I overheard mention of a particular problem on a recent British Mathematical Olympiad, namely the following:
A number written in base 10 is a string of 3^2013 digit 3s. No other digit appears. Find the highest power of 3 which divides this number.
Personally, I bemoan such problems that are trivialised by the knowledge of advanced theorems, as it enables competitors to gain an unfair advantage by rotelearning many results rather than demonstrating creative mathematical thought. In this case, the question is trivialised by a rather elegant but littleknown lemma called lifting the exponent.
So, what does the lemma state? Firstly, we establish the following definition:
Definition: the padic valuation v_p(n) of an integer n to be the highest power of p which divides n. For example, v_2(40) = 3, since 2^3 divides 40 but 2^4 does not.
Then the lemma is as follows:
Theorem (lifting the exponent): Let p be an odd prime, and a and b integers such that neither a nor b is divisible by p, but p divides their difference a − b. Then v_p(a^n − b^n) = v_p(a − b) + v_p(n).
Why is this true? The idea is that we factorise a^n − b^n like so, where n = k p^m and k is not divisible by p:
The factors in the final factorisation are highlighted. It is clear that it is sufficient to prove that the yellow factor is coprime to p (which is easy, since all of the terms are congruent modulo p and are nonzero) and each of the blue factors are divisible by p (easy for the same reason) but not by p², as we shall prove:

If a and b are congruent modulo p², this is again trivial for the same reason as before.
 Otherwise, we have to actually rely on the property that p is odd (if p = 2, we need a and b to be congruent modulo 4 rather than modulo 2). We let x and y be equal to a^p^i and b^p^i, respectively, for the obvious value of i, such that the factor is of the form:
Γ = x^(p1) + x^(p2) y + x^(p3) y^2 + … + y^(p1)
Then we set y = x + lp for some integer l (which we can do, since y and x are clearly congruent modulo p). Expand the factor Γ to produce a sum of binomial expansions; we can ignore all terms of order p² and higher since we’re only interested in the residue modulo p². This gives the following expression:
The rightmost term vanishes, leaving something that is clearly not divisible by p². Consequently, the proof is complete and the result follows immediately.
Zsigmondy’s theorem
Another useful fact concerning a^n − b^n is this: except in a few exceptional cases, it has a new prime factor p that does not occur in any of a − b, a² − b², a³ − b³, …, a^(n−1) − b^(n−1). The exceptions to the rule are the following:

a = 2, b = 1, n = 6: we have 2^6 − 1^6 = 63, whose prime factors are 3 and 7, which occur in 2^2 − 1^2 and 2^3 − 1^3, respectively.
 a + b is a power of 2, and n = 2: then a² − b² = (a + b)(a − b). The first factor is a power of 2 (so no new primes there), and the second factor is itself the previous term in the sequence (so, by definition, no new primes there either).
A related statement about Fibonacci numbers (where the integers a and b are replaced with irrational algebraic integers, and an extra factor of √5 slips in) is known as Carmichael’s theorem. Zsigmondy’s theorem and Carmichael’s theorem can be mutually generalised to other Lucas sequences.