## Le Monde puzzle [#1085]

Given 13 arbitrary relative integers chosen by Bo, Abigail can select any subset of them to be drifted by plus or minus one by Bo, repeatedly until Abigail reaches the largest possible number N of multiples of 5. What is the minimal possible value of N under the assumption that Bo tries to minimise it?

I got stuck on that one, as building a recursive functiion led me nowhere: the potential for infinite loop (add one, subtract one, add one, …) rather than memory issues forced me into a finite horizon for the R function, which then did not return anything substantial in a manageable time. Over the week and the swimming sessions, I thought of simplifying the steps, like (a) work modulo 5, (b) bias moves towards 1 or 4, away from 2 and 3, by keeping only one entry in 2 and 3, and all but one at 1 and 4, but could only produce five 0’s upon a sequence of attempts… With the intuition that only 3 entries should remain in the end, which was comforted by Le Monde solution the week after.

### 2 Responses to “Le Monde puzzle [#1085]”

1. T3 Says:

I’m having trouble understanding why this difficult and/or how it is nuanced.
If Bo wants to minimize then he could choose the integers from 1 to 13.
Abigail wants to maximize the sum, so she chooses the sum of all of them which sums up to 91.
Bo “drifts” down by 1 to 90, and 90 = 18*5. So N is 18.
What am I missing?

• xi'an Says:

The description is unclear I am afraid: Bo picks x¹,….,x¹³ which are integers between 0 and 4 without loss of generality. On each round, Abigail picks a subset of the 13 digits and Bo either adds or substract one to all digits in the subset. Those that are multiple of 5 get removed and Abigail goes at the remaining digits, unless it gets impossible to reach further multiples of 5, for instance when (x¹,x²,x³)=(2,3,4) or (x¹,x²,x³)=(4,4,4).

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