Problem Solving

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

m and n are two different numbers selected from the integers between 11 and 20, inclusive. What is the maximum value of mn / (m-n)?

A. 320
B. 340
C. 360
D. 380
E. 400

Beautiful problem, Max. Congrats!
$$m,n\,\,{\rm{distinct}}\,\, \in \left\{ {11,12, \ldots ,20} \right\}$$
$$? = \max \left( {{{mn} \over {m - n}}} \right)$$
$$\left( {\rm{I}} \right)\,\,m > n\,\,\,\,\left( {{\rm{otherwise}}\,\,{{mn} \over {m - n}} < 0\,\,,\,\,{\rm{not}}\,\,\max } \right)$$
$$\left( {{\rm{II}}} \right)\,\,0 < {{mn} \over {m - n}} \le \max \left( {{{mn} \over {m - n}}} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{1 \over n} - {1 \over m} = {{m - n} \over {mn}} \ge {\left[ {\max \left( {{{mn} \over {m - n}}} \right)} \right]^{ - 1}}$$
$$\left( {{\rm{III}}} \right)\,\,\,?\,\,\,\, \Leftrightarrow \,\,\,\,\min \left( {{1 \over n} - {1 \over m}} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\left( {m,n} \right) = \left( {20,19} \right)$$
$$\left( {{\rm{IV}}} \right)\,\,\,? = {\left( {{{20 - 19} \over {20 \cdot 19}}} \right)^{ - 1}} = 20 \cdot 19 = 380$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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To obtain the maximum value for mn/(m-n), m-n must be the smallest possible positive number and mn must be the largest possible product.
mn is largest when both m and n are largest, that is, when m = 20 and n - 19. In this case, m - n = 1 is as small as possible and mn = 20*19 = 380 is as large as possible.
Thus, the maximum value of mn/(m-n) is 20*19/(20-19) = 380/1 = 380.

Therefore, D is the answer.
Answer: D