AAPL wrote:Manhattan Prep

A certain game pays players in tokens, each of which is worth either \(m\) points or \(n\) points, where \(m\) and \(n\) are different positive integers whose greatest common factor is 1. In terms of \(m\) and \(n\), what is the greatest possible sum, in points, that can be paid out with only one unique combination of these tokens? (For example, if \(m = 2\) and \(n = 3\), then a sum of 5 points can be created using only one combination, \(m + n\), which is a unique combination. By contrast, a sum of 11 points can be created by \(4m + n\) or by \(m + 3n\). This solution does not represent a unique combination; two combinations are possible.)

A. \(2mn\)
B. \(2mn-m-n\)
C. \(2mn-m-n-1\)
D. \(mn+m+n-1\)
E. \(mn-m-n\)

OA B

Let's see each option taking m = 2 and n = 3.

A. \(2mn\):

2mn = 2*3*2 = 12

We can achieve the sum 12 as 4n = 4*3 = 12 (4 tokens) OR as 6m = 6*2 = 12 (6 tokens). So this is not a correct answer as the sum 2mn does not render only one unique combination of these tokens. Incorrect.

B. \(2mn-m-n\):

\(2mn-m-n\) = \(2*2*3-2-3\) = 7

We can achieve sum 7 only as 2m + n = 2*2 + 3 = 7 (3 tokens). No other combination is possible. So this could be the correct answer as the sum \(2mn-m-n\) renders only one unique combination of these tokens.

Note that keeping in view Option B, we can eliminate Option C and E as even if these two options qualify, the sum would be greater for Option B. So, let's go straight to Option D.

D. \(mn+m+n-1\):

\(mn+m+n-1\) = \(2*3+2+3-1\) = 10

Sum 10 can be achieved with either 5 m-tokens or with 2 tokens each of m-type and n-type. Not a correct option.

The correct answer: B

Hope this helps!

-Jay
_________________
Manhattan Review GRE Prep

Locations: GRE Classes Seattle | GMAT Prep Course Hong Kong | GRE Prep San Francisco | SAT Prep Classes NYC | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.