+0  
 
+2
36
3
avatar+114 

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!

 #1
avatar
0

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.

 Nov 6, 2022
 #2
avatar+114 
+1

That was incorrect, can you please try again?

 #3
avatar+114 
+2

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.

 Nov 13, 2022

15 Online Users