# Issue 2, 2017 | OFF AIR

## The cube conundrum

A pilot has two cubes in his cockpit. Every day he arranges both cubes so that the front faces show the current day of the month.

**What numbers are on the faces of the cubes to allow this?**

Note: You can't represent the day "7" with a single cube with a side that says 7 on it. You have to use both cubes all the time. So the 7th day would be "07".

The answer will be published in the next edition of ON AIR. Please send your solutions to [email protected]. All entries must be received by 30 August 2017. The first correct answer drawn at random after this date will win a pair of Helios noise-cancelling headphones. Good luck!

## And the winner is…

Congratulations to **Peter Douglas **of the Northern
Lighthouse Board, who sent the correct answer and came first in the draw. Find the answer to the puzzle "Saving your ATM System" below:

**ANSWER: **10

**JUSTIFICATION:**

For each computer, any number of programs can be installed. If the computer develops symptoms of the malware, then we know it is one of the programs which has been installed on that computer. The problem is then how many unique combinations of yes/no can be made with any given number of computer, and each program is assigned a combination.

1: Computer can only provide 2 combinations C1-Yes/C1-No

Therefore 1 computer can be used to test up to 2 programs: If C1 develops the malware then it is program 1, and if it does not then it is program 2.

2: Computers can provide 4 combinations C1-Yes, C2-Yes/ C1-Yes, C2-No/C1-no, C2-Yes/C1-No, C2-No

Therefore 2 computers can be used to test up to 4 programs: If C1 and C2 develop the malware then it is program 1, if C1 does and C2 does not then it is program 2, if C1 does not and C2 does then it is program 3, and if neither of them do it is program 4.

Etc…

This is effectively a question about binary combinations. The maximum number of unique combinations provided by n computers is 2^{n}. We need at least 1000 unique combinations (one for each program) the smallest power of 2 which is larger than 1000 is 2^{10} = 1024. Therefore the minimum number of computers needed is 10.