Examples of math contest problems that can be solved in a 'cheap' way?

What are some examples of math contest problems that can be solved by using a nonrigorous, 'cheap' shortcut?

For instance, a problem on the 2011 AMC went:

A raft and a motorboat left dock A and started downstream. The raft traveled at the speed of the current. The motorboat maintained a constant speed with respect to the river. The motorboat went to point B then immediately turned back, meeting the raft 9 hours after leaving dock. How long did it take the motorboat to travel from A to B?

An attentive student may notice that the question does not mention the speed of the current, so it must not affect the answer. Then setting the current to be 0, he gets 4.5 hours trivially.

Another example might be the trivial 'derivation' of the probability of a random fill of a Ferrers diagram is a Young tableau. Assume all probabilities are independent, then multiply the individual probability for each hook; this gives the correct formula, but the proof is completely wrong (the probabilities are obviously not independent).

I'm looking for problems in which an otherwise non-rigorous step, or a false intuition leads to the correct answer.

Solution 1:

For me, the canonical example of this is the 'mosquito between oncoming trains' problem: two trains initially 1 mile apart are traveling towards each other at 20MPH, with a mosquito starting at one train, traveling 60MPH towards the other train, then immediately reversing direction when it gets there and traveling back to the first, etc. back and forth until the two trains collide. How far does it go? The 'long' way of solving the problem is to figure out how far the mosquito travels on each leg of the trip and sum the (geometric) series that results. The short way, of course, is left as an exercise to the reader...

@muntoo adds the following spoiler:

@Reader

Assume one train is stationary, and the other travels at $40 \textrm{ mph} = \frac{2}{3} \textrm{ mi/min}$. Therefore, it takes $\frac{1 \textrm{ mi}}{\frac{2}{3} \textrm{ mi/min}} = 1.5 \textrm{ min}$ for the trains to collide.

Next, we know the mosquito will be travelling $1 \textrm{ mi/min}$ the whole time, so the distance the mosquito travels is $(1 \textrm{ mi/min} \cdot 1.5 \textrm{ min}) = 1.5 \textrm{ mi}$.

Solution 2:

In a very similar vein, Martin Gardner posed the question of drilling a hole along the diameter of a sphere. The length of the cut edge is 6 inches. What is the volume remaining? Again, since the diameter of the hole is not specified, the volume remaining must not depend on it (which one can prove). So imagine a zero diameter hole...

Solution 3:

Another common pattern is "find the minimum (or maximum) value of $f(x)$ such that $x$ satisfies (certain constraints)". The poor student can somehow find one $x$ that satisfies the constraints, and that turns out to be the correct one. Of course the real work, which the student left out, was to show that there is no better solution.