Problem:
A rectangular box measures a×b×c, where a,b, and c are integers and 1≤ a≤b≤c. The volume and the surface area of the box are numerically equal. How many ordered triples (a,b,c) are possible?
Answer Choices:
A. 4
B. 10
C. 12
D. 21
E. 26
Solution:
Note that for any natural number k, when Aaron reaches point (k,−k), he will have just completed visiting all of the grid points within the square with vertices at (k,−k),(k,k),(−k,k), and (−k,−k). Thus the point (k,−k) is equal to p(2k+1)2−1. It follows that p2024=p(2⋅22+1)2−1=(22,−22). Because 2024−2015=9, the point p2015=(22−9,−22)=(13,−22).
Because the volume and surface area are numerically equal, abc=2(ab+ac+bc). Rewriting the equation as ab(c−6)+ac(b−6)+bc(a−6)=0 shows that a≤6. The original equation can also be written as (a−2)bc−2ab− 2ac=0. Note that if a=2, this becomes b+c=0, and there are no solutions. Otherwise, multiplying both sides by a−2 and adding 4a2 to both sides gives [(a−2)b−2a][(a−2)c−2a]=4a2. Consider the possible values of a.
a=1:(b+2)(c+2)=4.
There are no solutions in positive integers.
a=3:(b−6)(c−6)=36.
The 5 solutions for (b,c) are (7,42),(8,24),(9,18),(10,15), and (12,12).
a=4:(b−4)(c−4)=16.
The 3 solutions for (b,c) are (5,20),(6,12), and (8,8).
a=5:(3b−10)(3c−10)=100.
Each factor must be congruent to 2 modulo 3, so the possible pairs of factors are (2,50) and (5,20). The solutions for (b,c) are (4,20) and (5,10), but only (5,10) has a≤b.
a=6:(b−3)(c−3)=9
The solutions for (b,c) are (4,12) and (6,6), but only (6,6) has a≤b.
Thus in all there are (B)10 ordered triples (a,b,c):(3,7,42),(3,8,24),(3,9,18), (3,10,15),(3,12,12),(4,5,20),(4,6,12),(4,8,8),(5,5,10), and (6,6,6).
The problems on this page are the property of the MAA's American Mathematics Competitions