Question

3^2013+3^2014+3^2016 это нужно поделить на 31

Answer 1

$\tt \dfrac{3^{2013}+3^{2014}+3^{2016}}{31}=\dfrac{3^{2013}(1+3+27)}{3} =\dfrac{3^{2013}\cdot 31}{31}=3^{2013}$

Answer 2

3^(2013)+3^(2014)+3^(2016)=

3^(2013)(1+3+3³)=31*3^(2013)

31*(3^(2013):31=3^(2013)

Answer 3

To simplify the expression and then find the remainder when divided by 31, we can use the properties of modular arithmetic.

First, let's rewrite the expression:

3^2013 + 3^2014 + 3^2016

Now, we can apply the properties of modular arithmetic to each term:

1. For any positive integer n and a, b ∈ Z (integers): If a ≡ b (mod n), then a^k ≡ b^k (mod n)

2. For any positive integer n and a, b ∈ Z (integers): If a ≡ b (mod n), then a + b ≡ b + b (mod n)

Now, let's calculate the remainders of each term when divided by 31:

1. 3^2013 % 31: Since 3^5 ≡ 3 (mod 31) (as 3^5 - 3 = 243 - 3 = 240 = 31 * 7), we can rewrite 3^2013 as (3^5)^402 * 3^3 ≡ 3^3 ≡ 27 (mod 31).

2. 3^2014 % 31: Similarly, 3^5 ≡ 3 (mod 31), so 3^2014 ≡ (3^5)^402 * 3^4 ≡ 3^4 ≡ 81 ≡ 19 (mod 31).

3. 3^2016 % 31: Again, 3^5 ≡ 3 (mod 31), so 3^2016 ≡ (3^5)^403 * 3^1 ≡ 3^1 ≡ 3 (mod 31).

Now, let's add these remainders together and find the remainder when the sum is divided by 31:

(27 + 19 + 3) % 31 = 49 % 31 = 18

So, the remainder when the expression (3^2013 + 3^2014 + 3^2016) is divided by 31 is 18.