In each issue of the Automation Notebook we feature a section of brainteasers. These are the brainteaser answers from Issue 25, 2013 of the Automation Notebook. The brainteaser questions are repeated in black. The answers to the brainteaser questions are highlighted in red with explanations. You can view the brainteasers from Issue 25, 2013 without the answers here: http://library.automationdirect.com/brainteasers-issue-25-2013/
1. Circular References
A factory has a circular work cell, which they wish to divide into four equal sections, using three safety curtains, each the same length. How might this be done?
Extra Credit: What is the length of the three safety curtains?
Length of the three safety curtains = 0.5πd (where d is the diameter of the work cell)
2. The Circling Cart
That same factory has a four wheel cart that rides on a pair of circular tracks. The outside wheels of the cart turn twice as fast as the inner wheels. The cart’s axles are 5 feet wide.
What is the length of the outside track?
The diameter of the outer track is twice the diameter of the inner (in order to cause the wheels to turn at twice the rate), and the inner track is 10 feet less than the outer track (i.e. twice the axel width), so the outer track must be 20 feet in diameter. Thus the length of the outer track is 20π, or about 20.8 feet.
3.) Ancient Riddle
This is a very old story – you may have heard it. It’s often told regarding 17 cows, but we couldn’t help taking a small liberty:
An eccentric automation tycoon passes away, leaving his empire (consisting of 17 factories) to his children in the following manner: his eldest is to receive half of his factories, his second child – one third of the factories, and his youngest gets one ninth of them. The children do not wish to share or ‘break up’ a factory, and thus they cannot figure out how to divide the empire.
They consult a wise friend (also an automation expert) who offers to loan them a factory, so that with a total of 18 factories; the eldest can take half (or 9 factories), the second child gets the one third share (6 factories), the youngest child receives the one ninth share (2 factories) AND they can give the ‘loaned’ factory back to the friend. The three children also realize that they are each better off in the end, than they would have been if they had subdivided one of the factories.
Can you explain the apparent paradox in simple mathematical terms?
The parent is either not very good at math – or they wanted a small portion of the inheritance to go un-inherited (to charity? or perhaps ‘to the state’?). The portions stipulated (½ + 1/3rd +1/9th) only add up to 17/18ths of the total fortune. There is still 1/18th of the fortune that is not bequeathed (almost an entire factory).
When the friend artificially increases the inheritance to 18 factories, then all the fractions work out to even numbers, and there is no ‘unused portion’ in that solution – it is effectively divvied-up amongst the children.
And yes, each child gets more than was originally stipulated. And whether the parent’s wishes were honored is a very good question. In the first reckoning the children would only have received 8.5, 5.7, and 1.9 factories respectively.
4.) Tooling Around
In one well-automated factory, the operators were getting bored, so the foreman offered them a challenge. The factory has 25 CNC machine tools arranged in a neat grid of 5 rows, and 5 columns. Each machine tool has an operator. The foreman offered to let each operator move to a new machine tool as long as they followed certain restrictions. Each operator could only move to a machine tool that was in the position directly in front, or directly behind, or directly to the right side, or directly to the left side – of that operator’s original position. All of the operators were required to move, and in the end there must be an operator at each station.
Were the operators able to meet the foreman’s restrictions? Why or why not?
There are many ways to prove that the foreman’s challenge cannot be accomplished. Here’s one:
Imagine that the machine tools were laid out on a checker board, with alternating red and black squares. It’s easy to see that each operator would have to change his/her color during the move (i.e. each would have to move from either red to black, or black to red). Now count the occurrences of each color. Because 25 is an odd number, there must be more of one color that the other – so it follows that not all the operators can change their color. There is no way for all of them to move given the foreman’s restrictions – but he did solve their boredom problem for at least a few minutes while they argued about it 😉