p² - 2pa + a = 0 ⇒ Let me solve for a:

3p² + a(1 - 2p) = 0 ⇒ a(1 - 2p) = -3p² ⇒ a = (-3p²)/(1 - 2p) = (3p²)/(2p - 1)

So a is expressed in terms of p. Now, substitute this into equation B:

a p² - 2p³ = 1 ⇒ [(3p²)/(2p - 1)] * p² - 2p³ = 1 ⇒ [3p^4 / (2p - 1)] - 2p³ = 1

Let me combine the terms:

Multiply both sides by (2p - 1):

3p^4 - 2p³(2p - 1) = 2p - 1

Compute left side:

3p^4 - 4p^4 + 2p³ = -p^4 + 2p³

Therefore:

-p^4 + 2p³ = 2p - 1 ⇒ -p^4 + 2p³ - 2p + 1 = 0 ⇒ Multiply by -1:

p^4 - 2p³ + 2p - 1 = 0

So need to solve p^4 - 2p³ + 2p - 1 = 0. Let me try to factor this equation. Let me try rational roots. Possible rational roots are ±1.

Testing p = 1: 1 - 2 + 2 - 1 = 0. Yes, p = 1 is a root. Therefore, factor out (p - 1):

Using polynomial division or synthetic division. Let me do synthetic division with root 1:

Coefficients: 1 (p^4), -2 (p³), 0 (p²), 2 (p), -1 (constant)

Wait, original polynomial is p^4 - 2p³ + 0p² + 2p - 1.

Divide by (p - 1):

Bring down 1.

Multiply by 1: 1. Add to next coefficient: -2 + 1 = -1.

Multiply by 1: -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by 1: -1. Add to next coefficient: 2 + (-1) = 1.

Multiply by 1: 1. Add to last coefficient: -1 + 1 = 0. Perfect.

Therefore, polynomial factors as (p - 1)(p³ - p² - p + 1). Now factor the cubic: p³ - p² - p + 1.

Group terms: (p³ - p²) + (-p + 1) = p²(p - 1) -1(p - 1) = (p² - 1)(p - 1) = (p - 1)(p + 1)(p - 1). Wait:

Wait, p³ - p² - p + 1 = p²(p - 1) -1(p - 1) = (p² - 1)(p - 1) = (p - 1)(p + 1)(p - 1) = (p - 1)^2(p + 1). Therefore, the original polynomial factors as (p - 1)(p - 1)^2(p + 1) = (p - 1)^3(p + 1). Therefore, roots are p = 1 (triple root) and p = -1. Since p is positive, p = -1 is rejected. Therefore, only p = 1 is valid. Therefore, p = 1. Therefore, from earlier, a = (3p²)/(2p - 1) = 3(1)/(2 - 1) = 3/1 = 3. Therefore, again, a = 3. Therefore, even in this case, the minimal a is 3. Therefore, even if we assume two variables equal, we still get a = 3. Therefore, seems like 3 is the minimal value. Therefore, maybe the answer is indeed 3. But let me check if there are other cases where p, q, k are not equal or two equal, but still give a smaller a. Suppose that all three variables are different. But how can I check that? Maybe using inequalities. Let me think.

We have p + q + k = a, pq + pk + qk = a, and p q k = 1. Let me consider that for positive real numbers p, q, k. Let me denote S = p + q + k = a, P = pq + pk + qk = a, and Q = p q k = 1. So S = P. Let me recall that for three variables, we have the relation between S, P, Q. Also, we can think of the cubic equation whose roots are p, q, k: x³ - S x² + P x - Q = 0 ⇒ x³ - a x² + a x - 1 = 0. Wait, but this is similar to the original equation. Wait, original equation was x³ + a x² + a x + 1 = 0. If we substitute x = -y, then (-y)^3 + a (-y)^2 + a (-y) + 1 = -y³ + a y² - a y + 1 = 0 ⇒ -y³ + a y² - a y + 1 = 0 ⇒ y³ - a y² + a y - 1 = 0. Which is the same as the cubic equation for p, q, k. Therefore, the roots p, q, k are the roots of y³ - a y² + a y - 1 = 0. Which is same as original equation with x replaced by -y. Therefore, if original equation has roots r, s, t, then this equation has roots -r, -s, -t. Therefore, since original equation had roots with product -1, then this equation has roots with product 1. Which matches Q = 1. Therefore, this seems consistent.

