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Japan Mathematical Olympiad

Algebra Difficulty 6.8 National Olympiad Prove it Japan

For all quadruples (x,y,z,w)(x, y, z, w) consisting of integers 1≤x,y,z,w≤10001 \le x, y, z, w \le 1000, we consider adding up the maximum value of xy+zwxy + zw, xz+ywxz + yw, xw+yzxw + yz and denote the sum by MM. Similarly, for all quadruples (x,y,z,w)(x, y, z, w) consisting of integers 1≤x,y,z,w≤10001 \le x, y, z, w \le 1000, we consider adding up the minimum value of xy+zwxy + zw, xz+ywxz + yw, xw+yzxw + yz and denote the sum by mm. Determine the number of positive divisors of M−mM - m.

Solution

20412 \boxed{20412}
For 3 real numbers a,b,ca, b, c, the difference of the maximum value and the minimum value of them is ∣a−b∣+∣b−c∣+∣c−a∣2\frac{|a-b|+|b-c|+|c-a|}{2}. Using ∣(xy+zw)−(xz+yw)∣=∣x−w∣∣y−z∣|(xy + zw) - (xz + yw)| = |x - w||y - z|, we have
M−m=12∑x,y,z,w=11000(∣x−w∣∣y−z∣+∣x−y∣∣z−w∣+∣x−z∣∣y−w∣). M - m = \frac{1}{2} \sum_{x,y,z,w=1}^{1000} \left( |x - w||y - z| + |x - y||z - w| + |x - z||y - w| \right).
Since
∑x,y,z,w=11000∣x−w∣∣y−z∣=(∑x,y=11000∣x−y∣)2=(2∑d=1999d(1000−d))2=(999⋅1000⋅1001)29, \sum_{x,y,z,w=1}^{1000} |x - w||y - z| = \left( \sum_{x,y=1}^{1000} |x - y| \right)^2 = \left( 2 \sum_{d=1}^{999} d(1000 - d) \right)^2 = \frac{(999 \cdot 1000 \cdot 1001)^2}{9},
we have
M−m=32⋅(999⋅1000⋅1001)29=25⋅35⋅56⋅72⋅112⋅132⋅372. M - m = \frac{3}{2} \cdot \frac{(999 \cdot 1000 \cdot 1001)^2}{9} = 2^5 \cdot 3^5 \cdot 5^6 \cdot 7^2 \cdot 11^2 \cdot 13^2 \cdot 37^2.
Therefore, the number of positive divisors of M−mM - m is 6⋅6⋅7⋅3⋅3⋅3⋅3=204126 \cdot 6 \cdot 7 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 20412.

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