Two people take turns picking a number from 1 to 5, where two numbers can’t be picked in a row (for example, “1 2 1 3” is legal, but “1 2 2 1” is not). Each number is added to the total of all numbers picked so far. The first person who is forced to pick a number that will make the total exceed 45 loses. What number should the first person pick to guarantee a win? There is only one correct choice.
IM me if you want my answer, and the logic behind it.
The puzzle solution from two weeks ago (there was no Sunday puzzle due to coverage of the Reagan funeral) was Prince Albert (anagram of “printer cable”).
In general, don’t post your solution here; it’s hard to avoid spoilers. Of course, for this particular puzzle, any given response is useless without the logic behind it.
*This puzzle, from the great American puzzlemaker Sam Lloyd, was first published in June 1904. It’s called “The Game of Matrimony,” and is played by two people. One person holds out a hand with fingers numbered from one to five. Player one up any finger and announces the number (for example, four). Player two picks out any other finger, announces the number, and gives the total: (for example, five, added to four, for a total of nine). The first player now picks any finger, other than the one that was just chosen, and gives a new total – say four again – to make 13. The object of the game is to force the opponent over 45 points. Only one move guarantees the first player a win. What is it? *
Working backwards from 45 is the easiest way to figure this one out. If you want your opponent to exceed 45, you yourself must be 45. The only way you can hit 45 is if you force your opponent into a number between 40 and 44, which means you have to be the one to hit 39. In order to ensure this, you must force your opponent into a number between 34 and 38, which leads us to try for 33. See the pattern here? Every six.
Provided you go second, you can always make your number plus your opponents number equal 6. Therefore, if you want to hit 45, you must also hit 39, 33, 27, 21, 15, 9, and lastly(or initially, in this case), 3.
Provided you go second, you can always make your number plus your opponents number equal 6. Therefore, if you want to hit 45, you must also hit 39, 33, 27, 21, 15, 9, and lastly(or initially, in this case), 3.
Let’s say you are person A and your opponent is B. The sum total is in parentheses
A flaw in that is that if you hit 39, I could use 3 (assuming you hadn’t just used it). You would be then forced to pick 1 or 2 or else go over and lose.
I think your solution would work if there wasn’t the rule that you can’t use the same number twice in a row.
