Puzzle of the Week #196 - Out and Back

Bakewell parkrun is a picturesque free, weekly, timed 5km run along the Monsal trail in Derbyshire. The course heads off 2.5km along the trail, then takes a u-turn, before heading back to the start/finish.

Two friends run the parkrun, both starting at the same time (Saturday 9am). They run at different paces, but their paces are consistent. Nicola, the faster of the two, reaches the turnaround point first, then exactly two minutes and twenty seconds later, passes Danny travelling the other way. At the finish, Nicola has exactly twelve minutes to wait before Danny crosses the finish line.

How long do Nicola and Danny take to complete their parkruns?

Puzzle of the Week #195 - Personal Bests

Runners will know, if they want to predict what time they should be aiming for in, say, a 10km race, based on what they can do at 5km, you should do more than merely doubling the time: the longer a race, generally the slower you need to go. One useful rule is known as the 6% rule. It isn’t unfortunately as simple as merely doubling the 5km time and adding 6% - instead it is raising the ratio of distances to a power of 1.06.

So (10/5)^1.06, would give you a factor of around 2.085. Multiply this by your typical 5km time and you should be looking at that for your 10km time.

Of course, this is only a rule of thumb, so will be more true for some people than for others. For one athlete, let’s call him Norm, he finds that his personal best times, at a variety of distances, precisely follow the rule (to the nearest second). His half-marathon best time is exactly 1 hour more than his 10km best time. (A half-marathon in metres is 21097.5).

What are his 10km and half-marathon personal bests?

Puzzle of the Week #194 - Double Birthdays

This conversation took place on a particular day of the (non-leap) year:


“Isn’t it curious,” remarked Izzie, “that if you take today’s date, double the date number and double the month number, you get my birthday?”

“Interesting!” replied Leila, “but if you consider today as the ‘n’th day of the year and work out what the ‘2n’th day of the year is, you get MY birthday!”


Izzie’s and Leila’s birthdays are in the same calendar month.


When did this conversation take place and when are the girls’ birthdays?


Puzzle of the Week #191 - One in Five Chance

In a betting game there are five cards placed face down. Each has a number on it.

You win the bet if you end up with the card with the highest number.

You can select any of the cards at random, and having seen the number you can decide whether to accept the card or reject it and choose another, and you may do this as many times as you like, however once you have rejected a card you cannot get it back.

The actual values on the card give you no indication of whether you might have the highest, for instance ‘1’ may be the highest number, or ’50,000,000’ may be the lowest.

Since there are five cards you might think a fair price for the bet would be 4 to 1 against (you place £1 against the house’s £4 and if you win you take all £5). Instead you are offered a measly 3 to 2 against (you place £2 against the house’s £3 and if you win you take all £5).

Is it possible to beat the house? If so how?

Number of Increases within a Random Permutation, and an Unexpected Link to a Familiar Series – Elliott Line

Consider a deck of n numbered cards. You play the following game: you turn over the first card and retain it. Then you turn the next one: if it ranks less than the first one, discard it; if it ranks greater then keep it and make it the new benchmark. Repeat this step for each subsequent card. At the end of the deck, count the number of cards you have retained. For a given ‘n’, what is the expected number of cards? If there were 11 cards? If there were 12,367 cards?

We’ll start with a few small values of n: trivially when n=0, the number of cards retained, x, will also be 0. For n=1, x will also be 1. For n=2, the two cards will either be in descending or ascending order, so x could be either 1 or 2, with equal probability, so X (big x, the average value of x) will be 1.5.

Since no cards will be retained after the maximum card has been seen, it is useful to consider where in the deck the maximum card might be and what the consequences are:

Consider the case with three cards. If the 3 card comes up first, which it will do in 1/3 of cases, x will be 1. If the 3 card comes up second, x will be 2. If the 3 card comes up last, x may be 2 or 3, depending on the order of the 1 and 2 cards.

In other words, when the maximum 3 card is first, there are 0 cards before it so the average number of cards retained before reaching the maximum card is 0, so the count will be 1 more than that, 1.

If the maximum card is second, there is 1 card before it so the average number of cards retained before reaching the maximum card is 1, so the count will be 1 more than that, 2.

If the maximum card is third, there are 2 cards before it so the average number of cards retained before reaching the maximum card is 1.5, so the count will be 1 more than that, 2.5.

