Explanation:
Okay, so I need to find the remainder when the sum of numbers from 1 to 12 is divided by 4. Let me think about how to approach this.

First, maybe I should just calculate the total sum and then divide by 4 to find the remainder. But adding all those numbers might take some time. Wait, there's a formula for the sum of the first n numbers, right? It's n(n + 1)/2. Let me check if I remember that correctly. Yeah, I think that's right. So if I use that formula, I can find the sum quickly.

Let me apply the formula here. The numbers go from 1 to 12, so n is 12. Plugging into the formula: 12 times (12 + 1) divided by 2. Let me compute that. 12 + 1 is 13, so 12 times 13 is... 12*10 is 120, 12*3 is 36, so 120 + 36 is 156. Then divide by 2: 156 divided by 2 is 78. So the total sum is 78.

Now, I need to find 78 modulo 4. That means I need to divide 78 by 4 and find the remainder. Let me do that division. 4 times 19 is 76, right? Because 4*20 is 80, which is too big. So 4*19 is 76. Subtract 76 from 78, and the remainder is 2. So the remainder is 2.

Wait, but maybe there's another way to do this without calculating the entire sum? Maybe by taking each number modulo 4 first and then adding them up? Let me try that method to verify.

Let me list the numbers from 1 to 12 and find each modulo 4:

1 mod 4 = 1

2 mod 4 = 2

3 mod 4 = 3

4 mod 4 = 0

5 mod 4 = 1 (since 5 - 4 = 1)

6 mod 4 = 2 (6 - 4 = 2)

7 mod 4 = 3 (7 - 4*1 = 3)

8 mod 4 = 0 (8 is divisible by 4)

9 mod 4 = 1 (9 - 8 = 1)

10 mod 4 = 2 (10 - 8 = 2)

11 mod 4 = 3 (11 - 8 = 3)

12 mod 4 = 0 (12 is divisible by 4)

So the remainders are: 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0.

Now let me add these up:

Starting from the beginning:

1 + 2 = 3

3 + 3 = 6

6 + 0 = 6

6 + 1 = 7

7 + 2 = 9

9 + 3 = 12

12 + 0 = 12

12 + 1 = 13

13 + 2 = 15

15 + 3 = 18

18 + 0 = 18.

So the total sum of the remainders is 18. Now, I need to find 18 modulo 4. Let me divide 18 by 4. 4*4 is 16, so 18 - 16 is 2. Therefore, the remainder is 2.

Okay, so both methods give me the same result, which is 2. That seems consistent. Let me check once more to be sure. Maybe I made a mistake in adding the remainders?

Let me add them again:

1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 + 1 + 2 + 3 + 0.

Grouping them:

(1 + 2 + 3) + (0 + 1 + 2 + 3) + (0 + 1 + 2 + 3) + 0.

Wait, maybe that's not the best way. Let me count how many of each remainder there are.

Looking at the remainders: 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0.

So for each group of four numbers, the remainders are 1, 2, 3, 0. Let me see how many complete groups there are. From 1 to 12, there are 12 numbers, so 12 divided by 4 is 3 groups. Each group has remainders 1, 2, 3, 0.

So each group adds up to 1 + 2 + 3 + 0 = 6. Three groups would be 3*6 = 18. Which matches the previous total. Then 18 mod 4 is 2.

Alternatively, since each group of four numbers (like 1-4, 5-8, 9-12) has a sum that is divisible by 4? Wait, let me check.

Sum of 1-4 is 10. 10 mod 4 is 2. Wait, but earlier I thought that the sum of the remainders is 6 per group. Wait, there's a discrepancy here. Wait, maybe I confused the sum of the numbers with the sum of their remainders.

Wait, the sum of numbers 1-4 is 1+2+3+4=10. But the sum of their remainders is 1+2+3+0=6. So they are different. So the total sum of numbers is 10 + 10 + 10 + 10? Wait, no. Wait, the numbers from 1-4, 5-8, 9-12. Each group of four numbers. Let me compute the sum of each group:

First group: 1+2+3+4=10

Second group: 5+6+7+8=26

Third group: 9+10+11+12=42

Total sum: 10 + 26 + 42 = 78. Which matches the previous total.

But if I take the sum of remainders, which is 6 + 6 + 6 = 18. But 18 mod 4 is 2. However, the total sum is 78, which mod 4 is also 2. So that's consistent.

