Originally Posted by

**Gamma** This is probably an Algebra or Discrete or number theory problem, but Ill help anyway.

I hope you know modular arithmetic if you were assigned this problem

Reduce mod 13 and see if it is 0 mod 13 or not.

$\displaystyle 54=13*4+2 \Rightarrow 54 \equiv 2 $ (mod 13)

$\displaystyle 69=13*5+4 \Rightarrow 69 \equiv 4 $ (mod 13)

Now I suggest calculating the orders of these guys mod 13.

$\displaystyle 2^4 \equiv 3 $

$\displaystyle 2^6 =2^42^2\equiv 3*2^2 = 12 \equiv -1 $

$\displaystyle 2^{12}=(2^6)^2 \equiv -1^2 = 1$

$\displaystyle 54^{103} =54^{12*8+7} \equiv (2^12)^8 2^7 \equiv 2^7 \equiv -2 \equiv 11$ (mod 13)

Similarly

$\displaystyle 4^2 \equiv 3$ (mod 13)

$\displaystyle 4^3 =4^24 \equiv 4*3 = 12 \equiv -1 $ (mod 13)

$\displaystyle 4^6=(4^3)^2 \equiv (-1)^2 \equiv 1 $ (mod 13)

$\displaystyle 69^{67}=69^{6*11+1}=(67^6)^{11}69 \equiv 4 $ (mod 13)

so Finally

$\displaystyle 54^{103} + 69^{67} \equiv 11 + 4 \equiv 2$ (mod 13)

but if it were divisible by 13 it should be 0 mod 13, so it is not divisible.

You need to check these calculations, I did them very quickly, so they may be unreliable, but that should at least give you an idea.