Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A magic square of order $N$ is an $ N \times N $ matrix with positive integral entries such that the elements of every row, every column and the two diagonals all add up to the same number. If a magic square is filled with numbers in arithmetic progression starting with $a \in N $ and common dierence $ d \in N$, what is the value of this common sum?

I am stuck on this problem. Can anyone help me please...

share|improve this question
    
You do remember the formula for the sum of an arithmetic progression, don't you? Or better yet, how to derive that formula! –  Jyrki Lahtonen Jan 5 '13 at 9:39
add comment

1 Answer 1

Let the commom sum be $C$. So $C$ is also the common row sum. There are $N$ rows. So the sum of all entries in the matrix is equal to $NC$. Hence $NC=a+(a+d)+(a+2d)+\ldots+(a+(N^2-1)d)$ and I will leave the rest to you.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.