But how does this help? Maybe not much. Let me think about inequalities. Since p, q, k are positive real numbers, and S = P. Let me recall that for positive real numbers, by AM ≥ GM:

(p + q + k)/3 ≥ (p q k)^{1/3} ⇒ a/3 ≥ 1^{1/3} ⇒ a/3 ≥ 1 ⇒ a ≥ 3. Equality holds when p = q = k. Therefore, this gives that a ≥ 3. Therefore, the minimal value of a is 3, achieved when p = q = k = 1. Therefore, this confirms our previous result. Therefore, regardless of the approach, the minimal a is 3. Therefore, the answer is 3. Therefore, the smallest possible value of a is 3. Therefore, I think the answer is \boxed{3}.

But just to be thorough, let me check if there's a case where the cubic has three real roots with a < 3. Suppose someone claims that maybe with some specific roots, even though discriminant of quadratic is negative, but... no, because if quadratic has complex roots, then cubic has only one real root. Therefore, not all roots real. Therefore, the minimal a is indeed 3. Therefore, I think the answer is 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 a root for any value of $ a $. Therefore, we can factor the cubic 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:

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

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: Verify the Minimal Value

Let’s test $ a = 3 $:

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

This has a **triple root** at $ x = -1 $, so all roots are real (with multiplicity). Therefore, $ a = 3 $ satisfies the condition.

Now, consider $ a < 3 $. For example, take $ a = 2 $:

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

The discriminant is:

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

So the quadratic has **complex roots**, and the cubic has only one real root (at $ x = -1 $). Thus, **not all roots are real**.

Therefore, the **smallest possible value** of $ a $ such that 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. Therefore, the cubic 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, if discriminant is negative, quadratic has complex roots, so cubic would have one real root and two complex roots. Therefore, to have all roots real, discriminant must be non-negative, 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 maybe 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 all roots of cubic are real. Therefore, minimal a is 3. But let me check with 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. So that's okay. If a is less than 3, then quadratic has complex roots, so cubic has one real root and two complex roots. Therefore, the minimal a is 3. But wait, the problem says "all the roots are real", so maybe there's another case where even if discriminant is negative, but... no, that's not possible. So maybe answer is 3? But I have a feeling that maybe there's something wrong here. Wait, let me check again.

Wait, but maybe there's another way to have all roots real? For example, maybe if the quadratic has a double root? But even so, discriminant is zero. So when discriminant is zero, quadratic has a double root. So for a = 3, discriminant is zero. So that's the minimal a? But maybe there's another case where the cubic has three real roots, but not necessarily with x = -1 as a root? Wait, but we already factored the cubic as (x + 1)(quadratic). So if x = -1 is a root, then the other roots come from quadratic. Therefore, if we want all roots real, quadratic must have real roots. Therefore, discriminant must be non-negative. Therefore, a ≥ 3. Therefore, minimal a is 3. But maybe there's a mistake here? Let me check with a different approach.

