Multiple surprises

Thank you to everyone who submitted a solution for this problem. There were too many good solutions to mention everyone, but well done to all! 
 
The submitted work involved several approaches, including work with numbers, algebra and using Python to find solutions using a computer. Many solutions used the suggestions of working on smaller, simpler versions of the problem, while some others worked directly on the challenging problem of finding ten consecutive numbers. You can scroll to near the bottom of this page to see the more direct approach. 

Many of you started off with a 'trial and error' approach for finding small examples and then started to work systematically, looking for patterns in what you had found. Barathraam and Nithilan, from The GYM Foundation in India, both found these sets of three numbers where the first is a multiple of 1, the second is a multiple of 2 and the third is a multiple of 3 and noticed that the first number of each set goes up by 6 each time. 

$1, 2, 3$

$7, 8, 9$ 

$13, 14, 15$

Moving on to the problem of finding sets of three consecutive numbers where the first is a multiple of 2, the second is a multiple of 3 and the third is a multiple of 4, many of you noticed that there was also a common difference between these sets of numbers. Here's some of Pinyi's work from Doha College in Qatar:

The first three consecutive numbers would be 2, 3 and 4. Then, I found the second three consecutive numbers, which were 14, 15 and 16. After, I compared those two groups of consecutive numbers.

$2,3,4$

$14,15,16$

$14-2=12$

$15-3=12$

$16-4=12$

Difference=12

I ended up with the result of the two groups’ difference was 12. That got me thinking: could there be a pattern of adding 12 each time? 

Pinyi's next two sets of numbers seemed to confirm this:

$14, 15, 16$
$26, 27, 28$
 

Pinyi's idea is spot on! Some people then looked into why adding $12$ seems to work. As Zac from Twyford School in UK commented, the least common multiple (L.C.M.) was key to solving the problem, because

by adding the L.C.M to each term in the sequence, you are adding something that is divisible by all the required numbers.

It was good to see several solutions explaining this clearly, often with examples, such as this explanation from Ananya, who was one of several students from Ganit Kreeda Vichar Vatika to illustrate this point clearly:

You need to add a multiple of 3 to a multiple of 3 to get a sum that is a multiple of 3.

Can you see how to extend Ananya's argument to explain why the L.C.M is a key idea for solving this problem? Akhil, from Cannon Lane Primary School in UK, explained it this way:

Find the L.C.M. of the three (or any other number) consecutive numbers whose multiples we are looking for. In the above scenario, the L.C.M. of 2, 3, 4 will be 12.

Now, if we add L.C.M. of these three numbers to any of the numbers, the result must be a multiple of that number and hence we can add the L.C.M. to the three numbers to get three consecutive numbers that are multiples of 2, 3 and 4 respectively.

A further crucial point was made by Avic, also from Ganit Kreeda Vichar Vatika. This is that 

If you add the same number (in this case the L.C.M) to a set of consecutive numbers you will get another set of consecutive numbers.

This is crucial, but you may have taken it for granted. Sometimes when making arguments in mathematics it can feel as if you're being asked to state the obvious,. However, pinning down these details is important for checking your argument and for communicating it clearly to others.

Reyansh from Garden International School in Malaysia, Ruhi from Ganit Kreeda Vichar Vatikar in India, Jack from Beths Grammar School, Ishan from Sir Thomas Rich's School and Thomas from Twyford School in the UK all used algebra in their explanations as they started to generalise. Here is Reyansh's argument.

Some of the consecutive numbers can be: $2, 3, 4$, $14, 15, 16$, $26, 27, 28$, and so on.

Let $N$ be a multiple of 2, $N+1$  a multiple of 3, and $N+2$ a multiple of 4.

After solving this it will show that $N$ is in the form of $12k+2$. Since $12k$ is a multiple of $(2, 3 \text{and} 4)$ these expressions will always satisfy the divisibility rules for any integer of $k$. The first example $(2, 3, 4)$ is for $k=0$ the second $(14, 15, 16)$ is $k=1$ and so on. Also the LCM (lowest common multiple) of $2, 3, 4$ is $12$ which is why the pattern goes up in $12$.

We think it's particularly interesting to see that Reyansh highlighted that their example of $2, 3, 4$ is a particular case of their general form. 

