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Many of you approached this problem in the same way to start with. Maddie and Alex from The Mount School wrote:

We found out that 1+2+3+4+5+6...+20=210. Because we needed 7 different subsets that when added together made 7 consecutive numbers, we divided 210 by 7. 210/7=30This told us that the seven numbers had to be around 30. The numbers turned out to be 27, 28, 29, 30, 31, 32 & 33.

We tried a variation of numbers but ended up using:

$$

\begin{align}

0+7+20&=27\\

6+10+12&=28\\

2+8+19&=29\\

5+9+16&=30\\

3+11+17&=31\\

1+13+18&=32\\

4+14+15&=33

\end{align}

$$

This is one of many solutions!

Well done, Maddie and Alex. In fact, not many of you mentioned that there are lots of solutions to this problem. Boris from Gresham's Preparatory School found the seven consecutive numbers in a similar way and then said:

So now we check which numbers from 0 to 20 can make them in groups of 3:

27 - 7,3,17

28 - 8,5,15

29 - 19,4,6

30 - 9,10,11

31 - 1,18,12

32 - 2,16,14

33 - 0,13,20

To find these it's easier to make two of the numbers end with a zero (e.g. 3+17=20, 5+15=20, 4+6=10, 9+11=20, 18+12=30, 16+14=30, 0+20=20) and then add the other number to finish it.

Thank you, Boris. Many of you suggested ways to make pairs of numbers and then add a third to total one of the consecutive numbers like Boris' method.

Ivo from Gresham's Prep School used a different method to work out which consecutive numbers to aim to make:

We can see that 0+1+2+3+.....+17+18+19+20=210.

Each consecutive number is made of three numbers between 0 and 20. Let the first of the consecutive number be made of A,B and C. That means that the next consecutive number will be A+B+C+1. This led me to my solution.

(A+B+C) + (A+B+C+1) + (A+B+C+2)+ (A+B+C+3) + (A+B+C+4) + (A+B+C+5) + (A +B+C+6) = 210

7A + 7B + 7C + 21 = 210

7(A+B+C) = 189

A+B+C=27

This means that the first of the consecutive numbers is 27. The next is 28 and so on. So the consecutive seven are 27, 28, 29, 30, 31, 32, 33.

Thank you, Ivo! Zoe, Andrew, Nikita and Ben from Aqueduct Primary School went about the problem in a slightly different way:

We looked at the numbers from 0 - 20. We used trial and error to solve this problem. The first thing we did was decide what our first set of three numbers was going to be. We looked at the "Nrich" example:

4+5+17= 26

These sets when added together obviously gave consecutive answers. We decided that for there to be consecutive answers there must be some relationship between each number in each set.

We separated the numbers into 3 columns: a + b + c = answer:

A | B | C | Answer |

2 | 7 | 16 | 25 |

4 | 5 | 17 | 26 |

We noticed that the pattern in A = adding 2 each time

We noticed that the pattern in B = subtract 2 each time

We noticed that the pattern in C = adding 1 each time

We followed the pattern on to see if we could come up with 7 sets however, we only managed to achieve 5.

Here are our results:

0 + 9 + 15 = 24

2 + 7+ 16 = 25

4 + 5 + 17 = 26

6 + 3 + 18 = 27

8 + 1 + 19 = 28

We decided that we needed to take a more systematic approach to solve this problem. The most logical number to start with was 0 because you can build on 0. We noticed that the numbers 0-20 were already organised into 3 rows of 7 numbers:

0 | 1 | 2 | 3 | 4 | 5 | 6 |

7 | 8 | 9 | 10 | 11 | 12 | 13 |

14 | 15 | 16 | 17 | 18 | 19 | 20 |

We listed the numbers consecutively. We made three columns. This gave us 7 sets of 3 numbers.