Alternatively, maybe I can use 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 the cubic as (x + 1)(x² + (a - 1)x + 1), then the discriminant of the cubic is the discriminant of the quadratic times something? Wait, maybe not. Alternatively, since we already have one real root, and the other two roots depend on the quadratic. Therefore, the condition is that quadratic has real roots, so discriminant ≥ 0. Therefore, leading to a ≥ 3. Therefore, answer is 3. But maybe I need to check if there are other possibilities where the cubic has three real roots without x = -1 being a root? But since we have x = -1 as a root for any a, so if the quadratic has real roots, then all roots are real. If quadratic has complex roots, then only one real root. Therefore, the only way for all roots to be real is discriminant of quadratic ≥ 0, so a ≥ 3. Therefore, minimal a is 3. But maybe I need to check if there's a case where even if discriminant is negative, but... no. So answer is 3? But I have a doubt because sometimes problems like this have more nuanced answers. Let me check with a = 3. Then equation is (x + 1)^3 = 0. All roots are -1. So that's okay. If a is slightly more than 3, say a = 4, then quadratic is x² + 3x + 1, which has roots [-3 ± sqrt(9 - 4)]/2 = [-3 ± sqrt(5)]/2, which are real. So that's okay. If a is less than 3, say a = 2, quadratic is x² + x + 1, discriminant 1 - 4 = -3 < 0. So complex roots. Therefore, only one real root. Therefore, seems like a must be ≥ 3. Therefore, minimal a is 3. But maybe the problem is trickier? Let me think again.

Wait, but maybe there's another way to have all roots real? For example, if the quadratic has two real roots, but maybe with some multiplicities? But even so, discriminant must be non-negative. So same condition. Therefore, answer is 3. But maybe I need to check with another method. Let me try using calculus. Let me consider the cubic function f(x) = x³ + a x² + a x + 1. To have three real roots, the function must have a local maximum and a local minimum, and the function must cross the x-axis three times. Therefore, first, find the critical points by taking derivative:

f'(x) = 3x² + 2a x + a.

Set derivative equal to zero: 3x² + 2a x + a = 0. For there to be two critical points, discriminant must be positive. Discriminant D' = (2a)^2 - 4 * 3 * a = 4a² - 12a = 4a(a - 3). Therefore, discriminant is positive when 4a(a - 3) > 0. Since a is positive, this is when a - 3 > 0 ⇒ a > 3. Therefore, when a > 3, there are two critical points. When a = 3, discriminant is zero, so one critical point (a double root). When a < 3, no critical points. Wait, but if there are no critical points, then the function is strictly increasing or decreasing? Since leading coefficient is positive, as x approaches infinity, f(x) approaches infinity, and as x approaches negative infinity, f(x) approaches negative infinity. Therefore, if there are no critical points, the function is strictly increasing, so it crosses the x-axis only once. Therefore, for a < 3, only one real root. For a = 3, derivative is 3x² + 6x + 3 = 3(x² + 2x + 1) = 3(x + 1)^2, so derivative is zero only at x = -1. Therefore, the function has a horizontal tangent at x = -1, but since it's a double root in the derivative, does that mean the function has a point of inflection? Wait, but if the derivative has a double root, then the function has a stationary point at x = -1, but since the derivative doesn't change sign there (since it's a square), the function is increasing everywhere? Wait, let me check. For a = 3, f'(x) = 3(x + 1)^2 ≥ 0 for all x, so the function is non-decreasing. Therefore, it can have only one real root (since it's strictly increasing except at x = -1 where derivative is zero). But when a = 3, the original equation is (x + 1)^3 = 0, so triple root at x = -1. So even though the derivative is zero there, the function still only crosses the x-axis once (with multiplicity three). Therefore, for a = 3, there is only one real root (with multiplicity three). Wait, but the problem says "all the roots are real". If we count multiplicity, then yes, all roots are real. But if we consider distinct roots, then there is only one distinct root. However, the problem says "all the roots are real", which usually includes multiplicity. So in that case, a = 3 is acceptable. However, if the problem requires three distinct real roots, then a must be greater than 3. But the problem doesn't specify distinct roots, just that all roots are real. Therefore, a = 3 is acceptable. Therefore, the minimal value of a is 3. But let me check again with the previous approach where we factored the cubic. If a = 3, then quadratic factor is x² + 2x + 1 = (x + 1)^2, so the roots are -1, -1, -1. So all real. Therefore, answer is 3. But maybe the problem expects a different answer? Wait, let me check with another approach. Let me suppose that all roots are real. Let me denote them as r, s, t. Then, as per Vieta:

r + s + t = -a

rs + rt + st = a

rst = -1

Since all roots are real, and a is positive. Let me try to find relations between these equations. Let me consider that 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. Let me think about possible roots. Since the product rst = -1, so either one negative and two positive roots, or three negative roots. But if all roots are negative, then their sum would be negative, which matches r + s + t = -a < 0. Also, rs + rt + st would be positive because product of two negatives is positive, and sum of positive terms. Alternatively, if there is one negative and two positive roots, then the product rst would be negative (since one negative and two positive), which matches rst = -1. So both cases are possible.

