How many steps on an escalator?

One of the joys (or burdens?) of being fascinated by mathematics is that you see interesting mathematics problems all around you in your everyday life!

I have recently acquired one of those little devices you attach to your belt that counts the number of steps you make in a day. The result of this is that you get obsessional about maximising your number of steps in every situation. There is even a temptation to mince along with very short steps rather than my usual manly stride, just to increase the step-count!

Today, I had to go up an escalator in a shopping centre. I never stand still on an escalator, unless it is totally blocked with other people. So, I found myself thinking about how many steps might I manage going up this escalator. It seemed intuitive to me that the faster I went the more steps I would get in before I reached the top. After all, if my speed was zero (i.e. standing still on one step) then my step-count for the escalator ride would also be zero.

Intuitively I imagined that there would be a linear relationship here. So, if I went twice as fast I would get in twice as many steps before I reached the top. But, wait a minute! That can't be true. Because I could never get more than the total number of visible steps on the escalator, however fast I went. In fact, to get in a number of steps equal to the number of visible steps on the escalator I would have to travel at an infinite speed!

This was getting so interesting, I forgot to do my shopping and just walked home puzzling it out.

It's a really nice mathematical problem. I managed to calculate, for example, that ...

• if there are 30 visible steps on the escalator (i.e. the number of steps you would see if the escalator were stationary), and
• the steps are disappearing at the top at the rate of 90 steps per minute, and
• I go up the escalator at the rate of 120 steps per minute (pretty fast),

then I will add 17 steps to my step-counter.

If I went twice as fast up this escalator, doing 240 steps per minute, I calculated that I would increase my step-count to 21 or possibly 22 steps. And at half the speed, i.e. 60 steps per minute, I would do 12 steps before reaching the top of the escalator.

I'll leave this as a challenge for readers ... can you produce a formula for the number of steps I add to my step-counter (m) if there are p visible steps; the steps are disappearing at the top at the rate of s steps per minute; and I go up at the rate of n steps per minute?

m  = ....?

And what would the graph look like, if you plotted m against n?

Solution in my next post.

Logic problem

Here's one my favourite problems requiring simple logic. Only the brightest children I have worked with in Year 6 have ever been able to get this right!

There are four cards on a table. Each card has a letter on one side and a number on the other. The cards are laid out so that we can see two of the letters and two of the numbers:

A B 8 5

Someone makes this assertion: 'If a card has a vowel on one side then it has an even number on the other side.'

Which cards must you turn over to find out whether this assertion is true or not?

Equality of the classes

Here's another little mathematical problem to give to upper primary school children. As with the previous problem, the actual calculations are simple. The challenge is working out what to do with all the information given here.

There are 60 children in a school, in two classes, A and B. The school moves 5 children from Class A to Class B. Then 3 children are moved from Class B to Class A. They then decide to make a new class, Class C. So, 12 children from Class A and some children from Class B are moved to Class C, so that there is now the same number of children in each of the three classes. How many children were there in Class A and Class B to begin with?

Solution to follow.

Token problem

Here's a little mathematical problem for Key Stage 2 children:

Two children, Mike and Nina, are collecting tokens for a free gift from a store. They both have some tokens, but Mike needs 49 more and Nina needs 2 more to get the gift. They decide to put their tokens together, but they still do not have enough to get the gift. How many tokens did they each have?

The arithmetic involved in this is really simple, yet many children in Key Stage 2 (and some adults!) struggle to organise the information given here.

How did you do? Answer and commentary in my next post.

Solution to last week's problem

What will be the first date from now which when written as XX/XX/XXXX (day/month/year) contains eight different digits?

That was the challenging little problem thrown at the students on University Challenge last week. It's not straightforward! Obviously (mathematicians use this word when we can't bothered to explain our thinking) we have to start by getting the earliest possible year, then the earliest month in that year and finally the earliest day in that month.

My immediate thought was that the earliest possible year, using four different digits, would be 2013. But 2013 will not work because the month must use either 0 or 1 for its leading digit, so we can't use both of these in the year. I then tried 2034. But the problem with that is that the leading digit of the day can be only 0, 1, 2 or 3, so I now realised that we need at least two of the digits 0, 1, 2 and 3 for the first digits in the day and month. So, I then went for the first possible year that uses only two of the digits, 0, 1, 2 and 3: 2045.

I was confident this would work! But, it doesn't. The only month that does not use 0 or 2 is November which repeats the digit 1! So, we can't have both 0 and 2 in the year.

OK, so surely the solution must be 2145? No, that doesn't work either! The month would have to use the 0. So, having now used 0, 1 and 2, the day would have to be 'thirty-something'. But the only possibilities are 30 or 31, which are not available because we have used the 0 and the 1.

