What are the last two digits of 4^2001?

Question

I think you have to use modular arithmetic.

Answer 1

4^11 mod 100
≡ 4 mod 100

4^2001 mod 100
≡ (4^11)^182 / 4 mod 100
≡ 4^181 mod 100
≡ 4^60 mod 100
≡ 4^5 mod 100
≡ 24 mod 100
The last two digits are 2 and 4.

Answer 2

The answer is 04. You can do it using the 'mod' method or create a table till a pattern develops. Here is the longer way. Suppose N is the power. Then 4^N will yield the following table:
N 4^N

1 ----------04
2 ----------16
3 ----------64
4 ---------256
5 -------1024

6 -------4096
7 ------16384
8 ------65536
9 ----262144
10 -1048576

11 - 4194304
12-17577216

etc.

If K is any non-negative integer(i.e. >=0) then in general the pattern of the power and the last two digits can be described as follows:
When the power is 10*(K)+1, the last 2 dgts are 04
10*(K)+ 2, 16
10*(K)+3, 64
10*(K)+4, 56
10*(K)+5, 24
10*(K)+6, 96
etc. etc. Also notice the last digit toggles between 4 and 6. The number system we are familiar with(base 10), have other such patterns(either addiiton or multiplications) with different numbers.

Answer 3

Yes.

4^10 = 1 (mod 25).

So 4^2000 = (4^10)^200 = 1^200 = 1 (mod 25)

So 4^2001 = 4 (mod 25)

On the other hand, 4^2001 = 0 (mod 4).

We know then that 4^2001 = 4 (mod 100)

and hence that the last two digits must be '04.'

In general, 4^(10x+1) ends in 04, for the same reasons.

Answer 4

4, 16, 64, 56, 24, 96, 84, 36, 44, 76, 04, 16
2001mod10 = 1,
so the last 2 digits of 4^2001 are 04

Answer 5

Look at the powers of 4:
4¹=04
4²=16
4³=64
4⁴=256
etc.

The final two digits start repeating after 4¹⁰, 4²⁰, etc.
2001mod10 = 1, so the last two digits of 4²⁰⁰¹ are 04