But let me check if there are constraints. Let me suppose that all roots are negative. Let me set r = -p, s = -q, t = -k, where p, q, k are positive real numbers. Then:

Sum: -p - q - k = -a ⇒ p + q + k = a

Sum of products: (-p)(-q) + (-p)(-k) + (-q)(-k) = pq + pk + qk = a

Product: (-p)(-q)(-k) = -p q k = -1 ⇒ p q k = 1

So we have p + q + k = a, pq + pk + qk = a, and p q k = 1, with p, q, k > 0.

So we need to find the minimal a such that there exist positive real numbers p, q, k with these three equations. So this seems like a symmetric system. Maybe we can assume that p = q = k? Let me check. If p = q = k, then:

3p = a

3p² = a

p³ = 1 ⇒ p = 1

Therefore, 3(1) = a ⇒ a = 3. So that gives a = 3. So that's the case where all roots are equal to -1. Which matches our previous result. So that's one case. But maybe there are other cases where p, q, k are not equal, but still satisfy the equations, leading to a smaller a? Let me check.

Suppose that two variables are equal, say p = q ≠ k. Let me set p = q. Then:

From p + p + k = a ⇒ 2p + k = a

From p² + 2pk = a

From p² k = 1

So we have three equations:

1) 2p + k = a

2) p² + 2pk = a

3) p² k = 1

Let me try to solve these equations. From equations 1 and 2, since both equal to a:

2p + k = p² + 2pk ⇒ 2p + k = p² + 2pk ⇒ Let me rearrange:

p² + 2pk - 2p - k = 0 ⇒ p² - 2p + 2pk - k = 0 ⇒ p² - 2p + k(2p - 1) = 0

Hmm, not sure. Let me try to express k from equation 1: k = a - 2p. Then plug into equation 2:

p² + 2p(a - 2p) = a ⇒ p² + 2pa - 4p² = a ⇒ -3p² + 2pa = a ⇒ -3p² + 2pa - a = 0 ⇒ Let me write it as:

-3p² + 2pa - a = 0 ⇒ Multiply both sides by -1:

3p² - 2pa + a = 0

Now, from equation 3: p² k = 1. Since k = a - 2p, then p²(a - 2p) = 1 ⇒ a p² - 2p³ = 1.

So now, we have two equations:

Equation A: 3p² - 2pa + a = 0

Equation B: a p² - 2p³ = 1

Let me try to solve these equations. From equation A: 3p² - 2pa + a = 0 ⇒ Let me solve for a:

3p² + a(1 - 2p) = 0 ⇒ a(1 - 2p) = -3p² ⇒ a = (-3p²)/(1 - 2p) = (3p²)/(2p - 1)

So a is expressed in terms of p. Now, substitute this into equation B:

a p² - 2p³ = 1 ⇒ [(3p²)/(2p - 1)] * p² - 2p³ = 1 ⇒ [3p^4 / (2p - 1)] - 2p³ = 1

Let me combine the terms:

Multiply both sides by (2p - 1):

3p^4 - 2p³(2p - 1) = 2p - 1

Compute left side:

3p^4 - 4p^4 + 2p³ = -p^4 + 2p³

Therefore:

-p^4 + 2p³ = 2p - 1 ⇒ -p^4 + 2p³ - 2p + 1 = 0 ⇒ Multiply by -1:

p^4 - 2p³ + 2p - 1 = 0

So need to solve p^4 - 2p³ + 2p - 1 = 0. Let me try to factor this equation. Let me try rational roots. Possible rational roots are ±1.