Welcome to the year 2345! This works. We can use the earliest month in that year, June (06) and the earliest day in that month (17). So, the solution is 17 June, 2345. But how on earth were those University-Challenged students supposed to work this out on the spot?

Challenging maths problem

This was asked by Jeremy Paxman to the puzzled students on University Challenge last Monday! Unsurprisingly in the time available none of them could do it. Not could I, to be honest. But then I don't suppose Jeremy Paxman could either. Here's the question:

What will be the first date from now which when written in the form XX/XX/XXXX (day/month/year) contains eight different digits?

I'll give the solution and explanation next week. It's not actually that difficult, you just need more than 30 seconds to think about it.

Change from £5

Here's a simple real-life problem for upper primary school children. I had to solve it rapidly last week when we were in Rochester, Kent. Our drive back to Norwich would involve paying the toll of £1.50 at the Dartford crossing, for which having the exact amount is a distinct advantage. I had only a £5 note – no change whatsoever.

So, I went into a post office and used my £5 note to buy some first class stamps for letters, which I knew cost 41p each. The mathematical problem was this: how many stamps to buy to guarantee that the change would include the £1.50 I needed?

Mathematical Problem 4: Solution

This is the solution to the problem set on 9 April.

We observed that if a daughter is born in '66 and her mother in '44 then in the year that the mother turns 66, the daughter turns 44.

Was this an unusual coincidence? Of course not! It happens for any two people! It would be fun to try this as a little investigation with some primary school children.

For example, if A was born in 1957 and B was born in 1971, then in the year that A turns 71 B turns 57. Why? Because the age of A minus the age of B (once they have both had their birthdays) is always equal to the birth year of B minus the birth year of A. If A is born 14 years earlier than B then B is always 14 years younger than A. So when A's age equals B's birth year (ignoring the 1900) then B's age (14 less than this) must equal A's birth year.

Why had I never noticed that before?!

There is a complication if they are born in different centuries, – this requires the older person to pass 100 before the phenomenon occurs. For example, Catherine was born in 1966 and her son, Jack, was born in 2001. So, the year Jack turns 66, Catherine will turn 101!

Mathematical Problem 3: Solution

This is the solution to the problem set on 1 April 2010. This was about the metric/imperial curiosity that 10 feet 3 inches converts to 3 metres 10 centimetres, rounded down to the nearest 5 cm. I wondered if there were other lengths like this where you just swap over the two numbers to convert from feet and inches to metres and centimetres. For reference purposes 1 metre is about 3.28084 feet.

Well, the obvious one to try next is 20 feet 6 inches – which works, because it converts to 6 m 20 cm, when rounded down to the nearest 5 cm. But 30 feet 9 inches does not work, because of the rounding error; it actually rounds down to 9 m 35 cm.

It's not too difficult to check all the possibilities. Because of the instruction to round down to the nearest 5 cm, we need check only imperial lengths that start 0 feet, 5 feet, 10 feet, 15 feet, and so on. Then because the number of inches cannot exceed 11, the number of metres cannot exceed 11. So we do not need to check lengths beyond 11 metres 95 cm, which is under 40 feet.

So, I've checked all the possibilities and all I can offer you are these two results:

10 ft 3 ins = 3 m 10 cm and 20 ft 6 ins = 6 m 20 cm

(both metric lengths rounded down to the nearest 5 cm).

Mathematical Problem 4: ages and birth years

It's my wife's birthday on Wednesday. Christina will be 66. In October, our daughter, Catherine, will be 44. Now, Christina was born in 1944 and Catherine was born in 1966.

So, daughter born in '66, mother in '44. And in the year that mother turns 66, daughter turns 44!

That's neat, isn't it! "What are the chances of that happening?" Mrs Haylock asked her mathematical husband in the car yesterday.

What do you think? Is it a rare coincidence?

Mathematical Problem 1: Solution

Here's a solution to Mathematical Problem 1, posted on 13 March 2010.

With 2 players, A wins a match of 29 rounds when A has won 15 rounds. Even if A loses all the remaining rounds B will only have won 14.

The principle is that A would still win even if all the remaining games were to be won by B (A's closest rival).

With 3 players, A wins when a is greater than b added to the number of rounds still to be played (assuming B has the second highest number of wins so far). The number of rounds still to be played is 29 – (a + b + c). So, algebraically:

a > b + [29 – (a + b + c)] which simplifies to 2a > 29 – c.

With 4 players, A wins when

a > b + [29 – (a + b + c + d)] which simplifies to 2a > 29 – (c + d).

The general solution , with any number of players and n rounds, is that A wins when

a > b + [n – (a + b + c + d + e + ...)] which simplifies to 2a > n – (c + d + e + ...)

The surprise in this result is that b does not appear in the simplified rule! Provided B has won more rounds than C, D, E and so on, the actual number that B has won does not make any difference to when A is declared the winner.