Thought Provokers to Start a Class With, VI
Here are a few engaging problems that many a student will be able to solve and, if not, would be able to appreciate the simplicity of a missed solution.

Without a measuring tape, is it possible to cut a halfmeter piece from the rope of 2/3 m length? Solution

What is the digital root of 100! Solution

Compute \frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\ldots\cdot\frac{2011}{2010}\cdot\frac{2012}{2011} Solution

Place between the numbers in the sequence 1\space 2\space 3\space \ldots \space 13\space 14 signs + or  to make the result of the algebraic sum 0. Solution

On a planet the oceans take less than one half of the total surface area of the planet. Prove that it is possible to dig a tunnel through the center of the planet with both ends on the dry land. Solution
Without a measuring tape, is it possible to cut a halfmeter piece from the rope of 2/3 m length?
Fold the rope (a paper band in the picture) twice and cut off one of the end quarter pieces. What remains is \frac{3}{4}\times \frac{2}{3}=\frac{2}{4}=\frac{1}{2}.
What is the digital root of 100!
Recollect the criterion of divisibility by 9: a number is divisible by 9 if and only if the sum of its digits is divisible by 9. Iteratively, a number is divisible by 9 if and only if its digital root is divisible by 9. There is no doubt that 100! is divisible by 9; this is also true of its digital root. But the digital root of any number is a single digit quantity, thus it is bound to be 9.
Compute \frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\ldots\cdot\frac{2011}{2010}\cdot\frac{2012}{2011}
Just proceed one step at a time:
\frac{3}{2}\cdot\frac{4}{3}=\frac{4}{2} \frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}=\frac{5}{2} \frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}=\frac{6}{2}and so on. You may observe that the final product is the fraction whose numerator coincides with the numerator of the last factor and the denominator is always 2. It follows that the answer to the question at hand is \frac{2012}{2}=1006.
(This problem is a yearly upgrade of the problem at the MAA MinuteMath site.)
Place between the numbers in the sequence 1\space 2\space 3\space \ldots \space 13\space 14 signs + or  to make the result of the algebraic sum 0.
This is not possible; the reason is that there is an odd number (7) of odd numbers: 1,\space 3,\space \ldots,\space 13. The parity of the sum a+b and the difference ab is always the same. It follows that, regardless of the signs that are put in, the result will always be odd. But 0 is an even number.
(This problem has been broadcast on twitter by @JohnAllenPaulos.)
There is a tacit assumption that the planet is centrally symmetric. Let A stands for the "dry land" set. Consider its centrally symmetric image A'. Both cover more than a half of the planet and, therefore, have a nonempty intersection. The intersection A\cap A' is also centrally symmetric so that if P\in A\cap A' then also P'\in A\cap A' where P' is the centrally symmetric image of P. Dig a tunnel from P to P'.