KARISHMA315 wrote:

monsoon1 wrote:

If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4

(2) ab ≤ 15

Is there any algebraic approach to be sure that B is sufficient

I am pretty sure I have an algebraic/number theoretic derivation of the general form of the solutions, but it's ugly and really not how you want to go about this. Still, for completeness and since you asked...

\(\frac{1}{c} = \frac{a+b}{ab}\)

\(c = \frac{ab}{a+b}\)

Now note that, since c is a positive integer, a+b must divide ab. This is only possible if a and b share a common factor greater than 1, based on the principle that "multiple + non-multiple = non-multiple"; unless a and b share some factor greater than 1, a+b will not share any factor greater than 1 with a and, by the same token, a+b will not share any factor greater than 1 with b.

So a and b have a greatest common factor larger than 1, and we will call this factor g. We will define a = gp and b =gq.

Now rewrite the expression as

\(c = \frac{gpgq}{gp+gq}\)

\(c = \frac{g^2pq}{g(p+q)}\)

\(c = \frac{gpq}{p+q}\)

Now recall that p and q by definition are positive integers that have no common factors greater than 1. So p+q shares no factors besides 1 with either p or q. This means that c will be an integer if and only if g is divisible by p+q.

We can rewrite one final time by taking g = k(p+q). This gives, at last,

\(a = k(p+q)p\)

and

\(b = k(p+q)q\)

so

\(ab = k^2(p+q)^2pq\)

It is now easy to verify that \(ab \leq 15\) is sufficient, since p = q = k = 1 gives \(1^2 * (1+1)^2 * 1 * 1 = 4\), but if p = q = 1 and k = 2 then a = 4, b = 4, and the product is 16, and if k = 1, either p or q = 1, and the other of q or p = 2, then a = 3, b = 6 (or vice versa), and the product is 18.

And this should work for any positive integers p and q that share no common factors greater than 1 and for any positive integer k, in which case the result will be c = kpq.

Try it!

For instance:

p = 3

q = 8

k = 4

gives

a = 132

b = 352

and, sure enough, c will be an integer:

c = 96

Another set:

p = 7

q = 11

k = 77

a = 9702

b = 15246

c = 5929

Now, aren't you glad you asked for algebra?