# a question......

• Aug 11th 2008, 05:11 AM
OPETH
a question......
\$\displaystyle a.b-c=4\$

\$\displaystyle b.c-a=3\$

in equality, a, b, c are positive integers.

What is total of \$\displaystyle a+b+c \$ minumum?
• Aug 11th 2008, 05:23 AM
Quick
\$\displaystyle a.b-c=4\$

\$\displaystyle b.c-a=3\$

add the two equations together:

\$\displaystyle ab+bc-a-c=7\$

move things around:

\$\displaystyle ab-a+bc-c=7\$

factor:

\$\displaystyle a(b-1)+c(b-1)=7\$

factor again:

\$\displaystyle (a+c)(b-1)=7\$

hmm, 7 is a prime number though and so cannot have two integer factors, unless \$\displaystyle b-1 = 1\$ and \$\displaystyle a+c=7\$ or \$\displaystyle b-1=7\$ and \$\displaystyle a+c=1\$ (which means that in this case either a or c must be zero, I don't know if you consider that a positive integer)

In the first case: \$\displaystyle b = 2\$ and \$\displaystyle a+c=7\$

So if we add those two together: \$\displaystyle a+b+c=9\$

In the second case: \$\displaystyle b = 8\$ and \$\displaystyle a+c=1\$

So if we add those two together: \$\displaystyle a+b+c=9\$

So the answer is: \$\displaystyle \boxed{a+b+c=9}\$