Puzzles

SAM is 781.

There are 81 two-digit numbers with no duplicate digits: there are 9 choices of tens digit (1 to 9), and for each of these there are 9 remaining choices for the units digit (0 to 9 but not the number already used).

There are 648 three-digit numbers with no duplicate digits: there are 9 choices of hundreds digit (1 to 9), and for each of these there are 9 remaining choices for the tens digit (0 to 9 but not the number already used) and 8 choices for the units digit (0 to 9 but neight number already used).

648+81 = 729.

There are 9 gaps between the numbers 1 to 10. We need to pick 5 of these gaps to split the sequence. There are \(\left(\begin{array}{c}9\\5\end{array}\right)\) ways to do this: this is 126.

Let's say there were originally \(t\) cards of each colour, and \(2a\) red cards and \(a\) black cards in pile A. After the move, there were \(t-2a+108\) red cards and \(t-a\) black cards in pile B. Two thirds of the card in pile B are red, so:

Therefore there were 108×2=216 cards.

Interestingly, it appears that this answer is independent of \(a\): any number of cards could be put in each pile and this situation would still work. However, only one situation could have happened unless you allow there to at some points have been a negative number of cards in each pile.

There are one three-digit even numbers whose digits add to 24: 798, 888 and 996. Of these, only 888 is a multiple of 24.

The product of the numbers in the red boxes is 144.

The Octingham resident's 11 is equal to our number 9. 9 squared is 81. 81 in base eight is 121. Interestingly, this is the same answer as "just" doing 11 squred in base ten.

Each number will appear 7 times: one time paired with the numbers six numbers 5 to 10, plus an extra appearance on the tile containing the same number twice. The total of all the numbers is therefore 7×(5+6+...+10)=7×45=315.