+186 How many ordered triples (a, b, c) where a, b, and c are from the set {1, 2, 3, ... , 17} satisfy the equation
a^3 + b^3 + c^3 + 2abc = a^2*b + a^2*c + b^2*c + a*b^2 + a*c^2 + b*c^2
First I tried to factor the equation but I can't figure out where to even start with that... I don't know where to start with this problem so help will be appreciated!
I wrote a computer program:
i = 0
for (a = 1..17)
for (b = 1..17)
for (c = 1..17)
if (a^3 + b^3 + c^3 + 2abc = a^2*b + a^2*c + b^2*c + a*b^2 + a*c^2 + b*c^2): i += 1
print (i);
The output was 382.
+186 The solution was 408:
We can factor this into (a - b - c ) * (b - c - a) * (b - c + a)=0, so a = b + c, b = a + c, or c = a + b.
Since each number is an integer from 1 to 17, we can choose 2 values for a and b and then use the formula a = b + c, then c will only have one possible value after that. We do 17 choose 2 = 136 for this, then multiply by 3 for all of the respective equations. Doing the math, 3 * 136 = 408.
