What is the last digit of 3^36th power? explain, i still do not understand why it is 0.

The answer is actually 1, and not 0, and here's why.

Write 3^36 into (3^3)^12. Note that the exponents 3 and 12 multiply to 36

Now focus on 3^3 which turns into 27. The last digit here is 7. It turns out that this is the only thing that matters when it comes to finding the final answer.

So instead of 27^12, we can look at 7^12

Break 7^12 into (7^3)^4, then note how 7^3 = 343. Again the units digit is all we care about. So instead of 343^4, we just look at 3^4

3^4 = 81 and the units digit here is 1 which is the final answer.

Here's an alternative approach using modular arithmetic. This is the set of tools that helps us find remainders. If we work with mod 10, then we can find the units digit of any complicated expression such as 3^36

3^36 = (3^3)^12   (mod 10)

3^36 = 27^12 (mod 10)

3^36 = 7^12 (mod 10)

3^36 = (7^3)^4 (mod 10)

3^36 = 343^4 (mod 10)

3^36 = 3^4 (mod 10)

3^36 = 81 (mod 10)

3^36 = 1 (mod 10)

We effectively follow the same steps as mentioned above, but we're using a more formal set of steps to show.

  The last digit is 1.

Step-by-step explanation:

3^36 = 150094635296999121

The last digit is 1.

The digit cannot be zero because there are no factors of 10 in 3^36.

Consider the last digits of powers of 3:

  3^0 = 1

  3^1 = 3

  3^2 = 9

  3^3 mod 10 = 7

  3^4 mod 10 = 3·7 mod 10 = 1

The last digit repeats in the pattern 1, 3, 9, 7, 1. So every 4th exponent will have the same result.

So, 3^36 mod 10 = 3^(36 mod 4) = 3^0 = 1.

the slope of the first function is -3. (3/1)

the slope of the second function is 1. (13-10/6-3).

since |-3| is greater than |1|, the function with the steepest slope is the first.

the terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. a recursive formula for a geometric sequence with common ratio r is given by an=ran−1 a n = r a n − 1 for n≥2 n ≥ 2 .

step-by-step explanation: