What digit does this end in? 2^5028?

Question

theres a pattern recognition to this problem.
explain what the pattern is.

Answer 1

2 4 6 8...



Divide by 5 and there's an odd 3, it seems it would be 6.

Answer 2

Well, 2 raised to the following powers give the following endings:



01 - 2

02 - 4

03 - 8

04 - 6

05 - 2

06 - 4

07 - 8

08 - 6

09 - 2

10 - 4

11 - 8

12 - 6

13 - 2



From observation, we can get that if 2 is raised to the power of a multiple of 4, the result will end in 6.

Now, 5028 is divisible by 4.

So the number 2^5028 will end in 6.

Answer 3

You're onto it. Raise 2 to various powers. 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, so now the pattern has started repeating itself; the last digit is 2,4,8,6... I'll give you a clue; 2^googleplex ends in a 6 because a googleplex is divisible by 4.

Answer 4

2^1 ends in 2. 2^5 ends in 2

2^2 ends in 4. 2^6 ends in 4

2^3 ends in 8. 2^7 ends in 8

2^4 ends in 6. 2^8 ends in 6



This pattern repeats itself.

So, any no. of the form 2^(4n) ends in 6

So, the last digit of 2^5028 is 6.

Answer 5

2^4 = 16 ends in 6



6^any number ends in 6



so 2^5028 mod 10 = (2^4)^1257 mod 10 = 6^1257 mod 10 = 6

so ans = 6

Answer 6

2^1 ^2 ^3....



2 4 8 16 32 64 128 256 512 1024...



2 4 8 6 2 4 8 6 2 4....



what do you think now

Answer 7

to know what digit does it ends, the pattern is



Power---ending----pattern

1--------------2------------2

2--------------4------------4

3--------------8------------8

4-------------16-----------6

5-------------32-----------2

6-------------64-----------4

7-----------128-----------8

8-----------256-----------6

9-----------512-----------2

10--------1024----------4



Now we have seen that there is a pattern of 2-4-8-6-2-4-8-6 repeating for every power of four so for 2^5028 it ends in:



5028 / 4 = 1257



the answer is 1257 with no remainder so the number 2^5028 will end 6