For the later simpler problems, such as looking for three consecutive numbers which are multiples of $3$, $4$ and $5$, or two consecutive numbers that are multiples of $9$ and $10$, most people continued to use the idea of lowest common multiples. 

A slightly different way to explore and represent these sets was sent in by Argentum from India, who represented these sets on a grid. This helped Argentum to see the patterns for sets of three numbers. (Click on the image to see a larger version.)

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Solving the main problem: Finding ten consecutive numbers that meet the criteria

Having worked on smaller, simpler examples to make sense of some of the underlying ideas in a problem, mathematicians will often then think about how to extend these ideas, working towards the original problem. Many submitted solutions went straight from a few simpler examples to the problem of ten consecutive numbers, using the underlying reasoning explained here by Abi, Sophie and Ella from Surbiton High School in England:

First you must identify the prime factors of each consecutive numbers given. Then you multiply all the prime factors together to find a common multiple. Once the lowest common multiple is found, you add the lowest common multiple to the consecutive numbers given. To get another set of consecutive numbers, you multiply the lowest common multiple by two and add this to the original consecutive numbers shown. Carry on multiplying the lowest common multiple by 3,4,5 etc to continue with the sequence.

So for the particular case of finding the set of ten consecutive numbers, here is the method suggested by Dan and Amtih from Garden International School in Malaysia:

First of all, to solve a problem like this, we can say that the first set of numbers that satisfy the question is of course 1, 2, 3, 4… 10. However, if we want to find the next sequence of numbers that satisfy the question, we must find the LCM, or the number that is divisible by all of the numbers, of 1, 2, 3, 4…10.

To show this in practice to find the set of ten consecutive numbers, here are the calculations that Avic sent in from Universal Wisdom School in India: 

The LCM of $1,2,3,4,5,6,7,8,9,10$ is

$2x3x2x5x7x2x3 = 2520$

So, the sets will be:

$1, 2, 3, 4, 5, 6,7, 8, 9, 10$

$2521, 2522, 2523, 2524, 2525, 2526, 2527, 2528, 2529, 2530$

$5041, 5042, 5043, 5044, 5045, 5046, 5047, 5048, 5049, 5050$

$7561, 7562, 7563, 7564, 7565, 7566, 7567, 7568, 7569, 7570$

You may have noticed or wondered about the method Avic used to calculate L.C.M. of $1,2,3,4,5,6,7,8,9,10$. There were several different methods of calculating this in the submitted solutions, often involving prime factorisation. Nithilan from Tampereen Klassillinen Lukio in Finland had previously found a set of eight consecutive numbers where the first is divisible by 1, second divisible by 2, and so on, and a set of two numbers where the first is divisible by 9 and the second is divisible by 10. Nithilan made the interesting observation that they could use their previous calculations:

To combine a working set of 1-8 and 9-10, we need the lowest common multiple (LCM) of 840 and 90, which is 2520.

A few solutions used the same idea for building many solutions from one solution, but with a step size between their sets of ten consecutive numbers that was much larger than the least common multiple of $1, 2, \ldots, 10$. For example, a couple of students used the much bigger number 3628800 to get these sets:

$3628801, 3628802, 3628803, 3628804, 3628805, 3628806, 3628807, 3628808, 3628809, 3628810$ and $7257601, 7257602, 7257603, 7257604, 7257605, 7257606, 7257607, 7257608, 7257609, 7257610$.

These sets satisfy the criteria in the problem, so the students had definitely solved the problem, but there are sets of numbers in between these that also satisfy the criteria, so a step size of 362880 won't allow you to find all solutions. How do you think this number 362880 was calculated?

It's tempting to think only about finding positive solutions to this problem, but Ethan from ESF King George V School, Hong Kong, also sought out negative solutions using the same ideas:

Next, we must look for negative solutions. This is simple to find, just remove multiples of 2520 from the first set (1 to 10) and you can find them. Since divisibility works the same with negative numbers, we can use the exact same logic as before.

So a set could be

$-2519, -2518, -2517, -2516, -2515, -2514, -2513, -2512, -2511, -2510$

$-5039, -5038, -5037, -5036, -5035, -5034, -5033, -5032, -5031, -5030$

So far we have discussed approaches that involve working 'by hand', either numerically or algebraically. But Juan from IES Maximo Laguno, Spain, and Abdullah from Doha College in Qatar tackled the problem using computer programming. Click here to read Juan's work and click here to read Abdullah's work. 

