Multiple Surprises

Thank you to everyone who submitted a solution for this problem. There were just too many to mention everyone, but a lot of good work was submitted, so well done all!

Many of you started off with a 'trial and error' approach and then looked for patterns in what you had found:

If I write the first few sequences I found in columns like this:

$2     3     4\\

14    15    16\\

26    27    28\\

38    39    40$

I can see that each number is $12$ more than the number above, and so adding $12$ to each sequence will work.

Some people noticed that adding $10$ to the last number in one sequence and starting the next sequence from there would work, which is also true since this is effectively the same as adding $12$ to each number in the sequence.

Some people then looked into why adding $12$ seems to work:

I noticed that $12$ is a multiple of $2$, $3$, and $4$ and, more than that, $12$ is in fact the lowest common multiple of $2$, $3$, and $4$.

That is why adding $12$ works, because adding a multiple of $2$ to $2$ gives a multiple of $2$ and adding a multiple of $3$ to $3$ gives a multiple of $3$ and the same for $4$. So, if we start from the obvious consecutive sequence, $2, 3, 4$ and keep adding $12$ we will get more and more consecutive sequences which fit the criteria.

Hannah also looked for lots of sequences but then tried a slightly different approach:

I found the sequences:

$2, 3, 4\\

14, 15, 16\\

26, 27, 28\\

38, 39, 40\\

50, 51, 52$

In order to generate one of these sequences, I had to find the nth term rule. Because there are three terms per sequence, I decided to add the terms together to give me one number that would represent that sequence.

$2+3+4=9,\\

14+15+16=45,\\

26+27+28=81,\\

38+39+40=117,\\$

and $50+51+52=153.$

So the new terms of the sequence are: $9, 45, 81, 117, 153$.

The difference between each of these numbers is $36$ so my expression for the new sequences is:

$U = 36d + a$

And since the first term in the new sequence is $9$, that means $a=9$ and so:

$U = 9 + 36d$

I checked that my expression for the sequence worked and it did!

Then I tried to work out how to get from my new sequence back to the original sequences of three consecutive numbers.

I noticed that if I did $\frac{U}{3}$ I would get the middle term from my original sequence of three consecutive numbers. I think this is because the middle term in three consecutive numbers is also the average of the three terms.

Hannah is correct! We can show this algebraically:

Let's call the middle term, $p$.

Now, since the three terms in our sequence are consecutive we can write them as $p-1, p, p+1$.

So when we add the three terms together to get the number Hannah has called $U$, we are adding $ (p-1) + p + (p+1) = 3p$.

This is why Hannah can use the $U$ sequence to find new sequences of three consecutive numbers which satisfy the criteria in the question.

For the second part of the question, looking for three consecutive numbers which are multiples of $3$, $4$ and $5$ respectively, most people continued to use the idea of lowest common multiples.

For the numbers $3$, $4$ and $5$, I did the column strategy again, and figured out how the same thing happens, but with the number 60, as that is the Lowest Common Multiple of $3$, $4$ and $5$.

$3,   4,   5,\\

63,  64,  65,\\

123, 124, 125,\\

183, 184, 185,\\$

The lowest common multiple of $4$, $5$ and $6$ is also $60$ and so the columns for those numbers look like this:

$4, 5, 6,\\

64, 65, 66,\\

124, 125, 126,\\$

and so on.

We can prove algebraically why the LCM method works:

Starting from our three consecutive numbers, let's say $a, b, c$ we want to find more sequences of three consecutive numbers which are multiples of $a$, $b$, and $c$ respectively. This means we want

$a + m = $ a multiple of $a$

and $b + m = $ a multiple of $b$

and $c + m = $ a multiple of $c$

Therefore $m$ must be a multiple of $a$, $b$ and $c$ and to make sure we don't miss any possible sequences $m$ must be as small as possible. This proves that $m$ must be the LCM of $a$, $b$ and $c$.

As a lot of you noticed, regarding the final part of this question:

The LCM method will work for $4$, or $5$ or any number of consecutive numbers.

For example Rianna from Dr Challoner's High School for Girls looked at sequences of seven consecutive numbers:

A slightly more challenging one is using multiples of seven consecutive numbers such as $2,3,4,5,6,7,8$. The lowest common multiple of these is $840$.

$2,3,4,5,6,7,8$

$842,843,844,845,846,847,848$

$1682,1683,1684,1685,1686,1687,1688$

And Malory from DCHS looked at sequences of nine consecutive numbers

I then wanted to challenge myself further! I used $2, 3, 4, 5, 6, 7, 8, 9$ and $10$! I tried to find nine consecutive numbers and I found out that the lowest common multiple is $2520$. I added $2520$ to the nine consecutive numbers and it made a set of consecutive multiples.

Well done again to everyone who had a go and submitted solutions!