Arrange cards with values ace through 9 in a row, in counting order, with the ace on the left.
Take up a card from one end of the row — left or right, your choice.
Do this twice more, each time taking up either the leftmost or the rightmost card in the remaining row.
Now you have three cards. Add their values, divide the total by six, and call the result n.
Consider the order of the cards that remain on the table.
The card in the nth position (counting from left to right) will be the 4.
Please explain why.
There are only four possible trios of cards that could be taken:
(A,2,3) sum=6 and the first card on the left is a 4
(A,2,9) sum=12 and the second card in is a 4
(A,8,9) sum=18 and the third card in is a 4
(7,8,9) sum=24 and the fourth card in is a 4.
The reason these sums increase by 6 is plain to see.
The follow-up:
How many consecutively numbered cards would you need to start with
so that the number of taken cards is m, the sum of the values is divided by f(m) and the card in the nth position is m+1?
(I hope the author does not mind me tacking this onto his problem instead of submitting this variation separately.)
|
|
Posted by Jer
on 2015-11-02 13:12:58 |
Solution and a follow-up
