by Mark Valentine
Published Friday June 21 2024
There are 84 marbles, numbered sequentially from 1, in a bag. Oliver (going first) and Pam take turns, each removing two marbles from the bag. The player finds the “most square” integer factorisation of each number. The difference between the two factors for each number is then added to the player’s score (eg removing marbles 1 and 84 scores 5 points since 1=1×1 and 84=7×12), and the game ends when all marbles are removed.
After each of the first four turns (two each), both players’ expected final score was a whole number. For example, if there were 90 points remaining, with 4 turns left for Oliver and 5 for Pam, Oliver would expect to score another 40 points. In addition, each player’s score for their second turn was the same as their first. Also, Pamela scored the lowest possible game total.
What was Oliver’s expected final score after Pamela’s second turn?
Sunday Times Teasers Discussion Group 
I wonder what is a “whole number”?
The factorisation imposes integer numbers. The example of 90 points remaining might suggest “whole number” = multiple of 10. Or has a “whole number” just more than one digit, like 83 ?
Generally “whole number” means “a number with no fractional part” (i.e. an integer).
Although it is sometimes used to refer to the non-negative or positive integers.
Thanks, Jim
the factors of a marble number are integers [ganze Zahl], so I thought that the sentence “…both players’ expected final score was a whole number” is redundant. I was puzzled about this sentence and wondered why Mark Valentine included it.
@Peter: The actual scores will be whole numbers, but the expected score (a sort of average of all possible outcomes) is not necessarily an integer.
In fact very unlikely to be an integer.
Assume at some stage Oliver would get a final score O and Pam would subsequently get a final score P. An average (O+P)/2 for “final score” to be a whole number would imply that each subsequent turn involves an even increment which is impossible.
So I still think that the final score is meant to be O + P rather than an average.
In my calculation Oliver never scores zero.
I used all available zero – scores for Pam’s minimal total, as well as all available scores 2, 3, 4, 5. Also two scores 6 out of three available.
Oliver and Pam know of course their initial scores from the turns 1,2,3,4 but nothing else. So they think they have equal opportunities.
I know what is a whole number. I was just wondering if Mark Valentine’s “whole number” might be a special one such as a multiple of 10.
In solving this, I assumed that a “most square” integer factorisation is one where the numbers are closest together (i.e., nearest to the square root) so minimising their difference.
Example: 70 = 7×10 (diff. = 3) rather than 5×14 (diff. = 9)
I made a list of all 84 points of nearest square factorisations. It gives a 4-digit total sum T of all scores.
Pamela’s lowest possible game total P is easily extracted. So Oliver’s total is O = T – P.
Unfortunately this does not answer the question [which, I believe, is meant to be Oliver’s score for his initial 3 turns ?].
The tedious part of this teaser is to accurately determine the total number of points. After that there are only three possibilities for the number of points after Oliver’s first turn. And so the solution space is very small.
I initially determined the points total by using a semi-automated spreadsheet. While looking for a simpler manually based solution, I found a synthesis method, by making lists of the factors for each of the possible lower factors. I then used a carefully constructed table of primes to reduce the computational load. It made a manual solution less prone to mistakes. It was not necessary to calculate the points score for each number.
A good teaser.
I had a couple of sticking points but got there in the end.
Firstly I didn’t read the question properly! So I thought I’d have too many answers but it does say “after EACH of the first four turns” (not just after the first four turns). When I read it properly it became easier.
Then I had 3 possible scores for Oliver’s first turn. Then 17 sets of scores for Oliver and Pam after one turn each. Then 2 sets of scores after Oliver then Pam then Oliver. Then after two turns each I had 2 answers, which was my second sticking point.
I then realised that I had to apply more carefully the part of the teaser that says a player’s second turn score is the same as the first. That reduced the problem to use just 2 possible scores for Oliver’s first turn and accordingly reduced the later possibilities.
So I found the unique answer.
Out of interest I calculated Oliver’s smallest possible final score for the case that the given constraints are observed and the expectations are ignored.
It is obtained when Oliver picks in his turns 1 and 3 the largest possible combination of 4 primes such that p1+p2 = p3+p4. One obtains S = T – P – (p1-1) – (p2-1) – (p3-1) – (p4 – 1).
T = sum of all 84 individual scores, P = Pam’s imposed minimum sum of scores.
As expected S is substantially larger than the value given by Brian in the hidden answer.