But if I take each group's sum mod 4:

First group sum 10 mod 4 is 2 (since 10 - 8 = 2)

Second group sum 26 mod 4: 26 divided by 4 is 6*4=24, remainder 2. So 26 mod 4 is 2.

Third group sum 42 mod 4: 42 divided by 4 is 10*4=40, remainder 2. So 42 mod 4 is 2.

So adding those remainders: 2 + 2 + 2 = 6. Then 6 mod 4 is 2. Which again gives the same result.

So that's another way to think about it. So regardless of the method, the remainder is 2.

Alternatively, maybe I can use properties of modular arithmetic. For example, since we are dealing with modulo 4, we can consider each term modulo 4 and add them up. Which is what I did earlier.

Alternatively, since the sum from 1 to n is n(n+1)/2, then taking that modulo 4. So for n=12, sum is 12*13/2 = 78. Then 78 mod 4.

But maybe I can compute 78 mod 4 without dividing. Let me think. 4*19=76, so 78-76=2. So remainder 2.

Alternatively, breaking down 78 into 76 + 2, and since 76 is divisible by 4, the remainder is 2.

Alternatively, maybe I can compute 12*13/2 mod 4. Let me see.

But 12 is divisible by 4, so 12 mod 4 is 0. But 12*13/2 is 78. But maybe breaking it down:

First, compute 12 mod 4 = 0, 13 mod 4 = 1. So 0 * 1 / 2 mod 4? Wait, that seems problematic because division isn't straightforward in modular arithmetic. Maybe that approach isn't the best.

Alternatively, compute n(n+1)/2 mod 4. Let me think. For n=12, which is even. So n=2k, so the formula becomes 2k*(2k +1)/2 = k*(2k +1). So for n=12, k=6. So 6*(13) = 78. Then 78 mod 4 is 2.

Alternatively, maybe compute k*(2k +1) mod 4. For k=6, 6 mod 4 is 2, 2k +1 = 13 mod 4 is 1. So 2*1=2 mod 4. Which again gives 2.

So that's another way to see it.

So regardless of the method, the answer is 2.

But just to make sure, let me try adding the numbers step by step and taking mod 4 at each step. Maybe that's another way.

Starting with 0.

Add 1: 0 + 1 = 1 mod 4 = 1

Add 2: 1 + 2 = 3 mod 4 = 3

Add 3: 3 + 3 = 6 mod 4 = 2

Add 4: 2 + 4 = 6 mod 4 = 2 (since 4 mod 4 is 0, so 2 + 0 = 2)

Add 5: 2 + 5 = 7 mod 4 = 3 (since 5 mod 4 is 1, so 2 + 1 = 3)

Add 6: 3 + 6 = 9 mod 4 = 1 (6 mod 4 is 2, so 3 + 2 = 5 mod 4 is 1)

Add 7: 1 + 7 = 8 mod 4 = 0 (7 mod 4 is 3, so 1 + 3 = 4 mod 4 is 0)

Add 8: 0 + 8 = 8 mod 4 = 0 (8 mod 4 is 0, so 0 + 0 = 0)

Add 9: 0 + 9 = 9 mod 4 = 1 (9 mod 4 is 1, so 0 + 1 = 1)

Add 10: 1 + 10 = 11 mod 4 = 3 (10 mod 4 is 2, so 1 + 2 = 3)

Add 11: 3 + 11 = 14 mod 4 = 2 (11 mod 4 is 3, so 3 + 3 = 6 mod 4 is 2)

Add 12: 2 + 12 = 14 mod 4 = 2 (12 mod 4 is 0, so 2 + 0 = 2)

So after adding all numbers step by step and taking mod 4 each time, the final remainder is 2.

So that's another confirmation.

Therefore, all methods lead to the same answer, which is 2.

I think that's thorough enough. I considered multiple approaches: direct calculation, using the formula, breaking into groups, step-by-step addition with mod at each step, and even checking the properties of modular arithmetic. All of them give the same result. So I can be confident that the remainder when the sum is divided by 4 is 2.

**Final Answer**
The modulo 4 remainder of the sum is \boxed{2}.

To determine the modulo 4 remainder of the sum:

$$
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12
$$

we can approach the problem in a few different ways, all of which lead to the same result.

