Okay, let's see. The problem is asking for a nonzero value of N, which is the units digit of the number 21420N. And this number needs to be divisible by 6. Hmm, okay. Let me try to break this down step by step.

First, I remember that a number is divisible by 6 if and only if it is divisible by both 2 and 3. So, I need to check the conditions for divisibility by 2 and 3 for the number 21420N. Also, since N is the units digit of 21420N, N also affects the divisibility by 2 and 3. Let me recall how divisibility by 2 and 3 works.

Divisibility by 2 is straightforward: the units digit has to be even. So, N must be an even number. That means N can be 0, 2, 4, 6, or 8. But the problem specifies a nonzero value of N, so we can exclude 0. Therefore, possible values for N are 2, 4, 6, 8.

Divisibility by 3 is a bit trickier. For a number to be divisible by 3, the sum of its digits must be divisible by 3. So, let's calculate the sum of the digits of 21420N. The digits are 2, 1, 4, 2, 0, and N. Adding those up: 2 + 1 + 4 + 2 + 0 + N = 9 + N. Therefore, the total sum is 9 + N. For the number to be divisible by 3, 9 + N must be a multiple of 3.

Since 9 is already a multiple of 3 (because 9 ÷ 3 = 3), adding N must keep the sum divisible by 3. That means N must also be a multiple of 3. Because 9 + N ≡ 0 mod 3 implies N ≡ 0 mod 3. So, N has to be 0, 3, 6, or 9. However, N is a single digit, so those are the possibilities. But remember, N is the units digit of 21420N, and we already established that N must be even (since the number must be divisible by 2). So, N has to be both even and a multiple of 3. Let's list the possible N values that satisfy both conditions.

From the divisibility by 2, N can be 2, 4, 6, 8. From divisibility by 3, N must be 0, 3, 6, 9. The intersection of these two sets is N = 6. So, 6 is the only value that satisfies both conditions. Wait, but let me check again.

Wait, if N is 6, then the sum of the digits is 9 + 6 = 15, which is divisible by 3. And 6 is even, so the number is divisible by 2. Therefore, 6 is the correct answer. But let me make sure there isn't another number that I missed. Let me check each possible N:

- N=2: Sum is 9 + 2 = 11. 11 is not divisible by 3. So, not divisible by 3. So, 2 is out.
- N=4: Sum is 9 + 4 = 13. 13 is not divisible by 3. So, 4 is out.
- N=6: Sum is 15, which is divisible by 3. So, 6 is good.
- N=8: Sum is 9 + 8 = 17. 17 is not divisible by 3. So, 8 is out.

Therefore, the only possible nonzero N is 6. Let me verify this by plugging it back into the original number. The number becomes 214206. Let's check divisibility by 6. Since 6 is even, it's divisible by 2. Sum of digits: 2+1+4+2+0+6 = 15. 15 is divisible by 3, so yes, it's divisible by 3. Therefore, 214206 is divisible by 6. So, N=6 is correct.

Hmm, but wait a second. Let me make sure that N is indeed the units digit of 21420N. So, the number is 21420 followed by N. So, for example, if N=6, the number is 214206. The units digit is 6, which matches N. So, that's correct. So, the logic holds.

Therefore, the answer should be 6. Let me just recap to ensure I didn't miss anything. The problem requires N to be a single digit, nonzero, such that 21420N is divisible by 6. Divisibility by 6 requires divisibility by both 2 and 3. Divisibility by 2 requires the last digit (N) to be even. Divisibility by 3 requires the sum of all digits to be a multiple of 3. The sum of the digits except N is 2+1+4+2+0=9. So, adding N, the sum becomes 9 + N. Therefore, 9 + N must be divisible by 3, which implies N must be congruent to 0 mod 3. Since N is even, the intersection is N=6. Therefore, N=6 is the correct answer. Yes, that seems solid.

I don't see any mistakes in this reasoning. Let me just check with another approach. Suppose I take each possible N and check if 21420N is divisible by 6.

For N=2: 214202. Divisible by 2? Yes, last digit 2. Divisible by 3? Sum of digits: 2+1+4+2+0+2=11. 11 isn't divisible by 3. So, no.

N=4: 214204. Divisible by 2? Yes. Sum of digits: 2+1+4+2+0+4=13. Not divisible by 3. So, no.

N=6: 214206. Divisible by 2? Yes. Sum: 15. Divisible by 3. Yes. So, yes.

N=8: 214208. Divisible by 2? Yes. Sum: 2+1+4+2+0+8=17. Not divisible by 3. So, no.

Therefore, only N=6 works. Therefore, the answer is 6. I think that's thorough. I considered all possible even digits, checked the divisibility by 3, and confirmed that only 6 works. So, I'm confident that the answer is 6.
Therefore, the nonzero value of $N$ that makes the number $21420N$ divisible by $6$ is $\boxed{6}$.
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