SOLUTION: DETERMINE ALL THE TRIPLES OF POSITIVE INTEGERS (a,b,c), SUCH THAT ab+bc+ca= abc, WHERE A IS LESS THAN B LESS THAN C

Algebra ->  Sequences-and-series -> SOLUTION: DETERMINE ALL THE TRIPLES OF POSITIVE INTEGERS (a,b,c), SUCH THAT ab+bc+ca= abc, WHERE A IS LESS THAN B LESS THAN C      Log On


   

Question 1138427: DETERMINE ALL THE TRIPLES OF POSITIVE INTEGERS (a,b,c), SUCH THAT ab+bc+ca= abc, WHERE A IS LESS THAN B LESS THAN C
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Given: ab%2Bac%2Bbc+=+abc

Move one of the terms on the left to the right side. It doesn't matter which one, because the original equation is symmetric in a, b, and c.

abc-ab+=+ac%2Bbc
ab%28c-1%29+=+c%28a%2Bb%29
ab%2F%28a%2Bb%29+=+c%2F%28c-1%29

That equation says that ab is an integer that is 1 more than (a+b). So

ab+=+a%2Bb%2B1

Now solve this equation for either variable in terms of the other. Again it doesn't matter which, because again this equation is symmetric in a and b.

ab-a+=+b%2B1
a%28b-1%29+=+b%2B1
a+=+%28b%2B1%29%2F%28b-1%29
a+=+%28%28b-1%29%2B2%29%2F%28b-1%29
a+=+1%2B2%2F%28b-1%29

1 is an integer, and a has to be an integer. That means 2/(b-1) has to be an integer; and that means (b-1) has to be a factor of 2.

So b-1 has to be either 1 or 2; that means b has to be either 2 or 3.

And now we can find all the triples a, b, and c for which the given equation is true.

(1) If b = 2 then

a+=+1%2B2%2F1+=+1%2B2+=+3

and then

%28ab%29%2F%28a%2Bb%29+=+c%2F%28c-1%29+=+6%2F5 which means c is 6.

The three numbers a, b, and c (in no particular order) are 2, 3, and 6.

(2) If b = 3 then

a+=+1%2B2%2F2+=+1%2B1+=+2

and then (as before)

%28ab%29%2F%28a%2Bb%29+=+c%2F%28c-1%29+=+6%2F5 which means c is 6.

So there is a single set of three integers for which the given equation is true.

Finally, since the problem specifies a < b < c, the single solution is

{a,b,c) = (2,3,6)