Testing p = 1: 1 - 2 + 2 - 1 = 0. Yes, p = 1 is a root. Therefore, factor out (p - 1):

Using polynomial division or synthetic division. Let me do synthetic division with root 1:

Coefficients: 1 (p^4), -2 (p³), 0 (p²), 2 (p), -1 (constant)

Wait, original polynomial is p^4 - 2p³ + 0p² + 2p - 1.

Divide by (p - 1):

Bring down 1.

Multiply by 1: 1. Add to next coefficient: -2 + 1 = -1.

Multiply by 1: -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by 1: -1. Add to next coefficient: 2 + (-1) = 1.

Multiply by 1: 1. Add to last coefficient: -1 + 1 = 0. Perfect.

Therefore, polynomial factors as (p - 1)(p³ - p² - p + 1). Now factor the cubic: p³ - p² - p + 1.

Group terms: (p³ - p²) + (-p + 1) = p²(p - 1) -1(p - 1) = (p² - 1)(p - 1) = (p - 1)(p + 1)(p - 1). Wait:

Wait, p³ - p² - p + 1 = p²(p - 1) -1(p - 1) = (p² - 1)(p - 1) = (p - 1)(p + 1)(p - 1) = (p - 1)^2(p + 1). Therefore, the original polynomial factors as (p - 1)(p - 1)^2(p + 1) = (p - 1)^3(p + 1). Therefore, roots are p = 1 (triple root) and p = -1. Since p is positive, p = -1 is rejected. Therefore, only p = 1 is valid. Therefore, p = 1. Therefore, from earlier, a = (3p²)/(2p - 1) = 3(1)/(2 - 1) = 3/1 = 3. Therefore, again, a = 3. Therefore, even in this case, the minimal a is 3. Therefore, even if we assume two variables equal, we still get a = 3. Therefore, seems like 3 is the minimal value. Therefore, maybe the answer is indeed 3. But let me check if there are other cases where p, q, k are not equal or two equal, but still give a smaller a. Suppose that all three variables are different. But how can I check that? Maybe using inequalities. Let me think.

We have p + q + k = a, pq + pk + qk = a, and p q k = 1. Let me consider that for positive real numbers p, q, k. Let me denote S = p + q + k = a, P = pq + pk + qk = a, and Q = p q k = 1. So S = P. Let me recall that for three variables, we have the relation between S, P, Q. Also, we can think of the cubic equation whose roots are p, q, k: x³ - S x² + P x - Q = 0 ⇒ x³ - a x² + a x - 1 = 0. Wait, but this is similar to the original equation. Wait, original equation was x³ + a x² + a x + 1 = 0. If we substitute x = -y, then (-y)^3 + a (-y)^2 + a (-y) + 1 = -y³ + a y² - a y + 1 = 0 ⇒ -y³ + a y² - a y + 1 = 0 ⇒ y³ - a y² + a y - 1 = 0. Which is the same as the cubic equation for p, q, k. Therefore, the roots p, q, k are the roots of y³ - a y² + a y - 1 = 0. Which is same as original equation with x replaced by -y. Therefore, if original equation has roots r, s, t, then this equation has roots -r, -s, -t. Therefore, since original equation had roots with product -1, then this equation has roots with product 1. Which matches Q = 1. Therefore, this seems consistent.

But how does this help? Maybe not much. Let me think about inequalities. Since p, q, k are positive real numbers, and S = P. Let me recall that for positive real numbers, by AM ≥ GM:

(p + q + k)/3 ≥ (p q k)^{1/3} ⇒ a/3 ≥ 1^{1/3} ⇒ a/3 ≥ 1 ⇒ a ≥ 3. Equality holds when p = q = k. Therefore, this gives that a ≥ 3. Therefore, the minimal value of a is 3, achieved when p = q = k = 1. Therefore, this confirms our previous result. Therefore, regardless of the approach, the minimal a is 3. Therefore, the answer is 3. Therefore, the smallest possible value of a is 3. Therefore, I think the answer is \boxed{3}.

