Question

find the remainder when 3¹+3²+3³+3⁴-------+3⁴0is divided by 13​

Answer 1

Question:

Find the remainder when 3 + 3² + 3³ + 3⁴ +.........+ 3⁴⁰ is divided by 13​.

Solution:

3 ≡ 3 (mod 13)

3² ≡ 9 (mod 13)

3³ = 27 ≡ 1 (mod 13)

3⁴ = 3³ × 3 ≡ 1 × 3 = 3 (mod 13)

So, here we get that the remainders given by the powers of 3 on division by 13 are 3, 9 and 1 only, in order. Algebraically,

$\bullet\ \ 3^{3n}=(3^3)^n=27^n\equiv 1^n=\mathbf{1}\pmod{13}\ \ \ \ \ [\because\ 27\equiv 1\pmod{13}]\\ \\ \bullet\ \ 3^{3n+1}=3^{3n}\cdot 3\equiv 1\cdot 3=\mathbf{3}\pmod{13}\\ \\ \bullet\ \ 3^{3n+2}=3^{3n}\cdot 3^2\equiv 1\cdot 9=\mathbf{9}\pmod{13}$

→  The least and the highest powers of 3 among the given sum whose exponents are in the form 3n + 1 are 3 and 3⁴⁰. No. of such terms is 14.

→  The least and the highest powers of 3 among the given sum whose exponents are in the form 3n + 2 are 3² and 3³⁸. No. of such terms is 13.

→  The least and the highest powers of 3 among the given sum whose exponents are in the form 3n are 3³ and 3³⁹. No. of such terms is 13.

So,

$\begin{aligned}\sum_{n=1}^{40}3^n&\equiv 14\cdot 3+13\cdot 9+13\cdot 1\\ &=42+13(9+1)\\ \\ &=42+13\cdot 10\\ \\ &=42+130\\ \\ &\equiv 3+0\\ \\ &=\mathbf{3}&\pmod{13}\end{aligned}$

Hence 3 is the remainder.