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For how many positive integers n less than 100 is 5^n + 8^(n + 1) + 13^(n + 2) + 14^(n + 3) a multiple of 6?

 Jun 23, 2021
 #1
avatar+287 
+2

If you make a table of congruences modulo 6 of

$$5^n, 8^{n+1}, 13^{n+2}, 14^{n+3)$$

for $n \equiv 0,1,2,3,4,5\pmod6$, you'll see a pattern.  According to my table, half of $$5^n + 8^{n+1} + 13^{n+2} + 14^{n+3)$$ for $1\le n< 100$ are $\equiv 0\pmod6$ and the other half are $\equiv 2\pmod6$.

 

For conveniece, when creating the table, replace every base $a$ by $a \bmod 6$.
 

 Jun 24, 2021
 #2
avatar+114936 
+1

 

For how many positive integers n less than 100 is a multiple of 6?

 

Lets simplify the expression (mod6)

\(5^n\mod6\\ \equiv(-1)^n\mod6\\ \)

which is -1 when n is odd and +1 when n is even

 

\(13^{(n+2)}\equiv 1^{(n+2)}\equiv1\mod6\)

 

 

\(8^{(n+1)}\equiv 2^{(n+1)}\equiv2*2^n\mod6\\~\\ 14^{(n+3)}\equiv 2^{(n+3)}\equiv8*2^n\equiv2*2^n\mod6\\~\\ 8^{(n+1)}+14^{(n+3)}=4*2^n\mod6\\~\\ \)

so

\(5^n + 8^{(n + 1)} + 13^{(n + 2)} + 14^{(n + 3)}\mod6 \\ \equiv 1+ (-1)^n+4*2^n \mod6\\~\\ \text{When n is odd}\equiv 4*2^n \mod6\\~\\ \text{When n is even}\equiv 2+4*2^n \mod6\\~\\ \)

consider 4*2^n

n 1 3 5 7   2 4 6 6
4*2^n 8 32 128 512   16 64 256 1024
4*2^n mod6 2 2 2 2   4 4 4 4
2+4*2^n mod6 (even only)           0 0 0 0

 

So that expression is a multiple of 6 for all even values of n.  So that is 49 values,  2 to 98 inclusive

 Jun 24, 2021

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