Problem:
Find the number of triples of nonnegative integers (a,b,c) satisfying a+b+c=300 and
a2b+a2c+b2a+b2c+c2a+c2b=6,000,000.
Solution:
Adding 3abc to the second equation gives
6,000,000+3abc=a2b+a2c+b2a+b2c+c2a+c2b+3abc=(a+b+c)(ab+bc+ca)=300(ab+bc+ca).
Hence 2,000,000+abc=100(ab+bc+ca). On the other hand,
(a−100)(b−100)(c−100)=abc−100(ab+bc+ca)+1002(a+b+c)−1003=abc−100(ab+bc+ca)+1002⋅300−1003=abc−100(ab+bc+ca)+2,000,000.
Combining the two obtained relations gives
(a−100)(b−100)(c−100)=0.
In particular, one of a,b, and c must be 100. Conversely, if one of the variables, say c, is equal to 100 and a+b=300−c=200, then
a2b+a2c+b2a+b2c+c2a+c2b=ab(a+b)+100(a2+b2)+c2(a+b)=100(a+b)2+200c2=6,000,000.
Therefore the triples that satisfy the given equation are precisely the triples for which one of the variables is equal to 100. There are 201 valid triples with c=100 and a+b=200. Likewise, there are 201 valid triples with each of a=100 and b=100. However, the triple (100,100,100) has been counted three times, so the requested number of triples is 3⋅201−2=601.
OR
Grouping terms of the second equation gives
6,000,000=a2b+a2c+b2a+b2c+c2a+c2b=(a2b+b2a)+(a2c+c2a)+(b2c+c2b)=ab(a+b)+ac(a+c)+bc(b+c)=ab(300−c)+ac(300−b)+bc(300−a)=300(ab+bc+ca)−3abc.
Hence abc=100(ab+bc+ca)−2,000,000.
Let D=ab+bc+ca. Then a,b, and c are solutions of the cubic equation
x3−300x2+Dx−(100D−2,000,000)=0.
By inspection, x=100 is a solution of this equation, and thus one of a,b, or c is equal to 100. The rest of the solution then proceeds as in the first solution.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions