<Question>
Let $N$ be the units digit of the number $21420N$. Which nonzero value of $N$ makes this number divisible by $6$?
</Question>

<think>
<Plan_1>
Okay, let's try to figure out this problem. So, we need to find a nonzero digit N such that the number 21420N is divisible by 6. And N is the units digit of that number. Hmm, right. Let me recall the rules for divisibility by 6. If a number is divisible by both 2 and 3, then it's divisible by 6. So, first, the number must be even (divisible by 2), and the sum of its digits must be divisible by 3 (divisible by 3). 

Alright, let's break this down step by step.

First, let's check the divisibility by 2.
Since the number ends with N, which is the units digit, N has to be even. Because for a number to be even, its last digit must be 0, 2, 4, 6, or 8. So possible values for N are 2, 4, 6, or 8. But the problem says N is a nonzero value, so 0 is excluded. Wait, but the question mentions N is the units digit of the number 21420N. So N is a single digit, right? So N can be 0-9, but since it's nonzero, we have 2,4,6,8.

Now, moving on to divisibility by 3. The rule for 3 is that the sum of all digits must be a multiple of 3. Let's compute the sum of the digits of 21420N. The digits are 2, 1, 4, 2, 0, and N. So adding those up: 2 + 1 + 4 + 2 + 0 + N.
Let me calculate that: 2+1 is 3, 3+4 is 7, 7+2 is 9, 9+0 is 9. So total sum is 9 + N. Therefore, 9 + N must be divisible by 3.

So, 9 + N ≡ 0 mod 3. Which means N ≡ 0 mod 3, because 9 is already 0 mod 3. Therefore, N must be a multiple of 3.

But from earlier, N has to be even (2,4,6,8). So now we need to find which of these even digits are also multiples of 3. Let's check each possible N:

<prompt_1.1> Determine which even digits (2, 4, 6, 8) are multiples of 3. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
- N = 2: 2 is not a multiple of 3. 2 ÷ 3 is 0.666...
- N = 4: 4 ÷ 3 is 1.333... , not a multiple.
- N = 6: 6 ÷ 3 = 2, which is an integer. So 6 is a multiple of 3.
- N = 8: 8 ÷ 3 ≈ 2.666... , not a multiple.

So the only digit that is both even and a multiple of 3 is 6.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, but let me double-check. The sum of the digits is 9 + N. If N is 6, then the sum is 15, which is divisible by 3 (15 ÷ 3 = 5). If N were 0, the sum would be 9, which is also divisible by 3, but N has to be nonzero. So 6 is the only possible digit here.

Let me confirm by plugging back into the original number. If N=6, the number is 214206.

<prompt_2.1> Verify that the number 214206 is divisible by 6. </prompt_2.1>
-----

Wait, but just to make sure there are no other possible digits. Let's see. Let me check if there are any other digits that could satisfy both conditions. For example, if N=3, which is a multiple of 3, but it's not even, so the number would be 214203.

<prompt_2.2> Check if other digits (3, 9) could satisfy the conditions for divisibility by 6. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's check divisibility by 2: the last digit is 6, which is even, so yes. Divisibility by 3: sum of digits is 2+1+4+2+0+6 = 15, which is divisible by 3. Therefore, 214206 is divisible by 6.
</execution_2.1>

<execution_2.2>
That's odd, so not divisible by 2. Similarly, N=9: 214209. Sum of digits 9+9=18, which is divisible by 3, but the number ends with 9, which is odd. So not divisible by 2. Hence, not divisible by 6. So those are excluded.

Therefore, the only possible digit is 6.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But wait, let me check if there's any other possible digit. For example, N=3: sum would be 9+3=12, which is divisible by 3, but N=3 is odd, so the number would end with 3, making it not divisible by 2. Similarly, N=9: sum is 18, divisible by 3, but N=9 is odd. So those are out. So yeah, 6 is the only one.
-----

Another angle: maybe I made a mistake in adding the digits. Let me verify the sum again. The number is 21420N.

<prompt_3.1> Verify the sum of the digits of 21420N and the conditions for divisibility by 6. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Digits: 2 (ten thousands place), 1 (thousands), 4 (hundreds), 2 (tens), 0 (units), N (units). Wait, hold on! Wait, the number is 21420N. Let me parse the digits correctly. The number is: 2 (first digit, ten thousands place), 1 (thousands), 4 (hundreds), 2 (tens), 0 (units), and then N? Wait, wait, that can't be. Wait, 21420N is a six-digit number. Let's break it down:

Position: Hundred thousands Ten thousands Thousands Hundreds Tens Units

Digits:     2         1       4      2      0       N

So yes, the digits are 2,1,4,2,0,N. So sum is 2+1+4+2+0+N = 9 + N. Correct. So sum is indeed 9 + N. Therefore, the logic holds. So 9 + N must be divisible by 3, so N must be divisible by 3. And N must be even. So N=6.
</execution_3.1>
</Execution_3>

<Final_answer>
Therefore, the nonzero value of $N$ that makes the number $21420N$ divisible by $6$ is $\boxed{6}$.
</Final_answer>
</think>