Those three cases are equally likely, so we can express X3 in terms of previous X values:

X3 = ((X0) + 1 + (X1) +1 + (X2) +1)/3

We can generalise this so that any Xn can be expressed as an equation involving previous X values by extending the idea of considering where the maximum card is and how many cards are retained prior to reaching it:

Xn = ((X0) + (X1) + … + (Xn-1) + n)/n

Now a very interesting thing happens if you compare the equation for Xn with that of Xn-1

Xn-1 = ((X0) + (X1) + … + (Xn-2) + n-1)/(n-1)

In the first equation you multiply both sides by n, and in the second equation by (n-1) to get the following two equations:

n(Xn) = (X0) + (X1) + … + (Xn-1) + n

(n-1)(Xn-1) = (X0) + (X1) + … + (Xn-2) + n-1

Now if you take the difference between those two equations almost all the terms cancel out:

n(Xn) - (n-1)(Xn-1) = (Xn-1) + 1

Which then simplifies to:

(Xn) – (Xn-1) = 1/n

This is great as it means that since X0 = 0, the sequence of Xn is merely the partial sum of the harmonic series:

1 + 1/2 + 1/3 + 1/4 +1/5 …

… a series about which we know a great deal.

We know for instance that the sequence does not converge but tends to infinity. This rings true when we consider the numbered cards, as we would expect Xn to keep creeping up as the number of cards increased and it would seem very odd for it to converge to an upper bound.

What is surprising I feel, when considering it in terms of the cards, is just how slowly Xn does increase.

It seems quite reasonable that Xn first exceeds 3 when n is 11, that is when there are 11 numbered cards, the average number of cards retained is around 3.

However, when we calculate when the expected number of retained cards first exceeds 10, the number of cards would be 12367 (!).

Puzzle of the Week #189 - Three Palindromes

I recently stumbled upon an intriguing theorem that stated that every positive whole number can be expressed as the sum of three palindromic numbers, although it is often far from straightforward to find them.

Can you express the number 6878592 as the sum of three palindromes?

It is the sum of three numbers of 7, 6 and 5 digits respectively, and no digit is repeated except where being a palindrome dictates it:


Puzzle of the Week #188 - Triple Indivisibility

We have a whole number, let’s call it ‘b’.


We also have a number of statements:

1 divides exactly into ‘b’

2 divides exactly into ‘b’

3 divides exactly into ‘b’

4 divides exactly into ‘b’

… continuing all the way to …

 ‘a’ divides exactly into ‘b’


We are told that three consecutive statements in the list are NOT true, while all of the others are true.


Question a: what is the MAXIMUM value that ‘a’ could possibly be?

Question b: given that value for ‘a’, what is the MINIMUM value that ‘b’ could possibly be?


Puzzle of the Week #186 - Prime of my Life

I celebrated my birthday this week. Right now, my age, and those of my wife, my daughter and my son, are all prime numbers. After my daughter celebrates her next birthday in a few months we will not all be prime ages until 30 years from now. And after that (if we are all still around) it would be another 30 years.

My daughter is about four and a half years older than my son, and I am about one and a half years older than my wife. How old are we all?


Puzzle of the Week #185 - How Long is a Piece of String?

I have a loop of string N centimetres long. I can form this loop into a square with n cm along each side (n being exactly a quarter of N). Not surprisingly I find that if I draw diagonals from opposite corners of this square, they cross at right angles. I also find that I can tilt the square into a rhombus shape, while still having n cm along each side, the diagonals still cross at right angles. Still not surprising.

However, I discover that if I move some of the corners so that instead of all being n cm long, the four edges of the quadrilateral are n, n-5, n+1 and n+4, then the diagonals STILL cross at right angles.

How long in my loop of string?

piece of string.JPG

Puzzle of the Week #182 - The Rickety Carousel

There is a rickety old carousel, with 24 horses equally spaced around the edge of a disc. There are at least two things wrong with it: one is that the horses in positions A, B, C, D, E, F and G are broken and not safe to ride; another is that the carousel needs to be perfectly balanced around the centre of the disc, otherwise it won’t go round. This means that when there are two children wishing to ride, they would have to be positioned in diametrically opposite positions, eg, A and M, in order that the centre of gravity is at the centre of the disc. If there were three children they could be at positions A, I and Q.

On this particular day, nine children wish to ride the carousel. Where do they need to be positioned such that the centre of gravity is at the centre of the disc, and that positions A to G are unoccupied?