Question

"Determine all positive n integers such that n divides by 3^n-2^n."

Halp! I don't know where to start. :c

Answer 1

looks like mathematical Induction

Step 1 prove true for n = 1 3^1 -2^1 = 1 and n =1 is a factor of 1.

Step 2: assume true for n= k then 3^k - 2^k = km where m is a factor

or m = (3^k - 2^k)/k

Step 3: Prove for n=k+1 then 3^(k+1) - 2^(k+1)
=3 x 3^k - 2x2^k

ummm.. a bit stuck

Answer 2

Assuming the question is to find all positive integer solutions to:\[p=\frac{n}{3^n-2^n}\]then we know one trivial solution (as pointed out by campbell above) is \(n=1\).
we can show that this is the ONLY solution by proving (using induction) that:\[3^n-2^n>n\hspace{2cm}\text{for all }n\ge2\]firstly we can show this is true for \(n=2\):\[3^2-2^2=9-4=5\gt2\]next we assume it is true for \(n=k\):\[3^k-2^k\gt k\implies3^k\gt k+2^k\tag{a}\]and now we just need to prove it is also true for \(n=k+1\):\[\begin{align}
3^{k+1}-2^{k+1}&=3\cdot3^k-2\cdot2^k\\
&\gt3(k+2^k)-2\cdot2^k\hspace{2cm}\text{using (a)}\\
&\gt3k+3\cdot2^k-2\cdot2^k\\
&\gt3k+2^k\\
&\gt k+(2k+2^k)\\
&\gt k+1
\end{align}\]
Hence the only solution is \(n=1\).

Answer 3

Clever. :o

Answer 4

thx