We had many correct answers to this problem, including some well argued solutions from Thomas Wong (Reed High School, Sparks, Nevada, USA) and Stelios Serghiou (G.C.School of Careers).

Peter Simpson from Castell Alun School argued as follows:

The number must be odd because the holiday makers split into
two group, with the car group one less than the walkers.

If N was 7 (or greater) the time delays would be greater than 59 minutes, the time for the car to travel to B (the walkers arrive in 60 minutes - N miles in an hour, the car arrives 1 minute earlier).

If N was 3 the distance would be 22km, but this would mean the holiday makers would split into 2 walkers and 1 car traveller, and as the problem states "they" for the car travellers the number of people in the car must be greater than 1.Therefore N is 5, and the distance by road is 20km.

Momtchil Iliev from Drayton Manor School argued in a very similar way and added an explanation for how he calculated the distance by road:

Calculating Distance of road travelled by carTime taken driving = 59 - (5)² - 2(5)

= 59 - 25 - 10 = 24 minutes = 0.4 hours

Speed of travelling by car = 10N = 10(5) = 50 km/h

Distance of road = Speed of car x time taken travelling

= 50 x 0.4 = 20km

Harry also sent in his work on this problem:

Because each group has at least 2 people in it, and there is one more walker than person in the car, there must be at least 5 people, so $N\geq 5$.

The walkers walk $N$ km at $N$ km/hr, so take 1 hour, which is 60 minutes.

The people in the car drive $d$ km at $10N km/hr$, which takes $\frac{60d}{10N}$ minutes, and stop for $N+N^2+N$ minutes. This takes 60-1=59 minutes. So $\frac{60d}{10N}+N+N^2+N=59$, so $\frac{6d}{N}+2N+N^2=59$, so $d=\frac{59N-2N^2 -N^3}{6}$. If $N=6$, this gives 11/6, which is less than $N$, so that's not allowed, and since it's just going to get smaller and smaller, we must have $N< 6$. But $N\geq 5$, so $N=5$. So $d=20km$.

Well done to you all