Le Monde puzzle [#940]

A rather different Le Monde mathematical puzzle with no simulation:

A student has x days to train before an exam and decides to take one day off after every nine consecutive days of training. She makes her planning and manages to fit one chapter of her textbook per day of training. She then realises she has forgotten one extra chapter, but cannot keep the pattern, even when taking one day off after 10 consecutive days of training. What is the maximum value of x?

If y denotes the number of chapters, this means that

10⌊(y-1)/9⌋+(y-1-9⌊(y-1)/9⌋)⁻ ≤ x < 11⌊y/10⌋+(y-10⌊y/10⌋)⁻

with  ⌊y/10⌋ denoting the integer part and with (z)⁻ equal to z when z>0 and to -1 when z=0. Defining both functions of y as

```curv1 <- function(y){
10*trunc((y-1)/9)+(y-1-9*trunc((y-1)/9))-(y-1==9*trunc((y-1)/9))}
curv2 <- function(y){
trunc(y/10)*11+(y-10*trunc(y/10))-(y==10*trunc(y/10))}
```

I get the following graph:

which sees both curves meeting at y=20 or x=21, which could be the answer to the puzzle. However, larger values of x see the above inequality satisfied as well. It is only after y=73 or x=80 that the above inequality is forever impossible since the lower bound is then larger than or equal to the upper bound [when removing the -1 from (z)⁻] or for y=81 or x=88 [when keeping the initial definition of (z)⁻].

One Response to “Le Monde puzzle [#940]”

1. don’t know why but (for me) it resembles the bound problem from Lagarias paper on RH. A bit.