[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
m and n are two different numbers selected from the integer
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- Max@Math Revolution
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Beautiful problem, Max. Congrats!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
$$m,n\,\,{\rm{distinct}}\,\, \in \left\{ {11,12, \ldots ,20} \right\}$$
$$? = \max \left( {{{mn} \over {m - n}}} \right)$$
$$\left( {\rm{I}} \right)\,\,m > n\,\,\,\,\left( {{\rm{otherwise}}\,\,{{mn} \over {m - n}} < 0\,\,,\,\,{\rm{not}}\,\,\max } \right)$$
$$\left( {{\rm{II}}} \right)\,\,0 < {{mn} \over {m - n}} \le \max \left( {{{mn} \over {m - n}}} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{1 \over n} - {1 \over m} = {{m - n} \over {mn}} \ge {\left[ {\max \left( {{{mn} \over {m - n}}} \right)} \right]^{ - 1}}$$
$$\left( {{\rm{III}}} \right)\,\,\,?\,\,\,\, \Leftrightarrow \,\,\,\,\min \left( {{1 \over n} - {1 \over m}} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\left( {m,n} \right) = \left( {20,19} \right)$$
$$\left( {{\rm{IV}}} \right)\,\,\,? = {\left( {{{20 - 19} \over {20 \cdot 19}}} \right)^{ - 1}} = 20 \cdot 19 = 380$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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- Max@Math Revolution
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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
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
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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
To maximize mn/(m - n), we should look for the greatest value of the numerator (mn) and the smallest positive value of the denominator (m - n). Therefore, the maximum value of mn/(m - n) occurs when m = 20 and n = 19. So the maximum value is:
(20 x 19)/(20 - 19) = 380/1 = 380
Answer: D
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