Problem:
How many ordered triples (x,y,z) of distinct nonnegative integers ≤8 satisfy xy>z, zx>y, and yz>x?
Answer Choices:
A. 36
B. 84
C. 186
D. 336
E. 486
Solution:
Without loss of generality, assume x<y<z. Now we count the complement, that is how many triples result in xy≤z or xz≤y or yz≤x. yz≤x is impossible and xz≤y only holds if x=0, which would imply xy≤z anyways. Therefore, it suffices to compute the number of triples where xy<z.
Case 1: x=0. There are (28)=28 possibilities for y and z.
Case 2: x=1. If x=1, for some given y then, z∈{y+1,y+2,…,y+8} all result in xy<z. There are 8−y options in the set, so summing across all y gives:
y=2∑78−y=21
Case 3: x=2. We see that y=3, z≥6 and y=4, z=8 work for 4 possibilities.
Case 4: x≥3. If x≥3, then we are guaranteed xy>z. This gives 0 possibilities.
There are (39) ways to pick x, y, and z in increasing order for a total of 6(84−28−21−4)=(C) 186 ordered triples.
The problems on this page are the property of the MAA's American Mathematics Competitions