But does a 100x100 anti-magical square exist?

Number Theory Level 5

A heterosquare contains positive consecutive integers starting from 1 such that the rows, columns, and diagonals add to all different values. If the sums resulting from a heterosquare form a consecutive sequence, the heterosquare is also called an anti-magical square .

Assume that a $100\times 100$ anti-magical square exists. It is already known that the minimum value of the sum of any row in this anti-magical square is 5050 and that the maximum value of the sum of any row is 995050, so there are 989800 sequences you have to consider: from $(5050,5051,...,5251)$ to $(994849,994850,...,995050)$ .

Without using brute force (i.e. testing all heterosquares using a computer) or calculating the value of any row, column, or diagonal, how many sequences could you eliminate?

If, after eliminating, you still have $n$ sequences left, and the smallest value of every sequence is $a_1,a_2,...,a_n$ , find $n+a_1+a_2+\cdots+a_n.$

Note : If your answer is different, feel free to report in any way.

P/S: Generalize, and don't try to find a $100\times 100$ anti-magic square despite its existence!

Inspiration

The answer is 999901.

1 solution

Mark Hennings
Dec 14, 2017

Suppose that the row sums of a $N\times N$ antimagic square are $r_1,r_2,...,r_N$ , that the column sums are $c_1,c_2,...,c_N$ , and that the column sums are $d_1,d_2$ . Suppose that the collection of sums forms the sequence $a,a+1,a+2,...,a+2N+1$ . Let $m_N = \tfrac12N(N^2+1)$ . Note that $r_1+r_2+\cdots+r_N \; = \; c_1 + c_2 + \cdots + c_N \; = \; 1 + 2 + 3 + \cdots + N^2 \; = \; \tfrac12N^2(N^2+1) \; = \; Nm_N$ and hence $2Nm_N + d_1 + d_2 \; = \; (r_1 + \cdots + r_N) + (c_1 + \cdots + c_N) + d_1 + d_2 \; = \; a + (a + 1) + \cdots + (a + 2N+1) \; = \; (N+1)(2a+2N+1)$ and so $2a+1 \; = \; a + (a+1) \; \le \; d_1 + d_2 \; = \; (N+1)(2a+2N+1) - 2Nm_N \; \le \; (a+2N) + (a + 2N+1) \; = \; 2a + 4N + 1$ so that $m_N - N - \tfrac32 \; \le \; a \; \le \; m_N - N + \tfrac12$ and hence $a = m_N - N - 1$ or $a = m_N - N$ .

Assuming that both possibilities are achievable, we obtain the answer $2 + (m_N-N-1) + (m_N - N) = 2m_N - 2N + 1$ With $N=100$ , we obtain the answer $\boxed{999901}$ .

But does a 100x100 anti-magical square exist?

The answer is 999901.

1 solution

0 pending reports