I’ve seen multiple Pascal’s Triangle related puzzles that look something like this. You have a 6 by 6 square grid and a counter. The counter can start in either of the ‘off-grid’ positions indicated below, and with each move it can either go vertically down any number of spaces (up to 6), or horizontally right any number of spaces (up to 6). How many different ways are there that the counter could travel to the bottom right hand corner?

If we start in the top left corner and record how many paths there are to each square (in the squares themselves) we get something that looks familiar:

Surprise, surprise…it’s Pascal’s triangle!

And we can see why: to get to any square we can either come from above, or from the left. So, the number of ways of getting to that square is the number of ways of getting to the square above it plus the number of ways of getting to the square to its left. When Pascal’s Triangle is in its usual orientation the two cells being summed are those above the cell we are trying to find.

The value in the bottom right corner can be found by continuing this pattern. But we could solve this problem using another commonly referenced property of Pascal’s Triangle – its link to ** ^{n}C_{r}**. There are

Recently I was shown a similar ‘how many ways?’ puzzle that drew my attention to a less commonly referenced Pascal’s pattern. It excited me!

Imagine you are standing at one end of a footpath made of ten paving stones. You are standing before the first paving stone (the 0^{th} stone if you like) and your aim is to end up on the tenth and final stone. You can make steps of different sizes. You could make the journey by stepping one stone at a time, or by stepping straight to the tenth stone in one step, or by a combination of different integer-sized steps. How many different ways are there of getting to the tenth stone?

We could start by considering how many ways there are of getting to the earlier stones. There is clearly just one way of getting to the first stone, and below I have written out all the different ways of getting to second, third and fourth stone.

Each row represents a journey, and the numbers in the row represent the stones that are stepped on in that journey. For example, the final journey in each list represents stepping on all the stones.

We can see that these totals are powers of two (there are **2 ^{n-1}** ways for

We can use this formula to find how many ways there are of getting to the tenth stone and solve the puzzle. But this is not the only interesting pattern we can find in our results. Let’s consider where these different ‘ways’ are coming from. The table below breaks down how many of these journeys took two steps, or three steps, etc.

And surprise, surprise…it’s Pascal’s Triangle!

This also explains the powers of two – the sum of the cells in each row of Pascal’s Triangle add up to successive powers of two. But why does this problem result in Pascal’s Triangle?

Does it make sense that summing two cells will give the result in the cell below them (as in the diagram below)? Unlike the previous puzzle, it did not seem immediately* obvious to me why this would be the case. It was a different connection between different cases in the problem that helped me understand why this puzzle resulted in Pascal’s Triangle.

Let’s consider how to get to the fifth stone in three steps. Our first two steps must get us to either the fourth, the third or the second stone. Thus, if we sum the number of ways of getting to each of these stones in two steps (the cells that are circled in the diagram below) then we have all the possible ways of getting to the fifth stone in three steps (the cell with the triangle).

This is the Hockey Stick identity! It looks more like a hockey stick when Pascal’s Triangle is in its usual presentation (below).

The Hockey Stick identity tells us that to find the value of any cell we can sum all the cells in the diagonal before it, up to the cell in the row before it.

We can show that if the Hockey Stick identity is true then the ‘two above sum to the one below’ rule must also hold. In the diagram below we know the first two circled cells add to give the square cell (by the Hockey Stick identity). Therefore, summing all three circled cells is the same as adding the two cells above the triangular cell (the square cell and the bottom circled cell). And this works for any length or position of hockey stick! It was pleasing to think about it this way around.

As with all good puzzles there are multiple ways of solving it. There is also an explanation for the appearance of Pascal’s Triangle from combinatorics.

Let’s imagine we are trying to find the number of ways of getting to the fifth stone in three steps. Our first two steps must have been taken over the first four stones. We landed on exactly two stones between stone 1 and 4. And there are ^{4}C_{2 }ways of choosing which two of the four stones to step on, so there are ^{4}C_{2 }ways of getting to the fifth stone in three steps. Generally, there are ** ^{p-1}C_{s-1}** ways of getting to the

*It wasn’t until later on, having found these two other ‘explanations’ for why we end up with Pascal’s Triangle, that an explanation for the ‘two cells above sum to the one below’ pattern dawned on me. I’m glad I didn’t spot it immediately as I may not have had the fun of unexpectedly stumbling upon the Hockey Stick pattern.

Interestingly, it didn’t become clear to me why this is the case until the problem was re-framed. Imagine instead you are a frog on one side of a pond, and you are trying to reach the other side, by stepping on some, all or none of the **n** lily pads on the way. (Having **n** lily pads is equivalent to having **n+1** stones in the problem above.) And if we want to increase the number of lily pads (to make the 3-lily pad case become the 4-lily pad case, for example) we imagine adding another lily pad in between the two banks.

The difference is that in this representation the nth thing (lily pad) that is being added is not also the destination (the bank). In the ‘stone’ version I was imagining that the nth stone that is being added then becomes the destination. My explanation for the ‘two above add to the one below’ pattern hangs on that idea that the additional lily pad or stone does not have to be stood on. It’s easier to come to this idea if we imagine that the additional thing is not the destination; by its nature the destination has to be stood on.