A | B | C | Answer |

0 | 7 | 14 | 21 |

1 | 8 | 15 | 24 |

2 | 9 | 16 | 27 |

3 | 10 | 17 | 30 |

4 | 11 | 18 | 33 |

5 | 12 | 19 | 36 |

6 | 13 | 20 | 39 |

This did not give us consecutive answers. The answers increased by 3 each time. We looked more closely at our lists of numbers and realised that although the numbers in each list were consecutive, the lists were not consecutive to each other. We rearranged the lists:

A | B | C | Answer |

0 | 13 | 14 | 27 |

1 | 12 | 15 | 28 |

2 | 11 | 16 | 29 |

3 | 10 | 17 | 30 |

4 | 9 | 18 | 31 |

5 | 8 | 19 | 32 |

6 | 7 | 20 | 33 |

This did give us consecutive answers.

We noticed that the numbers in columns A and B when added together all give the same answer, for example:

0 + 13 = 13

1 + 12 = 13

2 + 11 = 13

3 + 10 = 13

4 + 9 = 13

5 + 8 = 13

6 + 7 = 13

These are number bonds for 13.

We also noticed that the lists of numbers had to follow on consecutively from each other. For example: 6 and 7 had to be in the same set, 13 and 14 had to be in the same set. (see the chart of results)

We noticed that the second and third numbers in each set when added together also gave the same total.

Thank you Zoe, Andrew, Nikita and Ben. It is always good to receive solutions which take us all the way through the process that you followed to solve the problem. Your solution shows us that "playing" with a problem can be a very good way to start and will often lead to us finding something out that helps us go about a solution more systematically (in other words more logically).

Rhiannon was able to give a set where the totals were all the same, as well as ones where the totals went up in twos and threes.

The numbers have a total of $0+1+2+...+19+20=210$, so, if there are seven sets of three numbers with the same total, these must each have a total of $210 \div 7 = 30$.

If there is one number from the bottom row in each set, these are $14$, $15$, $16$, $17$, $18$, $19$ and $20$. This means the other numbers in the sets need to have totals of $16$, $15$, $14$, $13$, $12$, $11$ and $10$.

If $0$ is used to make the smallest of these totals, this means $0+10=10$.

To make $11$, this total needs to be increased by $1$. Increasing either of the numbers by $1$ is impossible, as the other number would be used twice. Therefore I tried increasing one of the numbers by $2$ and decreasing the other number by $1$. Since $-1$ is not available, the next pair becomes $2+9=11$.

This can be repeated to give $4+8=12$ and $6+7=13$.

Repeating this would give $8+6=14$, but $6$ and $8$ have already been used. The unused numbers are $1$, $3$, $5$, $11$, $12$ and $13$, so using the same idea gives $1+13=14$, $3+12=15$ and $5+11=16$.

Recombining these with the other numbers in order gives:

$0+10+20=30$

$2+9+19=30$

$4+8+18=30$

$6+7+17=30$

$1+13+16=30$

$3+12+15=30$

$5+11+14=30$

If instead the totals are going up in twos, then these must be $24$, $26$, $28$, $30$, $32$, $34$ and $36$, since the average value must still be $30$.

Again using one of the bottom row of numbers in each case and in the same order, the remaining totals needed are $24-14=10$, $26-15=11$, $28-16=12$, $30-17=13$, $32-18=14$, $34-19=15$ and $34-16=20$. These are the same totals that were used above with the numbers from the top two rows, so the sets become:

$0+10+14=24$

$2+9+15=26$

$4+8+16=28$

$6+7+17=30$

$1+13+18=32$

$3+12+19=34$

$5+11+20=36$

To make the totals increase by threes the columns of the grid can be used, as each of the three numbers is increased by $1$ for each set, so the total increases by $3$ each time. This means the sets are:

$0+7+14=21$

$1+8+15=24$

$2+9+16=27$

$3+10+17=30$

$4+11+18=33$

$5+12+19=36$

$6+13+20=39$

Thank you and well done to everyone who submitted solutions!