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15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?

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The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?

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What are the possible dimensions of a rectangular hallway if the number of tiles around the perimeter is exactly half the total number of tiles?

Always the Same

Age 11 to 14 Challenge Level:

Thank you to Jimmy, Natalie and Lucy at West Flegg GM Middle School for their work on the problem "Always the Same". Natalie is not the first student to add together all the numbers 1 through 16 and take the average.

Thus replacing the ascending numbers with 8.5 in every cell and circling four cells gives a total of 34. Or as Natalie did, she realised that "you pick numbers from each column and row" and took the average between the sum of the four columns:

i.e. (28 + 32 + 36 + 40)/4 = 34

A good solution with this method came from Melanie and Rachel of Flegg High School.

A proof of this problem could be as follows.

Let the first number be a.

Magic square

Then when choosing numbers from rows and column that do not coincide we have:

a + (a + 4) + (a + 8) + (a + 12) + 1 + 2 + 3 = 'Magic number'

We add 1 because there is one number in the second column, 2 because there is one number in the third column and 3 because there is one number in the fourth column.

Hence:

4a + (4 + 8 + 12) + ( 1 + 2 + 3) = 34
i.e. 4a + 30 = 34
i.e a = 1
and the array is 1 through 16 as set.

But suppose the 'magic number' had been 62 then

4a + 30 = 62
i.e a = 8
and the array would have been 8 through 23.

Hope that the explanation above helps especially Josh at Russell Lower School to "work out where we went wrong".

We could have used a 5 by 5 array of ascending numbers!