---

### **Method 1: Using the Sum Formula**

The sum of the first $ n $ natural numbers is given by the formula:

$$
\text{Sum} = \frac{n(n+1)}{2}
$$

For $ n = 12 $:

$$
\text{Sum} = \frac{12 \times 13}{2} = \frac{156}{2} = 78
$$

Now, compute $ 78 \mod 4 $:

$$
78 \div 4 = 19 \text{ remainder } 2
$$

So, the remainder is:

$$
\boxed{2}
$$

---

### **Method 2: Modulo 4 of Each Term**

We can compute the remainder of each number modulo 4 and then sum those remainders:

$$
\begin{align*}
1 &\mod 4 = 1 \\
2 &\mod 4 = 2 \\
3 &\mod 4 = 3 \\
4 &\mod 4 = 0 \\
5 &\mod 4 = 1 \\
6 &\mod 4 = 2 \\
7 &\mod 4 = 3 \\
8 &\mod 4 = 0 \\
9 &\mod 4 = 1 \\
10 &\mod 4 = 2 \\
11 &\mod 4 = 3 \\
12 &\mod 4 = 0 \\
\end{align*}
$$

Now sum the remainders:

$$
1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 = 18
$$

Now compute $ 18 \mod 4 $:

$$
18 \div 4 = 4 \text{ remainder } 2
$$

So, the remainder is again:

$$
\boxed{2}
$$

---

### **Method 3: Grouping by Modulo 4 Cycles**

Since the modulo 4 pattern repeats every 4 numbers, we can group the numbers into sets of 4:

- $ 1 + 2 + 3 + 4 = 10 $, and $ 10 \mod 4 = 2 $
- $ 5 + 6 + 7 + 8 = 26 $, and $ 26 \mod 4 = 2 $
- $ 9 + 10 + 11 + 12 = 42 $, and $ 42 \mod 4 = 2 $

Now sum the remainders of the groups:

$$
2 + 2 + 2 = 6
$$

Then compute $ 6 \mod 4 $:

$$
6 \mod 4 = 2
$$

Again, the remainder is:

$$
\boxed{2}
$$

---

### **Conclusion**

All methods consistently yield the same result. Therefore, the modulo 4 remainder of the sum is:

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to find the remainder when the sum of numbers from 1 to 12 is divided by 4. Let me think about how to approach this.

First, maybe I should just calculate the total sum and then divide by 4 to find the remainder. But adding all those numbers might take some time. Wait, there's a formula for the sum of the first n numbers, right? It's n(n + 1)/2. Let me check if I remember that correctly. Yeah, I think that's right. So if I use that formula, I can find the sum quickly.

Let me apply the formula here. The numbers go from 1 to 12, so n is 12. Plugging into the formula: 12 times (12 + 1) divided by 2. Let me compute that. 12 + 1 is 13, so 12 times 13 is... 12*10 is 120, 12*3 is 36, so 120 + 36 is 156. Then divide by 2: 156 divided by 2 is 78. So the total sum is 78.

Now, I need to find 78 modulo 4. That means I need to divide 78 by 4 and find the remainder. Let me do that division. 4 times 19 is 76, right? Because 4*20 is 80, which is too big. So 4*19 is 76. Subtract 76 from 78, and the remainder is 2. So the remainder is 2.

Wait, but maybe there's another way to do this without calculating the entire sum? Maybe by taking each number modulo 4 first and then adding them up? Let me try that method to verify.

Let me list the numbers from 1 to 12 and find each modulo 4:

1 mod 4 = 1

2 mod 4 = 2

3 mod 4 = 3

4 mod 4 = 0

5 mod 4 = 1 (since 5 - 4 = 1)

6 mod 4 = 2 (6 - 4 = 2)

7 mod 4 = 3 (7 - 4*1 = 3)

8 mod 4 = 0 (8 is divisible by 4)

9 mod 4 = 1 (9 - 8 = 1)

10 mod 4 = 2 (10 - 8 = 2)

11 mod 4 = 3 (11 - 8 = 3)

12 mod 4 = 0 (12 is divisible by 4)

So the remainders are: 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0.

Now let me add these up:

Starting from the beginning:

1 + 2 = 3

3 + 3 = 6

6 + 0 = 6

6 + 1 = 7

7 + 2 = 9

9 + 3 = 12

12 + 0 = 12

12 + 1 = 13

13 + 2 = 15

15 + 3 = 18

18 + 0 = 18.

So the total sum of the remainders is 18. Now, I need to find 18 modulo 4. Let me divide 18 by 4. 4*4 is 16, so 18 - 16 is 2. Therefore, the remainder is 2.

