2015 AMC 12B Problem 23

Attempt Problem 23 of the 2015 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 12B solutions, or check the answer key.

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23.

A rectangular box measures a×b×c,a \times b \times c, where a,a, b,b, and cc are integers and 1abc.1 \le a \le b \le c. The volume and the surface area of the box are numerically equal. How many ordered triples (a,b,c)(a, b, c) are possible?

44

1010

1212

2121

2626

Answer: B
Concepts:Diophantine Equationfactoringcasework
Difficulty rating: 2340
Solution:

Numerically equal volume and surface area means abc=2(ab+bc+ca).abc=2(ab+bc+ca). Dividing by abcabc gives 1=2/a+2/b+2/c6/a,1=2/a+2/b+2/c\le6/a, so a6.a\le6. The cases a=1a=1 and a=2a=2 give no positive solutions. For a2,a\ne2, set u=(a2)b2a,v=(a2)c2a. \begin{gathered} u=(a-2)b-2a,\\ v=(a-2)c-2a. \end{gathered} The equation then factors as uv=4a2.uv=4a^2.

For a=3,a=3, (b6)(c6)=36(b-6)(c-6)=36 gives (b,c)=(7,42),(b,c)=(7,42), (8,24),(8,24), (9,18),(9,18), (10,15),(10,15), (12,12).(12,12). For a=4,a=4, (b4)(c4)=16(b-4)(c-4)=16 gives (5,20),(6,12),(8,8).(5,20),(6,12),(8,8). For a=5,a=5, the congruence of the two factors modulo 33 leaves only the valid pair (b,c)=(5,10),(b,c)=(5,10), and for a=6,a=6, (b3)(c3)=9(b-3)(c-3)=9 leaves only (b,c)=(6,6).(b,c)=(6,6). Thus there are 5+3+1+1=105+3+1+1=10 triples.

Thus, the correct answer is B.

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