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Thank you to everybody who sent in their ideas about this activity.

Viren shared these tips with us:

I look for the pattern and relations between numbers. In case of Level 1 and Level 3, by looking at lowest and highest numbers I figure out the times tables they might belong to. And then I work out what shift (up/down) is applicable.

In case of Level 2 and Level 4, the big numbers look intimidating but I work out the difference between numbers first, then figure out which number they are multiples of.

Good ideas, Viren. I wonder why working out the difference between the numbers is useful for finding the times tables that they might belong to?

Harry from Illogan School in Cornwall, UK sent in this method for the Level 4 problems, explaining how to use the difference to work out the times table:

This time I have numbers 123, 59, 171, 219 and 91

1) Put the numbers in order and find the differences

59 (+32) 91 (+32) 123 (+48) 171 (+48) 219

2) Find the highest number that goes into all the differences

Factors of 32 = 1, 2, 4, 8, 12, 16 and 32

Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48

So 16 in this case. This gives you the times table that has been shifted

3) Write out the 16x table, which will help you find the shift

In this case it is either 16 minus 5 or 16 plus 11. You can check it works for each number you were given

4) There will always be two answers because you can either add, or minus from the number you were given to get to a number in the 16 x table

This is really clearly explained, Harry!

Shaunak from Ganit Manthan, Vicharvatika, India described a similar solution:

This problem can be split into 2 parts: Working out the Table and Working out the Shift.

Working out the Table:

I observed the differences between the sets of consecutive numbers in the shifted times table.

In level 1 and 2, all these differences are the same.

In level 3 and 4, all these differences are different. The table has to be the greatest possible, so we have to find the HCF (Highest Common Factor) of the differences.

If the given sequence is 37, 17, 22, 52, 77, then the differences are 20, 5, 30, and 25. The HCF of these differences is 5, so the table is 5.

This method works because taking any two numbers in a times table and subtracting them will give a difference which is a multiple of the table. This happens because a times table is an arithmetic progression, which means a constant number (the table) is repeatedly getting added to the previous number, creating a sequence.

If the table is 5, then creating the sequence looks like this: 5, 5 + 5, 5 + 5 + 5, 5 + 5 + 5 + 5, and so on.

If you take two numbers from the times table, you will see that the table is repeatedly getting added to the smaller number to obtain the bigger number. Therefore, the difference between two numbers in a times table is a multiple of the table. This having been said, we can say that the HCF of the differences will give us the greatest possible table.

Let us consider the example stated above. Arranging the numbers in ascending order, we get 17, 22, 37, 52, and 77. 5 is being added to the first number to get the second number (17 + 5 = 22), 15 is being added to the second number to get the third number (22 + 5 + 5 + 5 = 37) , 30 is being added to the third number to get the fourth number (37 + 5 + 5 + 5 + 5 + 5 + 5 = 52) , and 25 is being added to the second number to get the third number (52 + 5 + 5 + 5 + 5 + 5 = 77).

Working out the Shift:

I first observed the first number of the shifted times table and the closest multiple of the times table to this number.

If the shifted times table is 30, 50, 70, 90, 110, then the table is 20, because all the differences are 20. I looked at the first number, 30, and figured the closest multiple of 20 is 20.

If the multiple is smaller than the first number of the shifted times table, then the difference of the first number of the shifted times table and the multiple is the shift up.

Considering the above case, we can see that 20 is smaller than 30. So the difference between 20 and 30, which is 10, is the shift up.

If the multiple is greater than the first number of the shifted times table, then the difference of the first number of the shifted times table and the multiple is the shift down.

If the shifted times table is 2, 5, 8, 11, 14, then the table is 3. The closest multiple of 3 to 2 is 3. We can see that 3 is greater than 2. So the difference between 3 and 2, which is 1, is the shift down.

We can observe that the shifted times table is also an arithmetic progression, just like the times table. The differences have the same HCF, but the first, second, third, fourth, and fifth have a common difference from their closest multiple.

Thank you for sending in these ideas, Shaunak, this is a really clear explanation of how the shifted times tables work and how they can be solved.

Veyhant from Ganit Kreeda, Vicharvatika, India sent in these pictures explaining their solution for each of the four levels. These pictures can be clicked on to make them bigger:

We also received similar solutions from Nikhil, Abhyudh, Viyaan, Aditi, Mridula, Pushan, Larisa, Laira, Shravani, Anika, Kiaan, Kaira and Shreehari (Group 1) and from Ahana, Sehar, Saanvi, Dhanvin, Aariz, Ananthjith, Vivaan, Sai, Pranathi, Paavani, Utkarsh, Tavish and Dhruv (Group 2) from Ganit Kreeda, Vicharvatika, India. Thank you all for sending in your ideas - you've worked really hard on this problem. Their full solutions can be seen by clicking on these links: Group 1 Ganit Kreeda, Group 2 Ganit Kreeda.

Thank you as well to Jeffery and Ava from Australia and Luna from CHPS in Australia who also sent in some similar solutions to this problem.