Let’s imagine we want to find the number of ways of crossing the four lily pad pond in three jumps. The fourth lily pad that has just been added in can be jumped on to, or not. The number of ways of getting to the other side in three jumps if the fourth lily pad is NOT used, is equal to the number of ways of crossing the three lily pad pond in three jumps (we’ve just made the final jump a bit longer). And the number of ways of getting to the other side if the fourth lily pad is used is equal to the number of ways of crossing the three lily pad pond in two jumps (the second jump becomes the second and third jump when it is broken by the frog landing on the fourth lily pad). These two values that we need to sum are the two cells above the one we are looking for in Pascal’s Triangle!

And there we have it. Sometimes the framing of the puzzle allows us to more easily see particular solutions. For me the ‘stone’ representation of this puzzle lent itself to the hockey stick pattern. Whereas the ‘frog’ representation allowed me to see why the ‘two above sum to the one below’ pattern applied.

Much of the fun of maths comes from finding different ways to reach the same conclusion. And puzzles like these are full of opportunities for that!

]]>What are cycling numbers? Try this trick about the number 37:

• Take a three-digit multiple of 37 (for example 481). It will be more interesting if all the digits of your multiple are different.

• ‘Cycle’ it (re-write it so one of the other digits is first but so the digits are still in the same order). For the example above you could create 814 or 148.

• Now divide your new number by 37, and you will find you get an integer.

The number you created by ‘cycling’ your multiple of 37, is also a multiple of 37!

Wow! Why does this work? Can you prove why this works for all three-digit multiples of 37? (Spoilers below!)

A while ago a friend told me about this curious property of three-digit multiples of 37 and I started to play around with it, trying to work out why this happens. I found an algebraic proof, which led me to find other numbers whose multiples have this property too.

I am defining cycling numbers to be any numbers in this family; numbers for whom all of their multiples of a specific length (in the case of 37 the multiples have to be three-digits) can be ‘cycled’ any number of places and give a result that is still a multiple of that cycling number.

There are an infinite number of cycling numbers! Every prime number apart from 2 and 5 is a cycling number. But I have a personal favourite: 41. Make any five-digit multiple of 41 and ‘cycle’ it any number of places. You will find that the number you have created is also a multiple of 41.

My proof that all three-digit multiples of 37 can be ‘cycled’ is below.

We can express a three-digit number as:**100a + 10b + c** (where a, b and c are any positive integers)

As our number is a multiple of 37 the following is true:**100a + 10b + c = 37m** (m is some positive integer)

When we cycle the number just one place, we get the number:**100b + 10c + a**

Below, I have expressed the new number as the original number plus some difference:**100b + 10c + a = 100a + 10b + c + [−99a +90b + 9c]****new number = original number + [−99a +90b + 9c]****new number = 37m + [−99a +90b + 9c]**

If (and only if) the difference ‘-99a + 90b + 9c’ is a multiple of 37, then the new number will also be a multiple of 37.

We can show this is the case by writing the difference in terms of the original number:**−99a +90b + 9c = (900a + 90b + 9c) − 999a****= 9(100a + 10b + c) − 999a****= 9(37m) − 37(27a)= 37(9m − 27a)**

The difference is a multiple of 37, and therefore the new number is a multiple of 37.

My proof imagines the number is being ‘cycled’ only one place. But ‘cycling’ one place repeatedly will also result in a multiple of 37. And ‘cycling’ one place repeatedly is the same as ‘cycling’ multiple places in one go, so our proof applies to ‘cycling’ any number of places.

37 is not the only cycling number for three-digit multiples! In the proof we were able to show 37 is a factor of the difference because it is a factor of 999 (and because we made our original number a multiple of 37). We could show that any other factor of 999 is a cycling number if we start instead with an original number that is a multiple of that number. All factors of 999 (apart from 1!) are cycling numbers.

Writing 999 as a product of its primes gives:

We can see that the prime cycling numbers (for three-digit multiples) are 3 and 37. This prime decomposition dictates which composite numbers will be cycling numbers.

The discovery that 3 and 9 are cycling numbers is not astonishing; we would expect 3 and 9 to be cycling numbers given their digit sum property. Multiples of 3 or 9 have digit sums that are themselves multiples of 3 or 9, respectively. This property means we can do more impressive things than the cycling trick; we could write the digits of a multiple of 3 or 9 in any permutation at all (without the restriction that they stay in the same order) and we would end up with a multiple of 3 or 9 still, because the digit sum would remain unchanged.

Thus I declare 37 the most interesting cycling number for three-digit multiples! But it doesn’t stop there! What about four-digit multiples? Or n-digit multiples? The same proof can be applied to numbers with higher numbers of digits. The difference between the new number and the original number will have the same structure as above, but with additional terms.

For example, for a four digit number the difference can be expressed as:

where

And this time any factors of 9,999 are cycling numbers. As we would expect, we can see that 3 and 9 are cycling numbers for any number of digits, as a string of 9s of any length with always have 9 as a factor.

We can find which of the primes (other than 3) cycle for four digit numbers by just looking for the prime factors of 1,111.

We can re-write the difference as:

And for five-digits we’d be looking for prime factors of 11,111, and so on.

I challenge you to investigate cycling numbers further and make your own cycling number trick.

*According to me: if anyone else has a different name for the numbers this article is about, I’d like to know! (Not to be confused with cyclic numbers – which are connected, and arguably more interesting.)