Okay, so both methods give me the same result, which is 2. That seems consistent. Let me check once more to be sure. Maybe I made a mistake in adding the remainders?

Let me add them again:

1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 + 1 + 2 + 3 + 0.

Grouping them:

(1 + 2 + 3) + (0 + 1 + 2 + 3) + (0 + 1 + 2 + 3) + 0.

Wait, maybe that's not the best way. Let me count how many of each remainder there are.

Looking at the remainders: 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0.

So for each group of four numbers, the remainders are 1, 2, 3, 0. Let me see how many complete groups there are. From 1 to 12, there are 12 numbers, so 12 divided by 4 is 3 groups. Each group has remainders 1, 2, 3, 0.

So each group adds up to 1 + 2 + 3 + 0 = 6. Three groups would be 3*6 = 18. Which matches the previous total. Then 18 mod 4 is 2.

Alternatively, since each group of four numbers (like 1-4, 5-8, 9-12) has a sum that is divisible by 4? Wait, let me check.

Sum of 1-4 is 10. 10 mod 4 is 2. Wait, but earlier I thought that the sum of the remainders is 6 per group. Wait, there's a discrepancy here. Wait, maybe I confused the sum of the numbers with the sum of their remainders.

Wait, the sum of numbers 1-4 is 1+2+3+4=10. But the sum of their remainders is 1+2+3+0=6. So they are different. So the total sum of numbers is 10 + 10 + 10 + 10? Wait, no. Wait, the numbers from 1-4, 5-8, 9-12. Each group of four numbers. Let me compute the sum of each group:

First group: 1+2+3+4=10

Second group: 5+6+7+8=26

Third group: 9+10+11+12=42

Total sum: 10 + 26 + 42 = 78. Which matches the previous total.

But if I take the sum of remainders, which is 6 + 6 + 6 = 18. But 18 mod 4 is 2. However, the total sum is 78, which mod 4 is also 2. So that's consistent.

But if I take each group's sum mod 4:

First group sum 10 mod 4 is 2 (since 10 - 8 = 2)

Second group sum 26 mod 4: 26 divided by 4 is 6*4=24, remainder 2. So 26 mod 4 is 2.

Third group sum 42 mod 4: 42 divided by 4 is 10*4=40, remainder 2. So 42 mod 4 is 2.

So adding those remainders: 2 + 2 + 2 = 6. Then 6 mod 4 is 2. Which again gives the same result.

So that's another way to think about it. So regardless of the method, the remainder is 2.

Alternatively, maybe I can use properties of modular arithmetic. For example, since we are dealing with modulo 4, we can consider each term modulo 4 and add them up. Which is what I did earlier.

Alternatively, since the sum from 1 to n is n(n+1)/2, then taking that modulo 4. So for n=12, sum is 12*13/2 = 78. Then 78 mod 4.

But maybe I can compute 78 mod 4 without dividing. Let me think. 4*19=76, so 78-76=2. So remainder 2.

Alternatively, breaking down 78 into 76 + 2, and since 76 is divisible by 4, the remainder is 2.

Alternatively, maybe I can compute 12*13/2 mod 4. Let me see.

But 12 is divisible by 4, so 12 mod 4 is 0. But 12*13/2 is 78. But maybe breaking it down:

First, compute 12 mod 4 = 0, 13 mod 4 = 1. So 0 * 1 / 2 mod 4? Wait, that seems problematic because division isn't straightforward in modular arithmetic. Maybe that approach isn't the best.

Alternatively, compute n(n+1)/2 mod 4. Let me think. For n=12, which is even. So n=2k, so the formula becomes 2k*(2k +1)/2 = k*(2k +1). So for n=12, k=6. So 6*(13) = 78. Then 78 mod 4 is 2.

Alternatively, maybe compute k*(2k +1) mod 4. For k=6, 6 mod 4 is 2, 2k +1 = 13 mod 4 is 1. So 2*1=2 mod 4. Which again gives 2.

So that's another way to see it.

So regardless of the method, the answer is 2.

But just to make sure, let me try adding the numbers step by step and taking mod 4 at each step. Maybe that's another way.

Starting with 0.

