# CTK Insights

• ## Pages

19 Feb

### 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 half-meter 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 half-meter 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 $a-b$ 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.

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 $P'$.