But just to be thorough, let me check if there's a case where the cubic has three real roots with a < 3. Suppose someone claims that maybe with some specific roots, even though discriminant of quadratic is negative, but... no, because if quadratic has complex roots, then cubic has only one real root. Therefore, not all roots real. Therefore, the minimal a is indeed 3. Therefore, I think the answer is 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 a root for any value of $ a $. Therefore, we can factor the cubic 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:

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

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: Verify the Minimal Value

Let’s test $ a = 3 $:

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

This has a **triple root** at $ x = -1 $, so all roots are real (with multiplicity). Therefore, $ a = 3 $ satisfies the condition.

Now, consider $ a < 3 $. For example, take $ a = 2 $:

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

The discriminant is:

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

So the quadratic has **complex roots**, and the cubic has only one real root (at $ x = -1 $). Thus, **not all roots are real**.

Therefore, the **smallest possible value** of $ a $ such that 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. Let me distribute the 6 to both terms inside the parentheses. So, 6 times 1 is 6, and 6 times 2i is... hmm, 6*2 is 12, so that would be 12i. So, after multiplying, I get 6 + 12i.

Now, the next part of the expression is subtracting 3i. So, I need to take the result from the first part, which is 6 + 12i, and subtract 3i from it. Let me write that out: 6 + 12i - 3i.

Now, combining like terms here. The real part is just 6, and the imaginary parts are 12i and -3i. So, combining those, 12i - 3i is... let me calculate that. 12 minus 3 is 9, so that would be 9i.

Therefore, putting it all together, the entire expression simplifies to 6 + 9i. Let me check if I did that correctly.

First step: multiplying 6 by (1 + 2i). Yes, 6*1 is 6, 6*2i is 12i. Then subtracting 3i: 12i - 3i is indeed 9i. So, 6 + 9i.

Wait, is there any chance I messed up the signs? Let me verify. The original expression is (1 + 2i)6 - 3i. So, the multiplication is straightforward. Then subtracting 3i. So, yes, 12i - 3i is 9i.

Alternatively, maybe I can think of it as (1 + 2i)*6 - 3i. Let me compute it another way. Maybe convert the entire expression into a single complex number.

Alternatively, maybe I can write the entire expression as 6*(1 + 2i) - 3i. Which is the same as 6*1 + 6*2i - 3i = 6 + 12i - 3i = 6 + 9i. Yep, same result.

I think that's correct. Let me check if there's another way to approach this. Maybe by expanding the brackets first. But there are no brackets to expand except the multiplication by 6.

Alternatively, maybe I can think of the expression as 6*(1 + 2i) - 3i. If I factor out the i from the imaginary parts? Let me see.

But maybe that complicates things. Let me think. The original expression is straightforward. Multiply 6 by the complex number, then subtract 3i.

Alternatively, if I wanted to write it as a single complex number, it's 6 + 12i - 3i = 6 + 9i.

I think that's correct. Let me check with another method. Let me compute the real and imaginary parts separately.

The original expression: (1 + 2i)6 - 3i.

First, compute (1 + 2i)6. The real part of (1 + 2i) is 1, multiplied by 6 gives 6. The imaginary part of (1 + 2i) is 2, multiplied by 6 gives 12. So, the result of the multiplication is 6 + 12i.

Then subtract 3i. The imaginary part is 12i - 3i = 9i. So, the total is 6 + 9i.

Yes, same answer.

Alternatively, maybe I can think of the entire expression as a complex number. Let me recall that complex numbers are of the form a + bi, where a is the real part and b is the imaginary part.

So, if I have (1 + 2i) multiplied by 6, that's 6*1 + 6*2i = 6 + 12i. Then subtract 3i, which is subtracting 3i from the imaginary part. So, 12i - 3i = 9i.

Therefore, the final result is 6 + 9i.

I think that's correct. Let me check if there's any chance I made a mistake in arithmetic.

