# Ethan Brown’s magic squares

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# Fibonacci tartan

Brady who makes the Numberphile videos has had a tartan based on the fibonacci sequence created for him:

You can read about why it was made at Brady’s blog entry.

It’s “printed” tartan, but does look very nice.

# Magic Squares problems to be solved What are the smallest possible magic squares? €6900 still to win! (≈ \$8500)

# Coursera FB problem

Find 11 positive integers such that none of them add up to a multiple of 11, or prove that this cannot be done.

The above problem was posted on the FB page of the Coursera Introduction to mathematical thinking mooc.

To begin with I misunderstood the problem,  however reading posts by other members of the group  it became clear what the problem was and of how to proceed. This provided a very interesting experience of being in a virtual classroom.

I began the problem after my Japanese class on Tuesday night reading it on my iPhone in my car  — please note that the car was stationary:) The final insight was provided by a fellow classmates posting several hours later as I got out of bed.

I am adding my attempt at understanding this problem below for the sake of interest. As discussed on the group I am looking at a smaller number to begin with ie showing the case for 3 positive integers:

• a, b, and c are +ve ints
• none of a, b, or c are divisible by 3
• hence a, b, and c leave remainder 1 or 2 when divided by 3
• we want to form the sums a+b, a+c, b+c, and a+b+c and to see what the remainder is when divided by 3
• it suffices to add the remainders of a, b, and c
• eg consider 4 and 8
• 4 leaves remainder 1 when divided by 3
• 8 leaves remainder 2 when divided by 3
• 4 + 8 = 12 leaves remainder 0 when divided by 12
• however the remainders of 4 and 8 when we add them gives 1 + 2 = 3 but note that in terms of dividing by 3 a remainder of 3 is the same as a remainder of 0
• in mathematical notation we say 1+ 2 = 3 ≡ 0 (mod3)
• similarly consider 4 and 7, both leave remainder 1 when divided by 3
• 4 + 7 = 11 which leaves a remainder of 2 when divided by 3
• note that the remainders of 4 and 7 are both 1 and that when we add these we get 2
 a b c a+b a+c b+c a+b+c Remainder 1 1 1 2 2 2 3 ≡ 0 (mod3) Remainder 1 1 2 2 2 3 4 ≡ 1 (mod3) Remainder 1 2 2 3 3 4 5 ≡ 2 (mod3) Remainder 1 2 1 3 2 3 4 Remainder 2 2 2 4 4 4 6 ≡ 0 (mod3) Remainder 2 2 1 4 3 3 5 Remainder 2 1 1 3 3 2 4 Remainder 2 1 2 3 4 3 5
• notice that in each row there always occurs at least one cell in which there is a 3 or a 6 (≡ 0 (mod3) ie divisible by 3).
• with the necessary changes being made it would be possible to show that for any set containing n positive integers there must be at least one way of combining the numbers in the set to form a sum which is divisible by n.