+0

number theory

0
206
2

For how many positive integers n less than 100 is 5^n + 8^n + 13^n + 17^n a multiple of 6?

Aug 1, 2022

#1
+1

There is no integer solution for n < 100!! Make sure your question is correct.

Aug 1, 2022
#2
+118533
+1

$$5^n + 8^n + 13^n + 17^n \mod6\\ \equiv (-1)^n + 2^n + 1^n + (-1)^n \mod6\\ \equiv 2*(-1)^n + 2^n + 1 \mod6\\$$

$$\qquad 2^1 \mod6 \equiv 2 \mod 6\\ \qquad 2^2 \mod6 \equiv 4 \mod 6\\ \qquad 2^3 \mod6 \equiv 2*2 \mod 6\equiv 4 \mod6\\ \qquad 2^4 \mod6 \equiv 4*2 \mod 6\equiv 2 \mod6\\ \qquad etc\qquad so\\ \qquad 2^n \equiv 2\mod 6\qquad \text{If n is odd}\\ \qquad 2^n \equiv 4\mod 6\qquad \text{If n is even}\\$$

$$5^n + 8^n + 13^n + 17^n \mod6\\ \equiv (-1)^n + 2^n + 1^n + (-1)^n \mod6\\ \equiv 2*(-1)^n + 2^n + 1 \mod6\\ \text{If n is odd}\\ \qquad \equiv 2*(-1) + 2 + 1 \mod6\\ \qquad \equiv -2 + 2 + 1 \mod6\\ \qquad \equiv 1 \mod6\\ \text{If n is even}\\ \qquad \equiv 2*(1) + 4 + 1 \mod6\\ \qquad \equiv 2 + 4 + 1 \mod6\\ \qquad \equiv 1 \mod6\\ so\\ \text{For all positive integer values of n}\\ 5^n + 8^n + 13^n + 17^n \equiv 1 \mod6\\ \text{It will never have a zero remainder when divided by 6}$$

LaTex:

5^n + 8^n + 13^n + 17^n \mod6\\
\equiv (-1)^n + 2^n + 1^n + (-1)^n \mod6\\
\equiv 2*(-1)^n + 2^n + 1  \mod6\\
\text{If n is odd}\\
\qquad \equiv 2*(-1) + 2 + 1  \mod6\\
\qquad \equiv -2 + 2 + 1  \mod6\\