6 times 1 is 6. 6 times 2 is 12. Then 12i minus 3i is 9i. So, 6 + 9i.

Yes, that seems right.

Alternatively, maybe I can plug in the numbers into a different order. Let me think. Suppose I first subtract 3i from (1 + 2i) before multiplying by 6? Wait, but the original expression is (1 + 2i) multiplied by 6, then subtract 3i. So, the order of operations is important here.

If I were to do (1 + 2i - 3i) * 6, that would be different. But that's not the case here. The original expression is (1 + 2i) * 6 - 3i. So, multiplication comes before subtraction.

So, the correct order is to do the multiplication first, then the subtraction.

So, if I were to incorrectly do the subtraction first, I would have (1 + 2i - 3i) * 6, which would be (1 - i) * 6 = 6 - 6i. Which is different from the correct answer.

But since the original expression is (1 + 2i)6 - 3i, the correct order is multiplication first, then subtraction.

Therefore, my original calculation of 6 + 9i is correct.

Alternatively, maybe I can check with substituting i with sqrt(-1), but that might complicate things.

Alternatively, maybe I can use the distributive property again. Let me write the entire expression as 6*(1 + 2i) - 3i.

Expanding that, it's 6*1 + 6*2i - 3i = 6 + 12i - 3i.

Combine like terms: 6 + (12i - 3i) = 6 + 9i.

Yes, same result.

I think I've verified it multiple ways. So, I can be confident that the answer is 6 + 9i.

But just to make sure, let me write down all steps again:

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

Step 1: Multiply 6 by (1 + 2i):

6 * 1 = 6

6 * 2i = 12i

So, 6 + 12i

Step 2: Subtract 3i from the result:

6 + 12i - 3i

Combine the imaginary terms:

12i - 3i = 9i

So, the result is 6 + 9i

Yes, that's consistent.

Alternatively, if I think of the complex number in terms of real and imaginary components:

Let me denote z = (1 + 2i) * 6 - 3i

First, compute (1 + 2i) * 6:

Real part: 1 * 6 = 6

Imaginary part: 2i * 6 = 12i

Then subtract 3i:

Imaginary part becomes 12i - 3i = 9i

So, z = 6 + 9i

Yep, same answer.

I think that's solid. I don't see any mistakes here. So, the final answer should be 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 multiplying the complex number $1 + 2i$ by 6:

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

This gives us the result of the multiplication: $6 + 12i$.

---

### Step 2: Subtract $3i$

Now, we 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. Let me distribute the 6 to both terms inside the parentheses. So, 6 times 1 is 6, and 6 times 2i is... hmm, 6*2 is 12, so that would be 12i. So, after multiplying, I get 6 + 12i.

Now, the next part of the expression is subtracting 3i. So, I need to take the result from the first part, which is 6 + 12i, and subtract 3i from it. Let me write that out: 6 + 12i - 3i.

Now, combining like terms here. The real part is just 6, and the imaginary parts are 12i and -3i. So, combining those, 12i - 3i is... let me calculate that. 12 minus 3 is 9, so that would be 9i.

Therefore, putting it all together, the entire expression simplifies to 6 + 9i. Let me check if I did that correctly.

First step: multiplying 6 by (1 + 2i). Yes, 6*1 is 6, 6*2i is 12i. Then subtracting 3i: 12i - 3i is indeed 9i. So, 6 + 9i.

Wait, is there any chance I messed up the signs? Let me verify. The original expression is (1 + 2i)6 - 3i. So, the multiplication is straightforward. Then subtracting 3i. So, yes, 12i - 3i is 9i.

Alternatively, maybe I can think of it as (1 + 2i)*6 - 3i. Let me compute it another way. Maybe convert the entire expression into a single complex number.

Alternatively, maybe I can write the entire expression as 6*(1 + 2i) - 3i. Which is the same as 6*1 + 6*2i - 3i = 6 + 12i - 3i = 6 + 9i. Yep, same result.