Evan from Bishop of Llandaff Church in Wales High School, Cardiff, Wales, submitted a particularly technical approach to identifying least common multiples. This uses some quite advanced mathematical ideas and notation, but you might be interested to read Evan's work here.

Hannah also looked for lots of sequences but then tried a slightly different approach. This offers some interesting mathematical ideas for how to work with sequences of sequences:

Tackling the problem directly

Some solutions used features of the problem of consecutive numbers to find a way directly into solving this difficult problem. This is how Lukas from St John's College School and Imeth from Pristine Private School in the United Arab Emirates tackled the problem.

Lukas observed a critical key point about any solution: 

The first step is to realise that the last number in the sequence must end in a zero as it has to be a multiple of 10. This means that the first number must have a 1 in it. 

Imeth used algebra to take a similar approach: 

You can write all the numbers as: $x, x+1, x+2, x+3, \ldots, x+9$.

Work backwards from ones (the conditions in the problem) that narrow down $x$:

$x+9$ is a multiple of 10 - the only way this happens is if the last digit of $x+9$ is 0, meaning the last digit of $x$ is 1.

$x+8$ is a multiple of 9 - the sum of the rest of the digits plus nine must be a multiple of 9. Since 9 itself is a multiple of nine, the rest of the digits have to have a sum of a multiple of nine, meaning the rest of the digits are a multiple of 9.

$x+7$ is a multiple of 8 - a number is a multiple of 8 if the first 3 digits (starting from the ones place) are a multiple of 8. You could try all combinations to see what 3-digit numbers ending with 8 are multiples of 8, and you'd get 008, 048, 088, 128, 168, 208, 248, 288, 328, 368, 408, 448, 488, 528, 568, 608, 648, 688, 728, 768, 808, 848, 888, 928, 968, meaning the last 3 digits of x has to be any one of these, minus 7. (You may be wondering why this only considers the last three digits. If so, you might find it helpful to think about what you know about factors of 1000.)

$x+6$ is a multiple of 7 - this is a tricky one - the divisibility rule of 7 is to take the last digit of a number, double it, and subtract that result from the remaining digits. Here, the last digit is 1+6, which is 7, so, the remaining digits have to be a multiple of 7 (if $y-14$ is a multiple of 7, $y$ is a multiple of 7. (Proof: $7z-14 = 7(z-2)$)).

$x+5$ is a multiple of 6 - the divisibility test for this one is if it's divisible by both 2 and 3. This gives nothing whatsoever - we already know the sum of the digits minus the first one is a multiple of 9 (and, by extension, a multiple of 3), and adding the last digit, 6, still keeps it like that. As for 2, 5+1, 6 is already a multiple of 2.

$x+4$ is a multiple of 5 - 1 + 4 is 5, and any number ending with 5 is divisible by 5.

$x+3$ is divisible by 4 - this is 4 less than $x+7$, and any multiple of 8 minus four is a multiple of four, and this again tells us nothing. ($8z-4 = 4(2z-1)$)

$x+2$ is a multiple of 3 - as stated in $x+5$, we already know the sum of the digits minus the first one is a multiple of 9 (and, by extension, a multiple of 3), and adding the last digit, 3, still keeps it like that.

$x+1$ is a multiple of 2 - $1+1=2$, and any number ending with $2$ is even.

After this detailed analysis, Imeth put all these ideas together to obtain some sets of ten consectutive numbers that meet the required conditions, noting that:  

As it turns out these are all just multiples of 252 with 1 on the end for the first number!

$2521, 2522, 2523, 2524, 2525, 2526, 2527, 2528, 2529,$ and $2530$ is one. There's also $5041, 5042, 5043, 5044, 5045, 5046, 5047, 5048, 5049, 5050$. And let's not forget the trivial case of $1,2,3,4,5,6,7,8,9,10$.

Taking it further

The ideas in this question can be taken much further than the original problem. As one example of this, Ananya from Ganit Kreeda Vichar Vatika found a sequence of 20 consecutive integers that does not contain a prime. You might want to think about this before clicking here to read Ananya's work.

Once again, well done again to everyone who had a go at this problem and submitted solutions!