Add 1: 0 + 1 = 1 mod 4 = 1

Add 2: 1 + 2 = 3 mod 4 = 3

Add 3: 3 + 3 = 6 mod 4 = 2

Add 4: 2 + 4 = 6 mod 4 = 2 (since 4 mod 4 is 0, so 2 + 0 = 2)

Add 5: 2 + 5 = 7 mod 4 = 3 (since 5 mod 4 is 1, so 2 + 1 = 3)

Add 6: 3 + 6 = 9 mod 4 = 1 (6 mod 4 is 2, so 3 + 2 = 5 mod 4 is 1)

Add 7: 1 + 7 = 8 mod 4 = 0 (7 mod 4 is 3, so 1 + 3 = 4 mod 4 is 0)

Add 8: 0 + 8 = 8 mod 4 = 0 (8 mod 4 is 0, so 0 + 0 = 0)

Add 9: 0 + 9 = 9 mod 4 = 1 (9 mod 4 is 1, so 0 + 1 = 1)

Add 10: 1 + 10 = 11 mod 4 = 3 (10 mod 4 is 2, so 1 + 2 = 3)

Add 11: 3 + 11 = 14 mod 4 = 2 (11 mod 4 is 3, so 3 + 3 = 6 mod 4 is 2)

Add 12: 2 + 12 = 14 mod 4 = 2 (12 mod 4 is 0, so 2 + 0 = 2)

So after adding all numbers step by step and taking mod 4 each time, the final remainder is 2.

So that's another confirmation.

Therefore, all methods lead to the same answer, which is 2.

I think that's thorough enough. I considered multiple approaches: direct calculation, using the formula, breaking into groups, step-by-step addition with mod at each step, and even checking the properties of modular arithmetic. All of them give the same result. So I can be confident that the remainder when the sum is divided by 4 is 2.

**Final Answer**
The modulo 4 remainder of the sum is \boxed{2}.

To determine the modulo 4 remainder of the sum:

$$
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12
$$

we can approach the problem in a few different ways, all of which lead to the same result.

---

### **Method 1: Using the Sum Formula**

The sum of the first $ n $ natural numbers is given by the formula:

$$
\text{Sum} = \frac{n(n+1)}{2}
$$

For $ n = 12 $:

$$
\text{Sum} = \frac{12 \times 13}{2} = \frac{156}{2} = 78
$$

Now, compute $ 78 \mod 4 $:

$$
78 \div 4 = 19 \text{ remainder } 2
$$

So, the remainder is:

$$
\boxed{2}
$$

---

### **Method 2: Modulo 4 of Each Term**

We can compute the remainder of each number modulo 4 and then sum those remainders:

$$
\begin{align*}
1 &\mod 4 = 1 \\
2 &\mod 4 = 2 \\
3 &\mod 4 = 3 \\
4 &\mod 4 = 0 \\
5 &\mod 4 = 1 \\
6 &\mod 4 = 2 \\
7 &\mod 4 = 3 \\
8 &\mod 4 = 0 \\
9 &\mod 4 = 1 \\
10 &\mod 4 = 2 \\
11 &\mod 4 = 3 \\
12 &\mod 4 = 0 \\
\end{align*}
$$

Now sum the remainders:

$$
1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 + 1 + 2 + 3 + 0 = 18
$$

Now compute $ 18 \mod 4 $:

$$
18 \div 4 = 4 \text{ remainder } 2
$$

So, the remainder is again:

$$
\boxed{2}
$$

---

### **Method 3: Grouping by Modulo 4 Cycles**

Since the modulo 4 pattern repeats every 4 numbers, we can group the numbers into sets of 4:

- $ 1 + 2 + 3 + 4 = 10 $, and $ 10 \mod 4 = 2 $
- $ 5 + 6 + 7 + 8 = 26 $, and $ 26 \mod 4 = 2 $
- $ 9 + 10 + 11 + 12 = 42 $, and $ 42 \mod 4 = 2 $

Now sum the remainders of the groups:

$$
2 + 2 + 2 = 6
$$

Then compute $ 6 \mod 4 $:

$$
6 \mod 4 = 2
$$

Again, the remainder is:

$$
\boxed{2}
$$

---

### **Conclusion**

All methods consistently yield the same result. Therefore, the modulo 4 remainder of the sum is:

$$
\boxed{2}
$$

Solution:
Grouping residues helps make some series computations easier:  \begin{align*}
1 + 2 + 3 + 0 + 1 + 2& + 3 + 0 + 1 + 2 + 3 + 0\\&\equiv 3(1 + 2 + 3 + 0) \\
&\equiv 18\\
& \equiv \boxed{2} \pmod{4}.
\end{align*}