Let's hit a Sixer

The last digit of $6\times 6$ is $6$ , and $6\times6$ its last digit is $6$ and so on $\therefore$ the last digits of $6^n$ always is $6\Rightarrow 6^{666666}+1$ lasts digit's is $6+1=\boxed{7}$

Any thing to the power $6$ has a units digit of $6.$ Therefore the units digit of $6^{666666}+1$ is $6+1 =7.$

6^2=36,6^3=216. So when 6 is multiplied by 6,the last digit is always 6.So here the last digit is (6+1)=7.

Last digit of 6^6 is 6, hence 6+1=7

After I realized that 6^2 and 6^3 ended with 6, I assumed there was a pattern. I assumed correctly.

This is purely beauty of multiples of 6. 6 * 6 = 36 36 * 6 = 216 216 * 6 = 1296 1296 * 6 = 7776 7776* 6 = 46656.... & so on son 6^666666's last digit is 6. And hence 6 + 1 = 7. Normally my answers are wrong ;) This time I was lucky...

Here's a formal approach using modular arithmetic.

1. First, we establish $6^{k} \equiv 6 \textrm{ mod } 10$ for any natural number $k$ . This can be done by induction quite readily.

2. Then, we consider $(6^{666666}+1) \textrm{ mod } 10$ .

   $(6^{666666}+1) \textrm{ mod }10 \equiv 6^{666666} \textrm{ mod }10 + 1 \textrm{ mod }10$ .

   Applying our result from (1.), we see

   $6^{666666} \textrm{ mod }10 + 1 \textrm{ mod }10 \equiv 6 \textrm{ mod } 10 + 1 \textrm{ mod }10 \equiv 7 \textrm{ mod }10$ .

   Therefore, the last digit must be $\boxed{7}$ .

Multiplication of 6 with 6 and thereafter the result again with 6 any number of times always yield a figure ending with 6 only. Eg 6x6=36, 36x6=216, 216x6=1296 and so on. Hence when we add 1 to the figure thus arrived at it is ending with 7. Hence the answer is 7.

Just multiply the unit digits... They never change.in 6. And don't forget to add the 1 so ans - 7

Power of 6 leaves an unit digit with 6 , so adding 1 gives us 7 in unit place..

The exponet of 6 always have last digit 6. Adding 1 to that makes it 7

It was just logical

6^2=36, 6^3=216, .. The product always ends with 6 so 6+1 =7

We have multiples of 6 which will always have the units digit as 6.Therefore adding one to it yields 7

First, If we create a base pattern for 6 and the different powers of six. however, because six has only one recurring digit when multiplying itself, we can simply voild the power and leave the six on its own.

now that we solidified the last digit, we can then follow whatever expectations the question throws at us, in this case, it is adding one to the last digit.

the answer is 7 (6+1)

6*6 =36... 6multiplied any no. Of times ends up with 6 at the units place.. So 6+1=7

6 to any power has a last digit of 6 and adding that by 1 gives7.

Intuition!

Well when 6 is multiplied by 6 then last digit will be a 6 no matter how many time you do it. So ans will be 6+1=7

If you see the multiples of 6, it's 6, 12, 18, 24, 30, 36, 42, 48, 56, 60. And if you look carefully the last digits are in a pattern: 6, 2, 8, 4, and 0. So first of all you can take away the answers that are even since you're going to add 1 to it. But you can also find that 6^666665 has the last digit as 0 because it's the fifth pattern. So then if you find 6^666666 would have 6 as the last digit, so when you add 1, you would get 7. Unfortunately, I'm not good at explaining things so my solution may be hard to understand. Please forgive me with that.

6^any no. Always ends in 6. And last 6+1=7

6 raise to the nth power ended with a figure 6, then the last digit should be 6+1 = 7

last digit always 6 so answer is 7

6 x 6 is 36. Last digit 6. When you multiply this number again to get the power you always end up with last digit as 6. So the answer is 7.

Last digit of 6 power any number gives 6.

Stick to math math please.

I have observed that (10y+x)^5 ie power 5 of any integer has the same last digit

1^5 has last digit 1 2^5 has last digit 2 3^5 has last digit 3 4^5 has last digit 4 5^5 has last digit 5 6^5 has last digit 6 7^5 has last digit 7 8^5 has last digit 8 9^5 has last digit 9 10^5 has last digit 0

Same as ramirez

Nicomachus of Gerasa, who lived in Syria in the 1st Century, showed that all perfect numbers end in 6 or 8 alternately. 6 is the perfect of them all, because its prime factors add up to it. Multiples of 6 also end in 6. Therefore 6+1 = 7.

After multiplying 6with 6 you Will again get product having unit 6 and plus 1 will give unit number 7

6^x where x is any positive integer will produce an integer whose last digit is 6, and 6+1=7.

6 to the power any number last digit must b 6 only den adding 1 to 6 is 7..

666666/4 = 166666 i.e 166666*4 = 666664 =>666666-666664 = 2==> 6^2+1 ==>36+1 = 37==>7

Considering 78% of the people got this right I don't think I really need to explain it.

6 + 1 = 7, no math done here, author is bating you here is all.

7 because when you multiply 2 numbers finishing by 6, the result will always finish by 6 so 6+1=...7

The last digit of 6 raised to the power is always six; 6 times 6 is 36, 6^2 times 6 is 216, etc. Always the same. So, 6 to any integral positive integer will end in 6; 6+1=7

the last digit of 6 raised to any power is always six so in the question 6^666666+1, the last digit would be 6+1=7