# 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.

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I am reminded of the joke where a kid correctly concludes $$\frac{26}{65} = \frac{2}{5}$$ by "cancelling" the $6$'s from the numerator and denominator. –  Srivatsan Sep 9 '11 at 22:47
@Srivatsan: Seen this? :) –  Guess who it is. Sep 10 '11 at 0:20
I would argue that the step of ignoring the current is not false intuition nor non-rigorous. In fact you are effectively transforming to a reference frame moving with the current. –  Ross Millikan Sep 10 '11 at 0:26
You might be interested in meeting Lucky Larry, BTW... –  Guess who it is. Sep 10 '11 at 0:26
It seems to me that the question refers to two distinct kinds of problems (and corresponding solutions): on one hand those like Srivatsan Narayanan's fraction example, where the correct answer can be found by an incorrect method, and on the other hand those like the AMC problem, where merely knowing that the problem can be solved using only the information given makes finding the solution much easier than it would be if one did not know that in advance. –  Ilmari Karonen Sep 10 '11 at 9:17

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...

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 $(1 \textrm{ mi/min} \cdot 1.5 \textrm{ min}) = 1.5 \textrm{ mi}$.

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Just want to point out that the short way in this case is perfectly rigorous (unless you are thinking of an even shorter non-rigorous way :)). –  Srivatsan Sep 9 '11 at 22:53
@Srivatsan: true true - I was going by the format of the first example given in the question (which also seems rigorous to me). –  Steven Stadnicki Sep 9 '11 at 23:31
@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 $(1 \textrm{ mi/min} \cdot 1.5 \textrm{ min}) = 1.5 \textrm{ mi}$. –  Mateen Ulhaq Sep 10 '11 at 21:46
Isn't this part of the joke where John von Neumann solved this problem quickly and his friend assumed he found the trick above, but von Neumann said, "uh, no.. I summed the series." –  Fixee Sep 11 '11 at 3:36

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...

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This is a great question, but it's known since before Martin Gardner. Some sources here –  RoundTower Sep 10 '11 at 2:22
I can easily believe it was known before, but that site seems newer as Gardner retired from the column 30 years ago. –  Ross Millikan Sep 10 '11 at 4:36
RoundTower's linked pdf states on page 2: "Martin Gardner has found the earliest reference for this problem in Samuel I. Jones's *Mathematical Nuts, self-published, Nashville, 1932" –  anon Sep 10 '11 at 8:25

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.

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