I think that's correct. Let me check if there's another way to approach this. Maybe by expanding the brackets first. But there are no brackets to expand except the multiplication by 6.

Alternatively, maybe I can think of the expression as 6*(1 + 2i) - 3i. If I factor out the i from the imaginary parts? Let me see.

But maybe that complicates things. Let me think. The original expression is straightforward. Multiply 6 by the complex number, then subtract 3i.

Alternatively, if I wanted to write it as a single complex number, it's 6 + 12i - 3i = 6 + 9i.

I think that's correct. Let me check with another method. Let me compute the real and imaginary parts separately.

The original expression: (1 + 2i)6 - 3i.

First, compute (1 + 2i)6. The real part of (1 + 2i) is 1, multiplied by 6 gives 6. The imaginary part of (1 + 2i) is 2, multiplied by 6 gives 12. So, the result of the multiplication is 6 + 12i.

Then subtract 3i. The imaginary part is 12i - 3i = 9i. So, the total is 6 + 9i.

Yes, same answer.

Alternatively, maybe I can think of the entire expression as a complex number. Let me recall that complex numbers are of the form a + bi, where a is the real part and b is the imaginary part.

So, if I have (1 + 2i) multiplied by 6, that's 6*1 + 6*2i = 6 + 12i. Then subtract 3i, which is subtracting 3i from the imaginary part. So, 12i - 3i = 9i.

Therefore, the final result is 6 + 9i.

I think that's correct. Let me check if there's any chance I made a mistake in arithmetic.

6 times 1 is 6. 6 times 2 is 12. Then 12i minus 3i is 9i. So, 6 + 9i.

Yes, that seems right.

Alternatively, maybe I can plug in the numbers into a different order. Let me think. Suppose I first subtract 3i from (1 + 2i) before multiplying by 6? Wait, but the original expression is (1 + 2i) multiplied by 6, then subtract 3i. So, the order of operations is important here.

If I were to do (1 + 2i - 3i) * 6, that would be different. But that's not the case here. The original expression is (1 + 2i) * 6 - 3i. So, multiplication comes before subtraction.

So, the correct order is to do the multiplication first, then the subtraction.

So, if I were to incorrectly do the subtraction first, I would have (1 + 2i - 3i) * 6, which would be (1 - i) * 6 = 6 - 6i. Which is different from the correct answer.

But since the original expression is (1 + 2i)6 - 3i, the correct order is multiplication first, then subtraction.

Therefore, my original calculation of 6 + 9i is correct.

Alternatively, maybe I can check with substituting i with sqrt(-1), but that might complicate things.

Alternatively, maybe I can use the distributive property again. Let me write the entire expression as 6*(1 + 2i) - 3i.

Expanding that, it's 6*1 + 6*2i - 3i = 6 + 12i - 3i.

Combine like terms: 6 + (12i - 3i) = 6 + 9i.

Yes, same result.

I think I've verified it multiple ways. So, I can be confident that the answer is 6 + 9i.

But just to make sure, let me write down all steps again:

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

Step 1: Multiply 6 by (1 + 2i):

6 * 1 = 6

6 * 2i = 12i

So, 6 + 12i

Step 2: Subtract 3i from the result:

6 + 12i - 3i

Combine the imaginary terms:

12i - 3i = 9i

So, the result is 6 + 9i

Yes, that's consistent.

Alternatively, if I think of the complex number in terms of real and imaginary components:

Let me denote z = (1 + 2i) * 6 - 3i

First, compute (1 + 2i) * 6:

Real part: 1 * 6 = 6

Imaginary part: 2i * 6 = 12i

Then subtract 3i:

Imaginary part becomes 12i - 3i = 9i

So, z = 6 + 9i

Yep, same answer.

I think that's solid. I don't see any mistakes here. So, the final answer should be 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 multiplying the complex number $1 + 2i$ by 6:

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

This gives us the result of the multiplication: $6 + 12i$.

---

### Step 2: Subtract $3i$

Now, we 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 $(1+2i)6-3i=6+12i-3i=\boxed{6+9i}$.